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		<title>Understanding Basic Trigonometric Identities</title>
		<link>https://complete-concrete-concise.com/mathematics/understanding-basic-trigonometric-identities/</link>
		
		<dc:creator><![CDATA[richardsplanet]]></dc:creator>
		<pubDate>Fri, 01 Mar 2013 10:37:37 +0000</pubDate>
				<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[cos]]></category>
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					<description><![CDATA[<p>This tutorial assumes you are familiar with the trigonometric functions and their derivation from the unit circle. A tutorial on the trigonometric functions can be found here. A tutorial on the trigonometric functions and the unit circle can be found here. What is an Identity? An identity (in mathematics) is something that is true (more [&#8230;]</p>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/understanding-basic-trigonometric-identities/">Understanding Basic Trigonometric Identities</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
]]></description>
										<content:encoded><![CDATA[<div class="c1">
<p>This tutorial assumes you are familiar with the trigonometric functions and their derivation from the unit circle.</p>
<p>A tutorial on the trigonometric functions can be found <a href="//complete-concrete-concise.com/mathematics/understanding-sine-cosine-and-tangent">here</a>.</p>
<p>A tutorial on the trigonometric functions and the unit circle can be found <a href="//complete-concrete-concise.com/mathematics/understanding-trigonomtric-functions-using-the-unit-circle-advanced">here</a>.</p>
</div>
<h1>What is an Identity?</h1>
<p>An identity (in mathematics) is something that is true (more precisely: <em>a tautological relationship</em>).</p>
<p>For example, the following is an identity:</p>
<pre><code>1 + 1 = 2</code></pre>
<p>It says that <code>1 + 1</code> is <u>exactly</u> the same as <code>2</code>.</p>
<p>A common (and important) trigonometric identity is:</p>
<pre><code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</code></pre>
<p>All this identity says is that what is on the <em>left hand side</em> is <u>identical</u> to what is on the <em>right hand side</em>.</p>
<h1>Why Learn / Use Identities?</h1>
<p>Identities (in any branch of mathematics) help us to:</p>
<ul>
<li>solve</li>
<li>simplify</li>
<li>or gain insight into</li>
</ul>
<p>mathematical problems.</p>
<p>Identities are a lot like synonyms in language. Instead of using the word <em>red</em> to describe the colour of something we could use one of the following: <em>carmine, cherry, copper, crimson, ruby, ruddy, rust, scarlet</em>. In language, synonyms give spice to communication; in mathematics, it is used to look at things in a different way.</p>
<p>For example, if you were solving an trigonometric problem and ended up with the following expression:</p>
<pre><code>cos<sup>2</sup>&theta; × tan&theta; × sec&theta; × csc&theta;</code></pre>
<p>you might not think you had achieved much. However, by using basic trigonometric identities, you would see (or maybe have to work out) that the entire expression can be reduced to the value of <code>1</code>.</p>
<h1>How Many Trigonometric Identities are There?</h1>
<p>There are an infinite number of trigonometric identities. Only a few are useful, the remainder generally ending up as problems and exercises for students to solve.</p>
<h1>Reference Image</h1>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2013/02/basic-trig-identities-0.png" alt="" border="0" class="centered"/></p>
<div class="c1">
<p>Basic trigonometric identities fall into four groups:</p>
<ol>
<li>Reciprocals</li>
<li>Ratios</li>
<li>Products</li>
<li>Pythagorean Theorem</li>
</ol>
</div>
<h1>Reciprocal Identities</h1>
<p>These identities derive from the fundamental definitions of the trigonometric functions:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2013/02/basic-trig-identities-1.png" alt="" border="0" class="centered"/></p>
<p>Given a right angle triangle, we know that:</p>
<pre><code>sin&theta; = A / C</code></pre>
<p>We also know that:</p>
<pre><code>csc&theta; = C / A</code></pre>
<p>It should be obvious that <code>sin&theta; = 1 / csc&theta;</code> because the ratio <code>C / A</code> (cosecant) is the reciprocal of the ratio <code>A / C</code> (sine).</p>
<p>The same reasoning applies to all the <u>reciprocal</u> identities.</p>
<h1>Ratio Identities</h1>
<p>These identities also derive from the fundamental definitions of trigonometric functions and are very similar to the <u>reciprocal</u> identities, except that they consist of the ratio of two trigonometric functions:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2013/02/basic-trig-identities-2.png" alt="" border="0" class="centered"/></p>
<p>The most important of these is probably <code>tan&theta; = sin&theta; / cos&theta;</code> (highlighted in blue).</p>
<p>We know that</p>
<pre><code>tan&theta; = A / B     (from a right angle triangle)
tan&theta; = y / x     (from the unit circle)</code></pre>
<p>We also know that <code>tan&theta;</code> is the slope of the hypotenuse.</p>
<p>From the unit circle we know that:</p>
<pre><code>sin&theta; = y
cos&theta; = x</code></pre>
<div class="c2">
<p>It is correct to say that <code>sin&theta; = y</code> is an <u>identity</u> because both are equivalent or identical.</p>
</div>
<p>Using the identities <code>sin&theta;</code> and <code>cos&theta;</code> to substitute for <code>y</code> and <code>x</code>, we get the following:</p>
<pre><code>tan&theta; = y / x = sin&theta; / cos&theta;</code></pre>
<p>A similar derivation can be done using the definitions from a right angle triangle:</p>
<pre><code>sin&theta; = A / C
cos&theta; = B / C</code></pre>
<p>In order to be able to substitute our identities, we need to multiply both the numerator and denominator by <code>1 / C</code></p>
<div class="c1">
<p>One of the principles of algebra is that as long as you multiply or divide ALL the terms by the same amount, the equation remains the same)</p>
</div>
<pre><code>       A   (1 / C)     (A / C)     sin&theta;
tan&theta; = - × &mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;  =  &mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;  =  &mdash;&mdash;&mdash;&mdash;
       B   (1 / C)     (B / C)     cos&theta;</code></pre>
<p>The same sort of reasoning applies to the derivation of the remaining <u>ratio</u> identities.</p>
<div class="c3">
<p>Another class of <em>obvious</em> <u>ratio</u> identities is the ratio of the term to itself:</p>
<pre><code>sin&theta;
&mdash;&mdash;&mdash;&mdash;  =  1
sin&theta;</code></pre>
<p>In all cases, the answer is <code>1</code>.</p>
</p></div>
<h1>Product Identities</h1>
<p>While these can be derived from the fundamental trigonometric identities, it is often easier to use the <u>ratio</u> identities.</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2013/02/basic-trig-identities-3.png" alt="" border="0" class="centered"/></p>
<p>Taking <code>cos&theta; × tan&theta; = sin&theta;</code>, we can derive it as follows:</p>
<p><strong>1) Find</strong> an identity for <code>tan&theta;</code>:</p>
<pre><code>       sin&theta;
tan&theta; = &mdash;&mdash;&mdash;&mdash;
       cos&theta;</code></pre>
<p><strong>2) Substitute</strong> this <em>identity</em> for <code>tan&theta;</code> into the identity <code>cos&theta; × tan&theta; = sin&theta;</code>:</p>
<pre><code>       sin&theta;
cos&theta; × &mdash;&mdash;&mdash;&mdash;  =  sin&theta;
       cos&theta;</code></pre>
<p><strong>3) Cancel</strong> the <code>cos&theta;</code> terms:</p>
<pre><code>       sin&theta;
<s>cos&theta;</s> × &mdash;&mdash;&mdash;&mdash;  =  sin&theta;
       <s>cos&theta;</s></code></pre>
<p>The same sort of reasoning applies to the derivation of the remaining <u>product</u> identities.</p>
<div class="c3">
<p>Another class of <em>obvious</em> <u>product</u> identities is the product of the term to its <u>reciprocal</u> identity:</p>
<pre><code>sin&theta; × csc&theta; = 1</code></pre>
<p>In all cases, the answer is <code>1</code>.</p>
</p></div>
<h1>Pythagorean Theorem Identities</h1>
<p>Because the trigonometric functions are derived from a right angle triangle and because the trigonometric functions make use of the sides of a right angle triangle, they can be used in the Pythagorean Theorem:</p>
<pre><code>A<sup>2</sup> + B<sup>2</sup> = C<sup>2</sup></code></pre>
<h3>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</h3>
<p>This is the most common (or most famous) example of a Pythagorean Theorem identity using trigonometric functions.</p>
<p>It can be arranged in many different ways, the three most common would be:</p>
<pre><code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1
sin<sup>2</sup>&theta; = 1 - cos<sup>2</sup>&theta;
cos<sup>2</sup>&theta; = 1 - sin<sup>2</sup>&theta;</code></pre>
<p>Looking at a right angle triangle, we see that it has 3 sides, normally labelled A, B, and C. With C being the hypotenuse.</p>
<p>Using these sides in the Pythagorean Theorem, we get:</p>
<pre><code>A<sup>2</sup> + B<sup>2</sup> = C<sup>2</sup></code></pre>
<p>To get it to look like <code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</code>, we see that we need to get <code>C<sup>2</sup></code> to be equal to <code>1</code>. This can be done by dividing the entire equation by <code>C<sup>2</sup></code>:</p>
<pre><code>A<sup>2</sup>    B<sup>2</sup>    C<sup>2</sup>
&mdash;  +  &mdash;  =  &mdash;
C<sup>2</sup>    C<sup>2</sup>    C<sup>2</sup></code></pre>
<p>Which gives us:</p>
<pre><code>A<sup>2</sup>    B<sup>2</sup>
&mdash;  +  &mdash;  =  1
C<sup>2</sup>    C<sup>2</sup></code></pre>
<p>This can be rewritten as:</p>
<pre><code>A   A     B   B
&mdash; &times; &mdash;  <strong>+</strong>  &mdash; &times; &mdash;  =  1
C   C     C   C</code></pre>
<p>We know that <code>A/C = sin&theta;</code> and <code>B/C = cos&theta;</code>, so we can substitute them in the equation giving us:</p>
<pre><code>sin&theta; &times; sin&theta;  +  cos&theta; &times; cos&theta;  =  1</code></pre>
<p>Which, when we collect the terms, gives us the familiar:</p>
<pre><code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</code></pre>
<h3>1 + cot<sup>2</sup>&theta; = csc<sup>2</sup>&theta;</h3>
<p>Just as we obtained <code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</code> by dividing the terms in the Pythagorean Theorem by <code>C<sup>2</sup></code>, we can divide all the terms in the Pythagorean Theorem by <code>A<sup>2</sup></code>, which gives us:</p>
<pre><code>A<sup>2</sup>    B<sup>2</sup>    C<sup>2</sup>
&mdash;  +  &mdash;  =  &mdash;
A<sup>2</sup>    A<sup>2</sup>    A<sup>2</sup></code></pre>
<p>By using similar steps as above, we get the identity:</p>
<pre><code>1 + cot<sup>2</sup>&theta; = csc<sup>2</sup>&theta;</code></pre>
<p>Three more common arrangements would be:</p>
<pre><code>cot<sup>2</sup>&theta; = csc<sup>2</sup>&theta; - 1
csc<sup>2</sup>&theta; = cot<sup>2</sup>&theta; + 1
csc<sup>2</sup>&theta; - cot<sup>2</sup>&theta; = 1</code></pre>
<h3>tan<sup>2</sup>&theta; + 1 = sec<sup>2</sup>&theta;</h3>
<p>Just as we obtained <code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1</code> by dividing the terms in the Pythagorean Theorem by <code>C<sup>2</sup></code>, we can divide all the terms in the Pythagorean Theorem by <code>B<sup>2</sup></code>, which gives us:</p>
<pre><code>A<sup>2</sup>    B<sup>2</sup>    C<sup>2</sup>
&mdash;  +  &mdash;  =  &mdash;
B<sup>2</sup>    B<sup>2</sup>    B<sup>2</sup></code></pre>
<p>By using similar steps as above, we get the identity:</p>
<pre><code>tan<sup>2</sup>&theta; + 1 = sec<sup>2</sup>&theta;</code></pre>
<p>Three common arrangements would be:</p>
<pre><code>tan<sup>2</sup>&theta; = sec<sup>2</sup>&theta; - 1
sec<sup>2</sup>&theta; = tan<sup>2</sup>&theta; + 1
sec<sup>2</sup>&theta; - tan<sup>2</sup>&theta; = 1</code></pre>

<h1>Trigonometric Identity Problems</h1>
<p>All of the following are trigonometric identities. Feel free to verify / prove using the identities described in the tutorial:</p>
<div class="c1">
<p>Note, I have used different terms to indicate the angles. This is because the actual term used does not matter.</p></div>
<pre><code>  csc<sup>2</sup>&theta;
&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;  =  cot<sup>2</sup>&theta;
tan<sup>2</sup>&theta; + 1</code></pre>
<pre><code>(1 - sin<sup>2</sup>&alpha;)(1 + tan<sup>2</sup>&alpha;) = 1</code></pre>
<pre><code>sec<sup>2</sup>&phi; - 1
&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;  =  sin<sup>2</sup>&phi;
  sec<sup>2</sup>&phi;</code></pre>
<pre><code>sin&beta; + cos&beta;cot&beta; = csc&beta;</code></pre>
<pre><code>1 - sin<sup>4</sup>&gamma; - 2sin<sup>2</sup>&gamma;cos<sup>2</sup>&gamma; = cos<sup>4</sup>&gamma;</code></pre>
<p>It should be obvious that you can create as many identities as you like. It should also be obvious that most (all?) of these are not as useful as the fundamental identities described in this tutorial.</p>
<div class="c4">
<p>Is the following an identity?</p>
<pre><code>sin<sup>2</sup>&theta; + cos<sup>2</sup>&beta; = 1</code></pre>
<p>It is an identity <u><strong><em>if and only if</em></strong></u> &theta; = &beta;, otherwise, it is not an identity.</p>
</div>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/understanding-basic-trigonometric-identities/">Understanding Basic Trigonometric Identities</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
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			</item>
		<item>
		<title>How to Solve Trigonometric Problems</title>
		<link>https://complete-concrete-concise.com/mathematics/how-to-solve-trigonometric-problems/</link>
		
		<dc:creator><![CDATA[richardsplanet]]></dc:creator>
		<pubDate>Thu, 15 Nov 2012 13:45:51 +0000</pubDate>
				<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[cosecant]]></category>
		<category><![CDATA[cosine]]></category>
		<category><![CDATA[cotangent]]></category>
		<category><![CDATA[how to]]></category>
		<category><![CDATA[secant]]></category>
		<category><![CDATA[sine]]></category>
		<category><![CDATA[solving]]></category>
		<category><![CDATA[tangent]]></category>
		<category><![CDATA[trigonometric functions]]></category>
		<category><![CDATA[trigonometry]]></category>
		<guid isPermaLink="false">http://complete-concrete-concise.com/?p=2555</guid>

					<description><![CDATA[<p>This tutorial offers advice on how to solve trigonometric problems and provides several problems worked through in detail. It assumes you are familiar with the trigonometric functions sine, cosine, tangent, secant, cosecant, and cotangent. A basic tutorial can be found here. A more advanced tutorial can be found here. If an explanation / walkthrough is [&#8230;]</p>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/how-to-solve-trigonometric-problems/">How to Solve Trigonometric Problems</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
]]></description>
										<content:encoded><![CDATA[<div class="c1">
This tutorial offers advice on how to solve trigonometric problems and provides several problems worked through in detail.<br />
It assumes you are familiar with the trigonometric functions sine, cosine, tangent, secant, cosecant, and cotangent. A basic tutorial can be found <a href="//complete-concrete-concise.com/mathematics/understanding-sine-cosine-and-tangent">here</a>. A more advanced tutorial can be found <a href="//complete-concrete-concise.com/mathematics/understanding-trigonomtric-functions-using-the-unit-circle-advanced">here</a>.
</div>
<div class="c2">
If an explanation / walkthrough is not clear, please let me know in a comment and I will try to improve the answer.<br />
Identify what it is that is not clear &#8211; just saying &#8220;I didn&#8217;t understand it&#8221; isn&#8217;t very helpful.
</div>
<div class="c1">
If you have a problem you would like to see solved, leave a comment. <strong>Note:</strong> (1) I don&#8217;t do homework, (2) I don&#8217;t promise I will solve it, (3) I don&#8217;t promise to solve it quickly.
</div>
<h1>Don&#8217;t Panic</h1>
<ol>
<li>Don&#8217;t panic or freak out. No one is trying to trick you or trip you up (although, often times, word problems are poorly worded)</li>
<li>Read and understand the problem.</li>
<li>Trigonometric problems are always about right angle triangles. So the trick is to try and break the problem down into right angle triangles.</li>
<li>Trigonometry is about the ratios of the lengths of two sides. If we know the length of two sides of a right angle triangle we can calculate the ratio between those two sides. Those ratios are given special names: sine, cosine, tangent, cosecant, secant, cotangent. Because of the properties of right angle triangles, if we know one of the angles then we also know the ratios of all the sides.</li>
<li>Look for what ratios or partial ratios you can find. Then work from there.</li>
<li>Similar triangles are triangles that have the same shape that have the same angles inside. Often times, we don&#8217;t solve the problem directly but look for a similar triangle. The triangle may be bigger or smaller, but it has the same shape as the triangle we are studying.</li>
</ol>
<h3>Basic Use of Trigonometric Definitions</h3>
<div class="c1">
<div class="c3">
Consider the following right angle triangle having sides of length 3, 4, and 5 units and two angles called alpha and beta. What is sin&alpha;, cos&alpha;, tan&alpha;, sin&beta;, cos&beta;, tan&beta; and the angles &alpha; and &beta;?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/3-4-5-triangle.png" alt="" border="0" /><br />
This problem is about understanding the definition of the trigonometric functions sine, cosine, and tangent.<br />
The triangle is displayed in a familiar way and we can easily calculate the trigonometric relations for &alpha; (remember sin&alpha; is shorthand for &#8220;<em>given a right angle triangle containing an angle &alpha;, what is the ratio of the side opposite the angle to the hypotenuse</em>&#8220;):<br />
We know that <code>sin&alpha; = opposite ÷ hypotenuse</code>, which is <code>3 / 5</code> which is <code>0.6</code>.<br />
We know that <code>cos&beta; = adjacent ÷ hypotenuse</code>, which is <code>4 / 5</code> which is <code>0.8</code>.<br />
We know that <code>tan&alpha; = opposite ÷ adjacent</code>, which is <code>3 / 4</code> which is <code>0.75</code>.<br />
Calculating the trigonometric relations for &beta; is no different, although, we need to correctly identify the opposite and adjacent sides:<br />
We know that <code>sin&beta; = opposite ÷ hypotenuse</code>, which is <code>4 / 5</code> which is <code>0.8</code>.<br />
We know that <code>cos&beta; = adjacent ÷ hypotenuse</code>, which is <code>3 / 5</code> which is <code>0.6</code>.<br />
We know that <code>tan&beta; = opposite ÷ adjacent</code>, which is <code>4 / 3</code> which is <code>1.333...</code><br />
We don&#8217;t know what the angles &alpha; and &beta; are, but we can find out. We know that <code>sin&alpha; = 0.6</code> and <code>sin&beta; = 0.8</code>. All we need to do is find out what angles &alpha; and &beta; have a sine corresponding to 0.6 and 0.8, respectively.<br />
Using the inverse sin function (sin<sup>-1</sup>) on a calculator, we see that &alpha; must be 36.87° for sin&alpha; to be 0.6 and &beta; must be 53.13° for sin&beta; to be 0.8.</p>
<div class="c2">
Before calculator use was wide spread, you would look up the value in a sine table. You can see a sine table with a resolution of 1 degree over <a href="http://www.grc.nasa.gov/WWW/k-12/airplane/tablsin.html">here</a>.
</div>
</div>
<h3>Basic Use and Application of Trigonometric Definitions and Properties of a Right Angle Triangle</h3>
<div class="c2">
<div class="c3">
Given a triangle having sides of length 1, 2, and 3 units, is this a right angle triangle?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/1-2-3-triangle.png" alt="" border="0" /><br />
The simplest way is to see if the Pythagorean Theorem true &#8211; if this is a right angle triangle, we know that: <code>A<sup>2</sup> + B<sup>2</sup> = C<sup>2</sup></code><br />
<code>1<sup>2</sup> + 2<sup>2</sup> = 3<sup>2</sup></code><br />
<code>1 + 4 = 9</code> &#8211; the left hand side and right hand side are not the same, so the triangle is not a right angle triangle<br />
Using trigonometric functions, we can determine the angles &alpha; and &beta; and then check that all the angles sum up to 180°: <code>&alpha; + &beta; + 90° = 180°</code><br />
sin&alpha; = opposite ÷ hypotenuse = 1 / 3 = 0.333&#8230;<br />
sin&beta; = opposite ÷ hypotenuse = 2 / 3 = 0.666&#8230;<br />
Using the inverse sin function (sin<sup>-1</sup>) on a calculator, we determine that &alpha; is 19.47° and &beta; is 41.81°.<br />
Consequently, the sum of the angles is:<br />
<code>19.47° + 41.81° + 90° = 180°</code><br />
<code>151.28° = 180°</code> &#8211; the left hand side and right hand side are not the same, so the triangle is not a right angle triangle</p>
<div class="c1">
The only thing that might seem strange in this example are expressions of the form:<br />
<code>4 = 9</code><br />
These types of expressions are perfectly normal and valid in mathematics. This is because the symbol <code>=</code> does not have the same meaning we are usually used to: in mathematics, <code>=</code> means <u>is the same as</u>.<br />
Written in English, the expression <code>4 = 9</code> would be &#8220;<em>The value of 4 <u>is the same as</u> the value of 9</em>&#8220;. This statement is <u>false</u>. Don&#8217;t get confused and think you have just proven that 4 is the same as 9 because it isn&#8217;t.<br />
In this example, we are saying:</p>
<ol>
<li>The angles in a triangle sum up to 180°</li>
<li>Since this is supposed to be a right angle triangle, we know one of the angles is 90°</li>
<li>Since this is supposed to be a right angle triangle, we can look up the angles &alpha; and &beta; once we know the ratios of the sides</li>
<li>If the sum of all the angles is 180° then we know this is a right angle triangle</li>
<li>If the sum of all the angles is not 180° then we know this is not a right angle triangle.</li>
</ol>
</div>
</div>
<h3>Calculating the Height of a Tree</h3>
<div class="c1">
Potiphar&#8217;s out walking his cat. He notices that he casts a shadow 2 m long. He also notices that a tree in the garden casts a shadow 7 m long. If Potiphar is 1.5 m tall how tall is the tree? What is the angle of elevation of the sun?<br />
<img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/potiphar-1.png" alt="" border="0" /><br />
This is a basic trigonometric problem involving similar triangles. Potiphar and his shadow form a right angle triangle. The tree and its shadow form another right angle triangle. The rays of the sun (which are invisible) form the hypotenuse of these right angle triangles.<br />
<img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/potiphar-2.png" alt="" border="0" /><br />
We know that similar triangles have the same shape and the same angles. We know that the angle or slope of the hypotenuse is the same for both the <u>tree-triangle</u> and the <u>Potiphar-triangle</u> (because rays of sunlight are parallel to each other). Since both triangles are similar, we also know that the ratios of their sides must be the same.<br />
We know the lengths of the sides of the <u>Potiphar-triangle</u> and can use them to calculate a ratio:<br />
<code>A / B = 1.5 / 2 = 0.75</code><br />
Of the tree-triangle, we know the length of one of the sides but not of the other. However, we know that the ratio of the unknown length to the known length (i.e. shadow) must be the same as for the Potiphar-triangle:<br />
<code> a / b = A / B = 0.75</code><br />
Plugging in the values we know (and removing the A / B for clarity):<br />
<code> a / 7 = 0.75</code><br />
Rearranging we get:<br />
<code>a = 0.75 × 7 = 5.25</code><br />
The tree is 5.25 m tall.<br />
To calculate the elevation of the sun, we know the ratio of height to length. In trigonometric terms, this is the definition of tangent.<br />
<code>tan&alpha; = opposite ÷ adjacent = A / B = a / b = 0.75</code><br />
Looking up (or using the inverse key on the calculator) we see that for tan&alpha; to equal 0.75, &alpha; must be 36.87°.</p>
<div class="c2">
Before calculator use was wide spread, you would look up the value in a tangent table. You can see a tangent table with a resolution of 1 degree over <a href="http://www.grc.nasa.gov/WWW/k-12/airplane/tabltan.html">here</a>.
</div>
</div>
<h3>Calculating the Height of a Pyramid and Elevation of the Sun</h3>
<div class="c2">
<div class="c3">
Thales of Miletus measures the length of the shadow cast by the great pyramid and discovers that it is 17 m to the base of the pyramid. Using his 2 m measuring stick he notices that it casts a shadow of 1.8 m. If the great pyramid measures 230 m on a side, how high is the great pyramid? What is the angle of elevation of the sun?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/thales-1-thumb.png" alt="" border="0" /><br />
This is another basic trigonometric problem involving similar triangles, but this time a little trickier to solve. We know the dimensions of the height and length of the <u>measuring-stick-triangle</u>, but we don&#8217;t know the height or length of the <u>pyramid-triangle</u>. We know the length of the shadow to the base of the pyramid, but we need to know the length of the shadow to the center of the pyramid (from which the height is measured).<br />
<img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/thales-2-thumb.png" alt="" border="0" /><br />
We know the length of a side of the pyramid (this 230 m), we can use this to determine what the length of the shadow to the center of the pyramid should be. The peak of the pyramid is in the center, so it is 115 m from the base of the pyramid. This gives us a corrected shadow length of 17 m to the base + 115 m from the base to the center which is 132 m.<br />
We now know the length of one of the sides of the <u>pyramid-triangle</u>. Since the <u>pyramid-triangle</u> is similar to the <u>measuring-stick-triangle</u>, and we know that similar triangles have the same ratios for their sides, we can calculate the height of the pyramid in the same way as the previous example:</p>
<div class="c1">
In the previous example (Calculating the HEight of a Tree), I wrote the equation as:<br />
<code>a / b = A / B = ratio</code><br />
Equations in mathematics can be written in many different ways. In this example, I don&#8217;t bother to write the ratio and instead write it as:<br />
<code>a / b = A / B</code>
</div>
<p>Substituting the known values we get:<br />
<code>2 / 1.8 = height / 132</code><br />
Rearranging we get:<br />
<code>(2 × 132) / 1.8 = height</code><br />
Simplifying everything we get:<br />
height = 146.7 m<br />
To calculate the elevation of the sun, we know the ratio of height to length. In trigonometric terms, this is the definition of tangent.<br />
tan&alpha; = opposite ÷ adjacent = 2 / 1.8 = 1.111&#8230;<br />
Looking up (or using the inverse key on the calculator) we see that for tan&alpha; to equal 1.111&#8230;, &alpha; must be 48°.
</div>
<h3>Calculating the Length of a Shadow</h3>
<div class="c1">
The Peace Tower in Ottawa, Canada is 92.2 m tall. On July 1st, 2012, at noon, the sun&#8217;s angle of elevation was 67.6°. How long was the shadow cast by the Peace Tower?<br />
<img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/peace-tower.png" alt="" border="0" /><br />
We know the height of the tower and we know the angle of the sun (relative to the ground).<br />
Unlike the previous two problems, we only have one right angle triangle &#8211; the <u>tower-shadow-triangle</u>. We know if we have two similar triangles, the ratios of their sides must be the same.<br />
We know the angle the sun makes with the ground, so we know one of the angles of the triangle (actually, we know two of them &#8211; the other is 90° &#8211; so, if we need to, we can find the third angle because all angles in a triangle must sum up to 180°).<br />
All right angle triangles with the same angles are similar triangles &#8211; it doesn&#8217;t matter if we don&#8217;t know the lengths of the sides because we know the ratios those sides will have.<br />
We know that the following must be true:<br />
<code>length of the opposite side ÷ length of the adjacent side = tan(&alpha;)</code><br />
Plugging in the values we know, we get:<br />
<code>92.2 / x = tan(67.6°)</code><br />
Evaluating (looking up) <code>tan(67.6°)</code> we get:<br />
<code>9.2. / x = 2.43</code><br />
Rearranging we get<br />
<code>x = 92.2 / 2.43</code><br />
Simplifying we get:<br />
<code>x = 37.94</code><br />
The length of the shadow is 37.94 m (and it is probably safe to round it up to 38 m).
</div>
<h3>Calculating the Height of a Cliff</h3>
<div class="c2">
<div class="c3">
A surveyor measuring the height of the cliff, determines that the angle to the top of the cliff is 60° and the distance to the base of the cliff is 120 m how high is the cliff?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/cliff-60.png" alt="" border="0" /><br />
This is similar to calculating the length of the shadow from two examples earlier.<br />
The distance from the cliff, with the height of the cliff and the straight line to the top of the cliff form a right angle triangle.<br />
We know one of the angles. We also know that all right angle triangles with the same angle are similar triangles &#8211; this means we also know the ratios the various sides of the triangles have.<br />
As before, we know that:<br />
<code>opposite ÷ adjacent = tan(&alpha;)</code><br />
Plugging in the values we know, we get:<br />
<code>x / 120 = tan(60°)</code><br />
Evaluating <code>tan(60°)</code> we get:<br />
<code>x / 120 = 1.732</code><br />
Rearranging we get:<br />
<code>x = 1.732 × 120</code><br />
Simplifying we get:<br />
<code>x = 207.84</code><br />
The cliff is 207.84 m high (it is probably ok to round it to 208 m).</p>
<div class="c3">
How would the problem and solution change if we were given the other angle instead?<br />
In this case, we would have:<br />
<code>120 / x = tan(30°)</code><br />
When solved it would give us the same answer.<br />
The relationship remains the same, the only things that change are which side we call <u>opposite</u> and <u>adjacent</u>.
</div>
</div>
<h3>Calculating the Height of a Lighthouse</h3>
<div class="c1">
<div class="c3">
A surveyor measuring the height of the cliff, determines that the angle to the top of the cliff is 60° &#8211; this angle was measured at a height of 1.7 m (i.e. the surveying equipment is on a tripod and stands 1.7 m above the ground). The angle measured to the top of the lighthouse is 65°. The distance to the cliff is 120 m. How tall is the lighthouse? How high is the cliff?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/cliff-lighthouse.png" alt="" border="0" /><br />
In this case, we are given two triangles, but they are not similar. One triangle is the <u>cliff-triangle</u>, the other is the <u>cliff-lighthouse-triangle</u>.<br />
If we find the height of both triangles, we can subtract the height of the cliff from the height of the cliff + lighthouse and get the height of the lighthouse.</p>
<h3>Cliff Height</h3>
<p>This is basic trigonometric ratios:<br />
<code>cliff height ÷ distance to cliff = tan(60°)</code><br />
Plugging in the known values:<br />
<code>x / 120 = tan(60°)</code><br />
Solving:<br />
<code>x = 207.84</code><br />
We&#8217;re not finished with the height of the cliff, the angle to the top of the cliff was measured from a height of 1.7 m, this means we need to add 1.7 m to the height of the cliff:<br />
<code>height = 207.84 + 1.7 = 209.54 m</code></p>
<div class="c2">
<div class="c3">
If we measured from the base of the tripod, what would the angle be to the top of the cliff?
</div>
<p>In this case, we know the length of two sides of the triangle: 120 m and 209.54 m.<br />
We can calculate the ratio of these two sides:<br />
<code>ratio = 209.54 / 120 = 1.74616...</code><br />
We know from trigonometric definitions that <code>tan(&alpha;) = opposite ÷ adjacent</code>, so all we need to find is what &alpha; gives us the necessary ratio. Using tan<sup>-1</sup> on the calculator we see that:<br />
<code>tan<sup>-1</sup>(1.74616...) = 60.2°</code>
</div>
<h3>Height of the Cliff + Lighthouse</h3>
<p>This is basic trigonometric ratios:<br />
<code>(cliff height + lighthouse height) ÷ distance to cliff = tan(&alpha;)</code><br />
Plugging in the known values we get:<br />
<code>x / 120 = tan(65°)</code><br />
Solving:<br />
<code> x = 257.34</code><br />
We&#8217;re not finished with the height of the cliff + lighthouse, the angle to the top of the cliff + lighthouse was measured from a height of 1.7 m, this means we need to add 1.7 m to the height of the cliff + lighthouse:<br />
<code>x = 257.34 + 1.7 = 259.04 m</code></p>
<h3>Height of the Lighthouse</h3>
<p>It is a simple matter of subtracting the height of the cliff from the height of the cliff + lighthouse:<br />
<code>lighthouse height = (lighthouse height + cliff height) - cliff height</code><br />
Plugging in the known values we get:<br />
<code>x = 259.04 - 209.54</code><br />
Solving we get:<br />
<code>x = 49.5 m</code></p>
<div class="c2">
In this example, it was not necessary to correct the height of the cliff and cliff + lighthouse by adding 1.7 m to each because each was offset (measured) from the same point &#8211; but it is still good practice to find the true heights (dimensions) even if only the relative ones are good enough.
</div>
</div>
<h3>Calculating the Length of a Field #1</h3>
<div class="c2">
<div class="c3">
Two surveyors measure the length of the field. One surveyor stands at one end of the field with the measuring equipment, the other surveyor stands at the other end of the field with a measuring stick 2 m high. The surveyor measures an angle of 2° to the base of the measuring stick. The surveying equipment is at a height of 1.7 m. How long is the field?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/surveyor-1.png" alt="" border="0" /><br />
This is basic trigonometric ratios. The straight line (of sight) to the measuring stick is perpendicular to the measuring stick, therefore, it forms one side of a right angle triangle. The measuring stick (or part of it) forms another side. Finally, the line of sight to the base of the measuring stick forms the remaining side (which is also the hypotenuse):<br />
<img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/surveyor-2.png" alt="" border="0" /><br />
We know the height where the straight line of sight intersects the measuring stick must be 1.7 m. We also know the angle to the base of the measuring stick is 2°.<br />
We don&#8217;t know the length of the side adjacent to the measured angle, we do know the length of the side opposite the measured angle, and we know the angle. Two trigonometric ratios use the <u>opposite</u> and <u>adjacent</u> sides: <em>tangent</em> and <em>cotangent</em>. Since <em>tangent</em> is the more commonly used ratio (and appears on many calculators) we will use the definition of <em>tangent</em>:<br />
<code>tan(&alpha;) = opposite ÷ adjacent</code><br />
Plugging in the known values, we get:<br />
<code>tan(2°) = 1.7 / x</code><br />
Solving, we get:<br />
<code>x = 48.68</code><br />
The length of the field is 48.68 m (it is probably safe to round up to 48.7 m)
</div>
<h3>Calculating the Length of a Field #2</h3>
<div class="c1">
<div class="c3">
Two surveyors measure the length of the field. One surveyor stands at one end of the field with the measuring equipment, the other surveyor stands at the other end of the field with a measuring stick 2 m high. The surveyor forgot to measure the height of the equipment, but, fortunately, took two readings. The surveyor measures an angle of 2° to the base of the measuring stick and 1° to the top of the measuring stick.. How long is the field?
</div>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/surveyor-3.png" alt="" border="0" /><br />
We know the straight line of sight to the measuring stick must be perpendicular to the measuring stick &#8211; this forms one side of a right angle triangle (as in the previous example). In fact, it forms the side of two right angle triangles &#8211; since two different angles were measured. Unlike the previous example, we don&#8217;t know the length of the other side, but we do know the length of the measuring stick &#8211; it is 2 m. We know that the sum of the two opposite sides must be 2 m. We also know the angles for both right angle triangles, so we know the ratios the sides have:<br />
<img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/11/surveyor-4.png" alt="" border="0" /><br />
From trigonometric definitions we know that <code>tan(&alpha;) = opposite / adjacent</code><br />
We also know the <u>adjacent</u> side is the same length for both triangles.<br />
Finally, we know that <code>opposite<sub>top</sub></code> + <code>opposite<sub>bottom</sub></code> = 2 m.</p>
<h3>Top Triangle</h3>
<p>Solving for the top triangle we get:<br />
<code>tan(1°) = opposite<sub>top</sub> / adjacent</code><br />
Calculating what we can, we get:<br />
<code>0.017455 = opposite<sub>top</sub> / adjacent</code></p>
<h3>Bottom Triangle</h3>
<p>Solving for the bottom triangle we get:<br />
<code>tan(2°) = opposite<sub>bottom</sub> / adjacent</code><br />
Calculating what we can, we get:<br />
<code>0.034921 = opposite<sub>top</sub> / adjacent</code></p>
<h3>Combining the Equations</h3>
<div class="c4">
Fundamental to basic algebra is the ability to manipulate an equation any way we like &#8211; as long as we do the same thing on both sides.<br />
For example, consider <span class="i2">4 = 4</span>:<br />
We can add 8 to both sides and the relationship remains the same: <span class="i2">4 + 8 = 4 + 8</span>.<br />
However, it is also perfectly valid to do: <span class="i2">4 + 8 = 4 + 1 + 2 + 5</span> (this is one of those non-intuitive mathematical tricks &#8211; but it is very useful).<br />
We need to employ this algebraic trick of substituting <em>equivalent</em> things into our equations.
</div>
<p>Since we know that the <u>adjacent</u> sides of both right angle triangles are identical, we can use this to merge the two equations together.<br />
Rearranging the bottom triangle equation so the <u>adjacent</u> is alone, we get:<br />
<code>adjacent = opposite<sub>bottom</sub> / 0.034921</code><br />
We now replace the <u>adjacent</u> side in the top triangle with its equivalent which is <code>opposite<sub>bottom</sub> / 0.034921</code>:<br />
<code>0.017455 = opposite<sub>top</sub> / (opposite<sub>bottom</sub> / 0.034921)</code><br />
Rearranging we get:<br />
<code>0.017455 × opposite<sub>bottom</sub> = 0.034921 × opposite<sub>top</sub></code></p>
<h3>Solving for the <em>opposite<sub>bottom</sub></em> side</h3>
<p>Dividing both sides by by 0.017455, we get:<br />
<code>opposite<sub>bottom</sub> = 2 × opposite<sub>top</sub></code></p>
<h3>Remembering the 2 m Measuring Stick</h3>
<p>At this point, we need to remember that <code>opposite<sub>bottom</sub> + opposite<sub>top</sub> = 2</code></p>
<div class="c2">
It is common in mathematics that you have to remember several things at the same time as you proceed through the steps of solving a problem.
</div>
<h3>Algebraic Substitution Trick on the Measuring Stick</h3>
<p>We can do a similar <i>equivalent</i> substitution in this equation (in this case, we will substitute <code>2 × opposite<sub>top</sub></code> for <code>opposite<sub>bottom</sub></code>:<br />
<code>2 × opposite<sub>top</sub> + opposite<sub>top</sub> = 2</code><br />
Solving we get:<br />
<code>opposite<sub>top</sub> = 0.666...</code></p>
<div class="c2">
If the top portion of the stick us 0.666&#8230; m, then the bottom portion must be 1.333&#8230; m
</div>
<h3>Returning to the original equation:</h3>
<div class="c2">
It does not matter which of the lengths we use (as long as we use them in the right place), the answer will still be the same (any small differences in the answers are due to rounding errors &#8211; you can avoid / minimize these by using the full precision provided by your calculator.
</div>
<p>Now we can take this value for <code>opposite<sub>top</sub></code> and substitute it our original equation for the top right angle triangle:<br />
<code>tan (1°) = 0.666... / adjacent</code><br />
Solving we get:<br />
<code>adjacent = 38.19</code><br />
The distance of the field is 38.19 m (which you can, probably, safely round up to 38.2 m)
</div>

<p>The post <a href="https://complete-concrete-concise.com/mathematics/how-to-solve-trigonometric-problems/">How to Solve Trigonometric Problems</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
]]></content:encoded>
					
		
		
			</item>
		<item>
		<title>Understanding Trigonometric Functions using the Unit Circle (Advanced)</title>
		<link>https://complete-concrete-concise.com/mathematics/understanding-trigonomtric-functions-using-the-unit-circle-advanced/</link>
		
		<dc:creator><![CDATA[richardsplanet]]></dc:creator>
		<pubDate>Tue, 09 Oct 2012 21:13:24 +0000</pubDate>
				<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[cosecant]]></category>
		<category><![CDATA[cosine]]></category>
		<category><![CDATA[cotangent]]></category>
		<category><![CDATA[secant]]></category>
		<category><![CDATA[sine]]></category>
		<category><![CDATA[tangent]]></category>
		<category><![CDATA[trigonometric functions]]></category>
		<category><![CDATA[understanding]]></category>
		<category><![CDATA[unit circle]]></category>
		<guid isPermaLink="false">http://complete-concrete-concise.com/?p=2348</guid>

					<description><![CDATA[<p>While I call this advanced, it does not mean harder or more complicated, it just means more abstract. Understanding the trigonometric functions (sine, cosine, tangent) using right angle triangles is simply a special case of trigonometric functions using the unit circle. I strongly recommend first reading and understanding the article Understanding Sine, Cosine, and Tangent [&#8230;]</p>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/understanding-trigonomtric-functions-using-the-unit-circle-advanced/">Understanding Trigonometric Functions using the Unit Circle (Advanced)</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
]]></description>
										<content:encoded><![CDATA[<div class="c1">
<p>While I call this advanced, it does not mean harder or more complicated, it just means more abstract. Understanding the trigonometric functions (sine, cosine, tangent) using right angle triangles is simply a special case of trigonometric functions using the unit circle.</p>
<p>I strongly recommend first reading and understanding the article <a href="//complete-concrete-concise.com/mathematics/understanding-sine-cosine-and-tangent">Understanding Sine, Cosine, and Tangent</a> first, because it explains the history and reasoning behind the trigonometric functions.</p>
<p>This is the way trigonometric functions are generally understood and defined in mathematics.</p>
</div>
<p>Trigonometric functions were originally developed and understood from the study of right angle triangles. The problem with using right angle triangles is that trigonometric functions can only be defined for angles between 0° and 90°, but not for angles &le; 0° or &ge;90° because no such right angle triangles exist.</p>
<h1>The Unit Circle</h1>
<p>The Unit Circle is a common tool in mathematics. It is a circle with a radius that is one unit long.</p>
<p><strong><u>How long is a unit?</u></strong> It doesn&#8217;t matter. A unit could be 3cm, or 5 inches, or 27 light years. It doesn&#8217;t matter how long the radius is, as long as we agree that whatever the length of the radius we will call that length <u>one unit</u>.</p>
<p>Having a unit circle means that we know the diameter is two units long and the length of the circumference (perimeter) is 2&pi; units long.</p>
<p>The unit circle (almost) always has its centre at (0,0). This is certainly true for the trigonometric functions.</p>
<p>Because the unit circle is centred at (0,0) and has a radius of one unit, we know that it intersects (crosses) the X and Y axes at (1,0), (0,1), (-1,0), and (0,-1).</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/09/trig-functions-unit-circle-1.png" alt="" border="0" class="centered"/></p>
<h1>Circling the Triangle</h1>
<p>Trigonometry is about triangles. The word trigonometry comes from the Greek <em>trigonometria</em> (<em>trigono</em> = triangle and <em>metria</em> = measure). Going from a triangle to a circle may not seem obvious:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/10/trig-functions-unit-circle-2.png" alt="" border="0" class="centered"/></p>
<p><strong>1) Draw</strong> a circle</p>
<p><strong>2) Draw</strong> a line from the centre of the circle to the perimeter. You can draw the line at any angle you like, but make it greater than 0° and less than 90° (just for now)</p>
<div class="c2">
<p><strong>NOTE:</strong> Angles are <strong><u>ALWAYS</u></strong> measured counterclockwise from the x-axis:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/10/trig-functions-unit-circle-4.png" alt="" border="0" class="centered"/></div>
<p><strong>3)</strong> This line is <u>one unit</u> long and is the same as the <u>radius</u> of the circle</p>
<p><strong>4) Drop</strong> a vertical line from where the radius intersects the perimeter. This line will be perpendicular to the x-axis (i.e. it intersects / crosses the x-axis and forms a 90° angle)</p>
<p><strong>5)</strong> The vertical line, along with the x-axis form two sides of a right angle triangle (the two sides that join at 90° &#8211; traditionally called the <u>opposite</u> and <u>adjacent</u> sides of the right angle triangle)</p>
<p><strong>6)</strong> The radius forms the <u>hypotenuse</u> of the right angle triangle with an angle &alpha; relative to the <u>adjacent</u> side (alpha is math-speak for &#8220;whatever angle you chose&#8221;)</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/10/trig-functions-unit-circle-3.png" alt="" border="0" class="centered"/></p>
<div class="c2">
<p>It should be obvious that for angles between 0° and 90° we can go from a circle to an equivalent right angle triangle.</p>
<p>It is not obvious that this makes the mathematics of trigonometry easier or more useful.</p>
</div>
<h1>Trigonometric Functions Using the Unit Circle</h1>
<p>Looking at the triangle formed inside the unit circle, we see there is enough information to work out the trigonometric ratios:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/10/trig-functions-unit-circle-5.png" alt="" border="0" class="centered" /></p>
<p>To calculate sin(&alpha;) we divide the length of the <u>opposite</u> side by the length of the <u>hypotenuse</u>.</p>
<p>The length of the <u>opposite</u> side is: |y<sub>1</sub> &#8211; y<sub>2</sub>| (the absolute value of the difference between y-coordinates). This is either <span class="i1">|(y &#8211; 0)|</span> or <span class="i1">|0 &#8211; y|</span> (depending which way we take the difference of the coordinates), which is the same as the y-coordinate where the radius intersects the perimeter.</p>
<div class="c3">
<p>Lengths are always the absolute value of a difference because a triangle cannot have a side with a length less than zero. The absolute value is either explicitly stated using the absolute value operator <span class="i4">|a &#8211; b|</span> or implicitly done by subtracting the smaller value from the larger value.</p>
</div>
<p>The length of the <u>hypotenuse</u> is: <span class="i1">1</span> (because this is a unit circle and the <u>hypotenuse</u> has the same length as the radius which is exactly one unit).</p>
<p>Therefore, sin(&alpha;) = opposite / hypotenuse = <span class="i1">y / 1</span>, which is the same as <span class="i1">y</span>.</p>
<div class="c2">
<p><strong>Note:</strong> Most math text books will define it as:</p>
<p><code>sin(&alpha;) = y / r</code></p>
<p>I think adding the <code>r</code> only complicates it because <code>r = 1</code>, so there really isn&#8217;t much point in adding it (except for strict completeness).</p>
</div>
<p>By using a unit circle, we can immediately find the sine of an angle by looking at the y-coordinate where the radius intersects the perimeter.</p>
<p>A similar derivation applies for cosine &#8211; which is the same as the x-coordinate where the radius intersects the perimeter, i.e. <span class="i1">cos(&alpha;) = x</span></p>
<div class="c2">
<p><strong>Note:</strong> Most math text books will define it as:</p>
<p><code>cos(&alpha;) = x / r</code></p>
<p>I think adding the <code>r</code> only complicates it because <code>r = 1</code>, so there really isn&#8217;t much point in adding it (except for strict completeness).</p>
</div>
<p>A similar derivation shows that tangent is <span class="i1">y / x</span>.</p>
<div class="c1">
<p>Since a unit circle is not a triangle, it is possible to measure or mark out angles greater than 90° or less than 0°.</p>
<p>It is reasonable to ask if this new way of finding trigonometric ratios work for angles that can never exist in a right angle triangle</p>
</div>
<h1>Angles Greater than 90° or Less Than 0°</h1>
<p>Consider the following radius drawn at an angle greater than 90°:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/10/trig-functions-unit-circle-6.png" alt="" border="0" class="centered"/></p>
<p>Can we use the x and y coordinates the way we did for angles between 0° and 90°?</p>
<p>We can draw a vertical line from where the radius intersects the unit circle and create a right angle triangle as before, but will we get the same answers?</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/10/trig-functions-unit-circle-7.png" alt="" border="0" class="centered"/></p>
<p>Calculating sin(&alpha;) we see that sin(&alpha;) calculated as <span class="i1">opposite / hypotenuse</span> = <span class="i1">y</span> &#8211; just as we observed before.</p>
<p>Calculating cos(&alpha;) we see that cos(&alpha;) calculated as <span class="i1">adjacent / hypotenuse</span> &ne; <span class="i1">x</span>. The calculated value is positive, whereas the value of the x-coordinate is negative.</p>
<p>Calculating tan(&alpha;) we see that tan(&alpha;) calculated as <span class="i1">opposite / adjacent</span> &ne; <span class="i1">y / x</span>. Again, the first value is positive whereas the second value is negative.</p>
<p>We have the correct magnitude (numbers), but the wrong sign (positive / negative).</p>
<p>In the <a href="//complete-concrete-concise.com/mathematics/understanding-sine-cosine-and-tangent">original article</a> on trigonometric functions, I said that the <u>tangent</u> is the slope of the line. Clearly, the slope is negative, so using the unit circle gives us the right answer &#8211; whereas, the triangle way of calculating the <u>tangent</u> gives the wrong answer.</p>
<h3>What Does All This Mean?</h3>
<p>At the beginning of this article I said that angles are measured in a counterclockwise direction relative to the x-axis.</p>
<p>It is not possible to measure the angle &alpha; of the right angle triangle in a counterclockwise direction relative to the x-axis if the triangle is oriented this way (which is the case with the example above):</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/10/trig-functions-unit-circle-8.png" alt="" border="0" class="centered"/></p>
<p>If the triangle is flipped, then it is possible to measure the angle in a counterclockwise direction:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/10/trig-functions-unit-circle-9.png" alt="" border="0" class="centered"/></p>
<p>The two triangles are mirror images of one another. The one on the left is what results with the angle is between 90° and 180° on a unit circle. The one on the right is a mirror image and occurs when we multiply the x-coordinates of the left triangle by -1:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/10/trig-functions-unit-circle-10.png" alt="" border="0" class="centered"/></p>
<div class="c1">
<p><strong>NOTE:</strong> rotating or flipping triangles does not change their properties.</p>
</div>
<div class="c3">
<p><strong>NOTE:</strong> In mathematics and physics (and other sciences) when you get the <em>wrong</em> answer because the signs are different (positive / negative) but the magnitude (numbers) are the same, it is probably because some initial assumption was wrong.</p>
<p>For example, if you calculate the speed at which an asteroid is moving away from the Earth and you get an answer of -400m/s, then you know that your initial assumption (the asteroid is moving away from the Earth) is wrong &#8211; the asteroid is, in fact, moving towards the Earth.</p>
</div>
<p>When using the traditional (triangle) way of calculating trigonometric relations we (implicitly) orient the triangle so it is like this:</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/10/trig-functions-unit-circle-9.png" alt="" border="0" class="centered"/></p>
<p>When we get a negative value, it tells us that the triangle is oriented in a different way.</p>
<h1>Summary</h1>
<ol>
<li>The <u>unit circle</u> allows us to define trigonometric values for all angles &#8211; not just those between 0&deg; and 90&deg;</li>
<li>Using the <u>unit circle</u> allows us to correctly determine the slope (tangent) of a line</li>
<li><u>Sine</u> and <u>cosine</u> can be read directly off the <u>unit circle</u></li>
<table width="480" border="2">
<tr>
<td>Function</td>
<td>How to Compute</td>
<td>Comment</td>
</tr>
<tr>
<td>Sine</td>
<td>y</td>
<td>y-coordinate where the radius intersects the perimeter of the circle</td>
</tr>
<tr>
<td>Cosine</td>
<td>x</td>
<td>x-coordinate where the radius intersects the perimeter of the circle</td>
</tr>
<tr>
<td>Tangent</td>
<td>y / x</td>
<td>ratio of the y-coordinate to the x-coordinate where the radius intersects the perimeter of the circle. This is the same as the <u>slope</u> of the line.</td>
</tr>
<tr>
<td>Cosecant</td>
<td>1 / y</td>
<td>inverse of the y-coordinate where the radius intersects the perimeter of the circle</td>
</tr>
<tr>
<td>Secant</td>
<td>1/ x</td>
<td>inverse of the x-coordinate where the radius intersects the perimeter of the circle</td>
</tr>
<tr>
<td>Cotangent</td>
<td>x / y</td>
<td>ratio of the x-coordinate to the y-coordinate where the radius intersects the perimeter of the circle</td>
</tr>
</table>

<p>The post <a href="https://complete-concrete-concise.com/mathematics/understanding-trigonomtric-functions-using-the-unit-circle-advanced/">Understanding Trigonometric Functions using the Unit Circle (Advanced)</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
]]></content:encoded>
					
		
		
			</item>
		<item>
		<title>Understanding Sine, Cosine and Tangent</title>
		<link>https://complete-concrete-concise.com/mathematics/understanding-sine-cosine-and-tangent/</link>
		
		<dc:creator><![CDATA[richardsplanet]]></dc:creator>
		<pubDate>Sat, 02 Jun 2012 05:31:19 +0000</pubDate>
				<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[cosecant]]></category>
		<category><![CDATA[cosine]]></category>
		<category><![CDATA[cotangent]]></category>
		<category><![CDATA[secant]]></category>
		<category><![CDATA[sine]]></category>
		<category><![CDATA[tangent]]></category>
		<category><![CDATA[trigonometric functions]]></category>
		<category><![CDATA[understanding]]></category>
		<guid isPermaLink="false">http://complete-concrete-concise.com/?p=2164</guid>

					<description><![CDATA[<p>An article explaining trigonometric functions using the unit circle can be found here Using the unit circle is the standard way trigonometric functions are defined and understood in mathematics. I recommend reading and understanding this article first. Later, if you want to understand how trigonometric functions are defined for values greater than 90° or less [&#8230;]</p>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/understanding-sine-cosine-and-tangent/">Understanding Sine, Cosine and Tangent</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
]]></description>
										<content:encoded><![CDATA[<div class="c1">
<p>An article explaining trigonometric functions using the unit circle can be found <a href="//complete-concrete-concise.com/mathematics/understanding-trigonomtric-functions-using-the-unit-circle-advanced">here</a></p>
<p>Using the unit circle is the standard way trigonometric functions are defined and understood in mathematics.</p>
<p>I recommend reading and understanding <u>this</u> article first. Later, if you want to understand how trigonometric functions are defined for values greater than 90° or less than 0°, go and read the other article.</p>
</div>
<p>Sine is often introduced as follows:</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent.png" alt="" border="0" /></p>
<p>Which is accurate, but causes most people&#8217;s eyes to glaze over.</p>
<p>The problem is that from the time humans starting studying triangles until the time humans developed the concept of trigonometric functions (sine, cosine, tangent, secant, cosecant and cotangent) was over 3000 years.</p>
<h1>A Little History</h1>
<p>The ancients studied triangles. One of the things they did was to compare the lengths of the sides of triangles:</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-2.png" alt="" border="0" /></p>
<p>A triangle has three sides so there are 6 different ways to compare sides:</p>
<p>A to B, A to C, B to C, B to A, C to A and C to B</p>
<p>Normally, we would write these as fractions:</p>
<pre>A   A   B   B   C   C
-,  -,  -,  -,  -,  -
B   C   C   A   A   B</pre>
<p>What they discovered was that if two triangles have the same ratios for their three sides, then the triangles are the same shape &#8211; they have the same angles inside even though the size of the triangles may be different.</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-3.png" alt="" border="0" /></p>
<p>This turned out to be very, very useful because it meant that if you could find a smaller triangle that had the same shape as a larger triangle, then you could study the smaller triangle and find out things about the big triangle. It works the other way around as well, if you have a very small triangle and can find a larger, easier to work with triangle, then you can study the larger triangle and learn things about the smaller triangle. These types of triangles are called <u>similar triangles</u>.</p>
<p>One of the earliest uses of this was <em>shadow reckoning</em> &#8211; using shadows to measure things.</p>
<p>Suppose you want to measure the height of a tree, how would you do it?</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-4.png" alt="" border="0" /></p>
<p>You could climb the tree while carrying a length of rope. You would lower the rope until it reached the ground, mark the length of the rope and then measure the length of the rope (once you climbed down). Unfortunately this is (1) slow, (2) dangerous (there is always the risk of falling), and (3) possibly inaccurate since it is not always possible to climb to the very top of a tree (trees trunks and branches get thinner near the top and they may not be strong enough to support you climbing all the way to the top).</p>
<p>You could cut the tree down and then measure it. But if you want to keep the tree, cutting it down is not an option (it is also pretty slow).</p>
<p>What if you wanted to measure the height of a building or a cliff? Using a rope would work (assuming the rope is long enough), but cutting it down wouldn&#8217;t.</p>
<p>However, the ancients noticed that the tree (or building, or cliff) forms one side of a right angle triangle (we&#8217;ll assume the tree is growing vertically instead of at an angle), its shadow forms the base of a right angle triangle and the rays of sunlight form the hypotenuse.</p>
<p>If we can make a smaller right angle triangle with the hypotenuse at the same angle as the rays of sunlight, then we have a similar triangle. We can then measure the smaller (and similar) triangle and apply what we know about it to the larger triangle:</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-5.png" alt="" border="0" /></p>
<p>Because the large and small triangle are similar, it means the ratios of the sides are the same. If we have a stick of a known length and place it perpendicular to the ground (this is side <span class="i3">a</span> of the small triangle), it will cast a shadow and we can measure the length of that shadow (this is side<span class="i1">b</span> of the small triangle). We can easily measure the length of the shadow cast by the tree (this is side <span class="i1">B</span> of the large triangle. Because the two triangles are similar, we know that the ratio a/b must be the same as the ratio A/B.</p>
<div class="c1">
<p><strong>NOTE:</strong> it doesn&#8217;t matter if the ratio is calculated as a/b or b/a &#8211; as long as it is calculated the same way for both triangles.</p>
</div>
<p>If our measuring stick is 2m tall and it casts a shadow 1.25m long, the ratio a/b is 1.6. Because the triangles are similar, we know that the ratio A/B must be 1.6. If the tree casts a shadow that is 15m long we can plug that number into the equation and write A/15 = 1.6. Multiplying both sides by 15 gives us: (A/15)*15 = 1.6*15. Multiplying through we get A = 24, so the height of the tree is 24m.</p>
<div class="c1">
<p>Definitely try this at home. Calculate the height of your house by measuring the length of its shadow and relating that to the length of a known measure (a yardstick (or meter stick) is good &#8211; if you don&#8217;t have a yardstick or meter stick, you can always use a 12 inch (30cm) ruler). Or measure the heights of trees in your neighbourhood.</p>
</div>
<div class="c2">
<p><a href="http://en.wikipedia.org/wiki/Thales">Thales of Miletus</a> used this technique to measure the height of the Great Pyramid (which was already some 2000 years old by the time Thales measured its height):</p>
<p><img decoding="async" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-6.png" alt="" border="0" /></p>
<p>Thales waited until the shadow was inline with one face of the pyramid and then measured the length of the shadow plus half the length of the base to get the length of side <span class="i1">B</span> of the similar triangle. Obviously he had a measuring stick whose shadow he also measured.</p>
</div>
<p>Eventually, somebody realized that instead of always measuring the sides of a triangle and computing the ratios you could create a lookup table which had the ratios for right-angle triangles with various angles.</p>
<div class="c1">
<p>We know triangles are similar if:</p>
<ol>
<li>the ratios of the 3 sides are the same or</li>
<li>all the angles are the same or</li>
<li>two sides have the same ratio and one angle is the same (as long as it is not between the two sides)</li>
</ol>
<p>We can tell if two right-angle triangles are similar by simply measuring one angle &#8211; as long as it is not the square (90°) angle. Because all the angles in a triangle must add up to 180° we can easily find the 3<sup>rd</sup> angle (180° &#8211; 90° &#8211; <em>measured angle</em> = remaining angle)</p>
<p>Any other lookup table would require we measure at least two sides and calculate their ratio.</p>
</div>
<p>The earliest indication we have of someone doing this is Hipparchus of Nicaea who, in the 2<sup>nd</sup> century BCE, compiled a table of trigonometric ratios (these were chords and are related to, but not the same as our trigonometric ratios). The mathematics of the Greeks ended up in India and was worked upon by Indian mathematicians, their work was then taken and worked further by the Arabs who (by the 9<sup>th</sup> century) had developed the modern notion of the six trigonometric functions we use and had lookup tables for right-angle triangles with different angles. Around the 12<sup>th</sup> or 13<sup>th</sup> century the work of the Arabs arrived back in Europe and was translated from Arabic into Latin, the term <u>sinus</u> (from which we get <em>sine</em>) was the translation used for the Arabic word <u>jiba</u> &#8211; both of which mean <u>fold</u>. Had the scholars not translated the word, we might be using <em>jiba</em> or <em>jib</em> or something similar instead.</p>
<div class="c1">
<p>You can read more about the history of trigonometric functions <a href="http://www-history.mcs.st-and.ac.uk/HistTopics/Trigonometric_functions.html">here</a>, <a href="http://nrich.maths.org/6843">here</a>, and <a href="http://en.wikipedia.org/wiki/History_of_trigonometry">here</a>. [<strong>NOTE:</strong> these are external links and were deemed to be working, safe, and on topic at the time of this article. While I double check links from time to time, I make no guarantee about them. I also appreciate being told if the links are not working or no longer valid.]</p>
</div>
<h1>Sine</h1>
<p>When we ask, <em>&#8220;What is the sine of 30°?&#8221;</em> the answer doesn&#8217;t really have much to do with the angle. The angle describes <u>which</u> right-angle triangle we are talking about and <u>which two sides</u> of the triangle we are interested in &#8211; but the answer is the <u>ratio of the lengths</u> of those two sides.</p>
<p>A better way to ask the question would be:</p>
<p>&#8220;Given a right-angle triangle that has an angle of 30°, what number do we get when we divide the length of the side furthest away from the angle by the length of the hypotenuse?&#8221;</p>
<p>Saying <em>sin(30)</em>, is a shorter way of expressing that previous (long) sentence:</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-7.png" alt="" border="0" /></p>
<p>Things to remember:</p>
<ul>
<li>the value of sine is always between 0 and 1 (more accurately, it is between -1 and 1)</li>
<li>one of the sides is always the hypotenuse (the longest side) and it is always the denominator</li>
<li>the opposite side can be <em>sliced</em> by a line from the angle in question (<strong>S</strong>ine and <strong>S</strong>liced both begin with the letter <strong>S</strong>)<img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-7a.png" alt="" border="0" /></li>
</ul>
<h1>Cosine</h1>
<p>When we ask, <em>&#8220;What is the cosine of 30°?&#8221;</em> the answer doesn&#8217;t really have much to do with the angle. The angle describes <u>which</u> right-angle triangle we are talking about and <u>which two sides</u> of the triangle we are interested in &#8211; and the answer is the <u>ratio of the lengths</u> of those two sides.</p>
<p>A better way to ask the question would be:</p>
<p>&#8220;Given a right-angle triangle that has an angle of 30°, what number do we get when we divide the length of the side closest to the angle by the length of the hypotenuse?&#8221;</p>
<p>Saying <em>cos(30)</em>, is a shorter way of expressing that previous (long) sentence:</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-8.png" alt="" border="0" /></p>
<p>Things to remember:</p>
<ul>
<li>the value of cosine is always between 0 and 1 (more accurately, it is between -1 and 1)</li>
<li>one of the sides is always the hypotenuse (the longest side) and it is always the denominator</li>
<li>the adjacent side is the one that is <u>co</u>nnected to the angle (<strong>Co</strong>sine and <strong>Co</strong>nnected both begin with <strong>Co</strong>)</li>
</ul>
<h1>Tangent</h1>
<p>When we ask, <em>&#8220;What is the tangent of 30°?&#8221;</em> the answer doesn&#8217;t really have much to do with the angle. The angle describes <u>which</u> right-angle triangle we are talking about and <u>which two sides</u> of the triangle we are interested in &#8211; and the answer is the <u>ratio of the lengths</u> of those two sides.</p>
<p>A better way to ask the question would be:</p>
<p>&#8220;Given a right-angle triangle that has an angle of 30°, what number do we get when we divide the length of the side furthest away from the angle by the length of the side closest to the angle?&#8221;</p>
<p>Saying <em>tan(30)</em>, is a shorter way of expressing that previous (long) sentence:</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-9.png" alt="" border="0" /></p>
<div class="c2">
<p><strong>NOTE:</strong> the <u>tangent</u> is the same as the <strong><u>slope</u></strong> of the hypotenuse.</p>
<p>This is important to know / remember when learning Calculus and you hear about the tangent line to a point on a curve. In fact, anytime you hear &#8220;tangent line&#8221; it means the &#8220;slope of the line&#8221;</p>
</div>
<p>Things to remember:</p>
<ul>
<li>the value of tangent is always between 0 and <em>undefined</em> (sometimes, inaccurately, called <em>infinite</em> ∞ &#8211; it is more accurate to say &#8220;the slope tends to infinity&#8221;). As for sine and cosine, it can be positive or negative.</li>
<li>the hypotenuse is not used in calculating the tangent</li>
<li>the tangent is the same as the <strong><u>slope</u></strong> of the hypotenuse</li>
<li>slope is always rise divided by run (height divided by length)</li>
<li>the run (length) is the side connected to the angle (if you were standing where the angle was, you would be standing on the side)</li>
<li>the rise (height) is the side furthest from the angle (if you were standing where the angle was and looked at the side, you would see it rising up)</li>
</ul>
<h1>Secant, Cosecant, and Cotangent</h1>
<p>Secant, cosecant, and cotangent are the other three trigonometric functions and they are the inverse of the first three trigonometric functions.</p>
<h3>Cotangent</h3>
<p>This one is easy, it is the inverse of tangent. Instead of calculating the slope of the hypotenuse, we calculate the inverse of the slope.</p>
<p>cot α = adjacent / opposite (or run / rise)</p>
<h3>Secant</h3>
<p>Secant is <strong>NOT</strong> the inverse of <u>sine</u>. Secant is the inverse of <u>cosine</u>.</p>
<p>We invert <u>both</u> the way we calculate and the way we name the function.</p>
<p>sec α = hypotenuse / adjacent</p>
<h3>Cosecant</h3>
<p>Cosecant is <strong>NOT</strong> the inverse of cosine. Cosecant is the inverse of sine.</p>
<p>We invert <u>both</u> the way we calculate and the way we name the function.</p>
<h1>Pay Attention to Detail</h1>
<p>Oftentimes, the right angle triangle is drawn this way:</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-10.png" alt="" border="0" /></p>
<p>However, the triangle and given angle can be drawn any which way, so it is important to properly identify which sides are which:</p>
<p><img decoding="async" class="centered" src="//complete-concrete-concise.com/wp-content/uploads/2012/06/understanding-sine-cosine-tangent-11.png" alt="" border="0" /></p>
<p>The <span class="i1">hypotenuse</span> is green.</p>
<p>The <span class="i2">adjacent</span> side is blue.</p>
<p>The <span class="i3">opposite</span> side is red.</p>
<h1>Mnemonic</h1>
<p>Nowadays, math seems to focus on teaching &#8220;math words&#8221; &#8211; acronyms or mnemonics that help in remembering the concept. For the trigonometric functions, the following is used: <code>SOH-CAH-TOA</code> (pronounce &#8216;soh kah toe ah&#8217;). It is supposed to help remember the trigonometric function and which sides give which answer.</p>
<p>SOH &#8211; <strong>S</strong>ine = <strong>O</strong>pposite ÷ <strong>H</strong>ypotenuse.</p>
<p>CAH &#8211; <strong>C</strong>osine = <strong>A</strong>djacent ÷ <strong>H</strong>ypotenuse.</p>
<p>TOA &#8211; <strong>T</strong>angent = <strong>O</strong>pposite ÷ <strong>A</strong>djacent.</p>
<p>To make it easier to remember, the letters are used as the first letters in a whimsical sentence. I created these two because they use homonyms (sound-a-like words) for the trigonometric functions:</p>
<div class="c1">
<p>Imagine you have just bought something and there is some paperwork to fill out:</p>
<pre><u>S</u>ign <u>O</u>ver <u>H</u>ere. <u>C</u>o-sign <u>A</u>t <u>H</u>ome. <u>T</u>ake <u>O</u>wnership <u>A</u>nytime.</pre>
<p>or</p>
<p>Imagine you are signing up for a vacation and get some friendly advice at the end. This one does have an extra (unused) word at the end, but it has a homonym for each trigonometric function:</p>
<pre><u>S</u>ign <u>O</u>ver <u>H</u>ere. <u>C</u>o-sign <u>A</u>t <u>H</u>ome. <u>T</u>an <u>O</u>nly <u>A</u>t night.</pre>
</div>
<div class="c3">
<p>This article uses modified tree and pyramid images from <a href="http://openclipart.org/">The Open Clip Art Library</a>. While the artwork is public domain and does not require attribution, I think it is important and polite to acknowledge sources.</p>
</div>
<p>The post <a href="https://complete-concrete-concise.com/mathematics/understanding-sine-cosine-and-tangent/">Understanding Sine, Cosine and Tangent</a> appeared first on <a href="https://complete-concrete-concise.com">Complete, Concrete, Concise</a>.</p>
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