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	<title>005 Sample Final A, Question 15 - Revision history</title>
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	<updated>2026-04-22T20:00:41Z</updated>
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		<title>MathAdmin: Created page with &quot;''' Question ''' Find an equivalent algebraic expression for the following, &lt;center&gt;&lt;math&gt; \cos(\tan^{-1}(x))&lt;/math&gt;&lt;/center&gt;  {| class=&quot;mw-collapsible mw-collapsed&quot; style = &quot;...&quot;</title>
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		<updated>2015-06-01T01:29:44Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot;&amp;#039;&amp;#039;&amp;#039; Question &amp;#039;&amp;#039;&amp;#039; Find an equivalent algebraic expression for the following, &amp;lt;center&amp;gt;&amp;lt;math&amp;gt; \cos(\tan^{-1}(x))&amp;lt;/math&amp;gt;&amp;lt;/center&amp;gt;  {| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;''' Question ''' Find an equivalent algebraic expression for the following, &amp;lt;center&amp;gt;&amp;lt;math&amp;gt; \cos(\tan^{-1}(x))&amp;lt;/math&amp;gt;&amp;lt;/center&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
! Foundations&lt;br /&gt;
|-&lt;br /&gt;
|1) &amp;lt;math&amp;gt;\tan^{-1}(x)&amp;lt;/math&amp;gt; can be thought of as &amp;lt;math&amp;gt;\tan^{-1}\left(\frac{x}{1}\right),&amp;lt;/math&amp;gt; and this now refers to an angle in a triangle. What are the side lengths of this triangle?&lt;br /&gt;
|-&lt;br /&gt;
|Answers:&lt;br /&gt;
|-&lt;br /&gt;
|1) The side lengths are 1, x, and &amp;lt;math&amp;gt;\sqrt{1 + x^2}.&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
! Step 1:&lt;br /&gt;
|-&lt;br /&gt;
|First, let &amp;lt;math&amp;gt;\theta=\tan^{-1}(x)&amp;lt;/math&amp;gt;. Then, &amp;lt;math&amp;gt;\tan(\theta)=x&amp;lt;/math&amp;gt;.&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
! Step 2:&lt;br /&gt;
|-&lt;br /&gt;
| Now, we draw the right triangle corresponding to &amp;lt;math&amp;gt;\theta&amp;lt;/math&amp;gt;. Two of the side lengths are 1 and x and the hypotenuse has length &amp;lt;math&amp;gt;\sqrt{x^2+1}&amp;lt;/math&amp;gt;.&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
! Step 3:&lt;br /&gt;
|-&lt;br /&gt;
| Since &amp;lt;math&amp;gt;\cos(\theta)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;\cos(\tan^{-1}(x))=\cos(\theta)=\frac{1}{\sqrt{x^2+1}}&amp;lt;/math&amp;gt;.&lt;br /&gt;
|-&lt;br /&gt;
| &lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
! Final Answer:&lt;br /&gt;
|-&lt;br /&gt;
| &amp;lt;math&amp;gt;\frac{1}{\sqrt{x^2+1}}&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
[[005 Sample Final A|'''&amp;lt;u&amp;gt;Return to Sample Exam&amp;lt;/u&amp;gt;''']]&lt;/div&gt;</summary>
		<author><name>MathAdmin</name></author>
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