Difference between revisions of "005 Sample Final A, Question 15"

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! Foundations
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|1) <math>\tan^{-1}(x)</math> can be thought of as <math>\tan^{-1}\left(\frac{x}{1}\right),</math> and this now refers to an angle in a triangle. What are the side lengths of this triangle?
 
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|1) The side lengths are 1, x, and <math>\sqrt{1 + x^2}.</math>
 
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|First, let <math>\theta=\tan^{-1}(x)</math>. Then, <math>\tan(\theta)=x</math>.
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| Now, we draw the right triangle corresponding to <math>\theta</math>. Two of the side lengths are 1 and x and the hypotenuse has length <math>\sqrt{x^2+1}</math>.
 
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| Since <math>\cos(\theta)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}</math>, <math>\cos(\tan^{-1}(x))=\cos(\theta)=\frac{1}{\sqrt{x^2+1}}</math>.
 
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| <math>\frac{1}{\sqrt{x^2+1}}</math>
 
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Latest revision as of 21:13, 21 May 2015

Question Find an equivalent algebraic expression for the following,

Foundations
1) can be thought of as and this now refers to an angle in a triangle. What are the side lengths of this triangle?
Answers:
1) The side lengths are 1, x, and
Step 1:
First, let . Then, .
Step 2:
Now, we draw the right triangle corresponding to . Two of the side lengths are 1 and x and the hypotenuse has length .
Step 3:
Since , .
Final Answer: