# Do not understand this question

• Aug 30th 2009, 10:48 AM
ruthvik
Do not understand this question
Find an algebraic expression equivalent to the given expression. (hint : use right triangles)

a. sin(tan inverse(^-1 it says in paper) x)
b. cos(tan inverse(^-1 it says in paper) x)

what does it want me to do? How can i approach this problem?

Thank You
• Aug 30th 2009, 11:35 AM
galactus
They want you to use a right triangle to express the inverse tangent functions in terms of x.

Like so

$\displaystyle sin(tan^{-1}(x))=\frac{x}{\sqrt{x^{2}+1}}$

It can also be done algebraically.
• Aug 30th 2009, 02:34 PM
ruthvik
Quote:

Originally Posted by galactus
They want you to use a right triangle to express the inverse tangent functions in terms of x.

Like so

$\displaystyle sin(tan^{-1}(x))=\frac{x}{\sqrt{x^{2}+1}}$

It can also be done algebraically.

How did you get that?
• Aug 30th 2009, 02:48 PM
o_O
Let $\displaystyle \theta = \tan^{-1} x \ \Leftrightarrow \ \tan \theta = \frac{x}{1} = \text{"} \frac{\text{opp}}{\text{adj}} \text{"}$

So construct a right triangle with angle $\displaystyle \theta$ such that the side opposite it has length $\displaystyle x$ and the side adjacent it is $\displaystyle 1$. By the Pythagorean theorem, the hypotenuse must be $\displaystyle \sqrt{x^2 + 1}$

So what does $\displaystyle \sin \theta$ equal to? What was $\displaystyle \theta$ in the first place?
• Aug 30th 2009, 02:58 PM
ruthvik
Quote:

Originally Posted by o_O
Let $\displaystyle \theta = \tan^{-1} x \ \Leftrightarrow \ \tan \theta = \frac{x}{1} = \text{"} \frac{\text{opp}}{\text{adj}} \text{"}$

So construct a right triangle with angle $\displaystyle \theta$ such that the side opposite it has length $\displaystyle x$ and the side adjacent it is $\displaystyle 1$. By the Pythagorean theorem, the hypotenuse must be $\displaystyle \sqrt{x^2 + 1}$

So what does $\displaystyle \sin \theta$ equal to? What was $\displaystyle \theta$ in the first place?

nice, got it...