Do not understand this question

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August 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
August 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

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

It can also be done algebraically.
August 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

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

It can also be done algebraically.

How did you get that?
August 30th 2009, 02:48 PM
o_O

Let $\theta = \tan^{-1} x \ \Leftrightarrow \ \tan \theta = \frac{x}{1} = \text{"} \frac{\text{opp}}{\text{adj}} \text{"}$

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

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

Quote:

Originally Posted by o_O

Let $\theta = \tan^{-1} x \ \Leftrightarrow \ \tan \theta = \frac{x}{1} = \text{"} \frac{\text{opp}}{\text{adj}} \text{"}$

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

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

nice, got it...

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