1. ## inverse trig indentities

hi all! i need help with the following. i'll show you how i approached them (sorry if this ends up lengthy!)

1) a) inverse sin sq rt 2/2 b) inverse cos sq rt 3/2 c) inverse cos - sq rt 3/2
2) a) inverse tan sq rt 3 b) inverse tan - sq rt 3 c) inverse sin sq rt 3
3) a) inverse sin 0 b) inverse cos 0 c) inverse cos -1/2
- ok i can't figure out if i know what im doing. an example in the book had a) sin-1= 1/2. so sin is 1/2, meaning it would be pi/6 (because pi/6= sq rt 3/2, 1/2). so for 1a.. would it be pi/4? i'm a little confused...

evaluate expression by sketching a triangle
4) tan(2 inverse tan 5/13)
5) cos(inverse sin 3/5 - inverse cos 3/5)
- i understand sketching triangles and using that information to find the basic trig functions. but, for example, when given tan=(2 inverse tan 5/13) how would u even set that up? i know tan = opposite/adjacent but the whole set up of the problem confuses me =[
rewrite the expression as an algebraic expression in x
6) sin(inverse tan x)
7) cos(inverse tan x)
the only way i would think of how to do these would be solving for x, but i'm not sure if im right. ex: sin(inverse tanx)... would you just do sin/inverse tan = x?

2. ## Re: inverse trig indentities

all of these values are on the unit circle ... review the domain and range of the inverse trig functions.

3. ## Re: inverse trig indentities

Yes, $\arcsin\left(\frac{\sqrt{2}}{2}\right)=\frac{\pi}{ 4}$
Can you be more exact about what you're confused?

To evaluate (for example):
$\tan\left[2\arctan\left(\frac{5}{13}\right)\right]$
Use the identity: $\tan(2\alpha)=\frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$

Honestly, I don't know what you mean by 'evaluete the expression by sketching a triangle'?

4. ## Re: inverse trig indentities

yeah, I'm not sure what it means by that either. And, I just don't understand how you got from tan [2 arctan (5/13)] to the identity you listed.

5. ## Re: inverse trig indentities

Originally Posted by butlerw2
yeah, I'm not sure what it means by that either. And, I just don't understand how you got from tan [2 arctan (5/13)] to the identity you listed.
you need to understand that the output of an inverse trig function is an angle.

$\alpha = \arctan\left(\frac{5}{13}\right)$

so ...

$\tan\left[2\arctan\left(\frac{5}{13}\right)\right] = \tan(2\alpha)$

you now use the double angle identity for tangent that Siron posted and evaluate the value of $\tan(2\alpha)$.