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Re: inverse trig indentities

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

Re: inverse trig indentities

Yes, $\displaystyle \arcsin\left(\frac{\sqrt{2}}{2}\right)=\frac{\pi}{ 4}$

Can you be more exact about what you're confused?

To evaluate (for example):

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

Use the identity: $\displaystyle \tan(2\alpha)=\frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$

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

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.

Re: inverse trig indentities

Quote:

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.

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

so ...

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

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