Simplifying Inverse Trigonometric Functions

**jhb** · July 10th 2009, 07:17 PM

Hello all,

well, I'm not sure how to go about simplifying the following problems

tan(2arctan(x))

and

cos(2arcsin(x))

I know I should let theta = 2arctan(x) => tan(theta) = 2x .... I think. However, from that point forward I'm not sure what to do.

I have several others, however these two are the ones that I believe will enable me to figure out the remainder of the problems.

Any and all help is greatly appreciated. Thank you for your time.

jhb

**Random Variable** · July 10th 2009, 07:56 PM

$\tan(2u) = \frac {2 \tan(u)}{1-\tan^{2}(u)}$

so $\tan(2 \arctan(x)) = \frac {2 \tan(\arctan(x))}{1 - \tan^{2}(\arctan(x))}$ $= \frac {2x}{1 -x^{2}}$

**simplependulum** · July 10th 2009, 08:05 PM

Originally Posted by jhb

Hello all,

well, I'm not sure how to go about simplifying the following problems

tan(2arctan(x))

and

cos(2arcsin(x))

I know I should let theta = 2arctan(x) => tan(theta) = 2x .... I think. However, from that point forward I'm not sure what to do.

I have several others, however these two are the ones that I believe will enable me to figure out the remainder of the problems.

Any and all help is greatly appreciated. Thank you for your time.

jhb

$<br /> \cos{2\sin^{-1}{x}} = \cos{2\theta} = 1 - 2\sin^2{\theta}$

since i let $\sin^{-1}{x} = \theta \implies x = \sin{\theta}$

$\cos{2\sin^{-1}{x}} = 1 - 2x^2$

**jhb** · July 10th 2009, 08:17 PM

awesome. thank you for clearing that up for me... I see what I was missing.

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