1. ## inverse trig function

simplify: $\displaystyle \tan\left(2\cos^{-1}(\frac{x}{4})\right)$

using the double angle foruma $\displaystyle tan(2x) = \frac{2tan(x)}{1-tan^2x}$
i got to this part:

$\displaystyle \frac{2tan(\cos^{-1}(\frac{1}{4x}))} {1-tan(\cos^{-1}(\frac{1}{4x}))^2}$

2. Originally Posted by viet
simplify: $\displaystyle \tan\left(2\cos^{-1}(\frac{x}{4})\right)$

using the double angle foruma $\displaystyle tan(2x) = \frac{2tan(x)}{1-tan^2x}$
i got to this part:

$\displaystyle \frac{2tan(\cos^{-1}(\frac{1}{4x}))} {1-tan(\cos^{-1}(\frac{1}{4x}))^2}$
Let $\displaystyle \theta = \cos^{-1} \left( \frac {x}{4} \right)$

$\displaystyle \Rightarrow \cos \theta = \frac {x}{4}$ (See diagram below)

Then $\displaystyle \tan \left( 2 \cos^{-1} \left( \frac {x}{4} \right) \right) = \tan (2 \theta )$

$\displaystyle = \frac {2 \tan \theta}{1 - \tan^2 \theta}$

Can you find the value of $\displaystyle \tan \theta$ from the diagram below? ......Of course you can! The problem is child's play from here. (You didn't have to use a triangle, you could have used the Pythagorean Identities for sine and cosine)