# inverse trig function

Printable View

• June 1st 2007, 01:40 PM
viet
inverse trig function
simplify: $\tan\left(2\cos^{-1}(\frac{x}{4})\right)$

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

$\frac{2tan(\cos^{-1}(\frac{1}{4x}))} {1-tan(\cos^{-1}(\frac{1}{4x}))^2}$
• June 1st 2007, 01:52 PM
Jhevon
Quote:

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

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

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

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

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

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

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

Can you find the value of $\tan \theta$ 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)