Equality of inverse trig. functions

Prove that:

$$\arccos \left ( \frac{1}{5} \right )= 2\arctan  \left ( \sqrt{\frac{2}{3}} \right )$$

Solution

We are using the formula:
$$2\arctan x = \arctan \frac{2x}{1-x^2}$$

which is easily extracted via the formula $\displaystyle \tan 2a = \frac{2\tan a}{1-\tan^2 a}$ we have that the RHs is equal to:

$$\frac{2\sqrt{\frac{2}{3}}}{1- \frac{2}{3}}= 6 \sqrt{\frac{2}{3}}$$

Now using the formula $\displaystyle \cos a = \pm \frac{1}{\sqrt{\tan^2 a+1}}$ we have that:

$$\cos C= \frac{a^2+b^2-c^2}{2ab}= \frac{5^2+6^2-7^2}{2\cdot 5\cdot 6}= \frac{1}{5} \Rightarrow C= \arccos \left(\frac{1}{5}\right)$$

and the result follows.

The exercise can also be found in mathematica.gr