# How do you evaluate cos[2 arcsin (-3/5) - arctan (5/12)]?

May 30, 2016

$= \pm \frac{24}{325} \mathmr{and} \pm \frac{144}{325}$.

#### Explanation:

Let $a = a r c \sin \left(- \frac{3}{5}\right)$. Then, $\sin a = - \frac{3}{5} < 0$.

a is in the 3rd quadrant or in the 4th.

So, $\cos a = \pm \frac{4}{5}$

Let $b = a r c \tan \left(\frac{5}{12}\right)$. Then, $\tan b = \frac{5}{12} > 0$.

b is in the 1st quadrant or in the 3rd.

So, $\sin b = \pm \left(\frac{5}{13}\right) \mathmr{and} \cos b = \pm \left(\frac{12}{13}\right)$.

Now, the given expression

$= \cos \left(2 a - b\right) = \cos 2 a \cos b + \sin 2 a \sin b$

$= \left(1 - 2 {\sin}^{2} a\right) \cos b + \left(2 \sin a \cos a\right) \sin b$

$= \left(1 - 2 {\left(- \frac{3}{5}\right)}^{2}\right) \left(\pm \frac{12}{13}\right) + 2 \left(- \frac{3}{5}\right) \left(\pm \frac{4}{5}\right) \left(\pm \frac{5}{13}\right)$

$= \pm \left(\frac{7}{25}\right) \left(\frac{12}{13}\right) \pm \left(\frac{12}{65}\right)$

$= \pm \frac{84}{325} \pm \frac{12}{65}$

$= \pm \frac{144}{325} \mathmr{and} \pm \frac{24}{325.}$