# How do you find the exact value of cos[2 arcsin (-3/5) - arctan (5/12)]?

Mar 28, 2016

Possible values are $\pm 0.1108$ or $\pm 0.6277$

#### Explanation:

arcsin(−3/5)=x means $\sin x = \left(- \frac{3}{5}\right) = - 0.6$.

As $\sin x = 0.6$ for $x = {36.87}^{o}$ and sine is negative in third and fourth quadrant, $x = {180}^{\oplus} {36.87}^{o}$ or ${216.87}^{o}$ and $x = {360}^{o} - {36.87}^{o} = {323.13}^{o}$.

$\arctan \left(\frac{5}{12}\right) = x$ means $\tan x = \left(\frac{5}{12}\right)$.

As $\tan x = \frac{5}{12}$ for $x = {22.62}^{o}$ and tan is positive in first and third quadrant, $x = {22.62}^{o}$ or $x = {180}^{o} + {22.62}^{o}$.or ${202.62}^{o}$.

Hence cos{2arcsin(−3/5)-arctan(5/12)]=cos[2xx216.87^o-22.62^o]=cos411.12^o=cos51.12^o=0.6277 or

cos{2arcsin(−3/5)-arctan(5/12)]=cos[2xx216.87^o-202.62^o]=cos411.12^o=cos231.12^o=-0.6277 or

cos{2arcsin(−3/5)-arctan(5/12)]=cos[2xx323.13^o-22.62^o]=cos623.64^o=-0.1108 or

cos{2arcsin(−3/5)-arctan(5/12)]=cos[2xx323.13^o-202.62^o]=cos443.64^o=0.1108

Mar 28, 2016

Values in exactitude are $\pm \frac{12}{125}$ and $\pm \frac{68}{125}$.

#### Explanation:

Let $A = a r c \sin \left(- \frac{3}{5}\right)$ and $B = a r c \tan \left(\frac{5}{12}\right)$.
Then, $\sin A = - \frac{3}{5} , \cos A = \pm \frac{4}{5}$.

$\cos 2 A = 1 - 2 {\sin}^{2} A = \frac{7}{25}$

$\sin 2 A = 2 \sin A \cos A = - \frac{24}{25} \mathmr{and} \frac{24}{25}$, respectively..

Also, $\tan B = \frac{5}{12} , \left(\sin B = \frac{5}{13} , \cos B = \frac{12}{13}\right) \mathmr{and} \left(\sin B = - \frac{5}{13} , \cos B = - \frac{12}{13}\right)$

The given expression is

cos(2A-B) = cos 2A cos B+sin 2A sin B

= $\pm \frac{84 \pm 120}{375}$.

Note that both sin B and cos B have the same sign + or $-$.