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

#arcsin(−3/5)=x# means #sinx=(-3/5)=-0.6#.

As #sinx=0.6# for #x=36.87^o# and sine is negative in third and fourth quadrant, #x=180^o+36.87^o# or #216.87^o# and #x=360^o-36.87^o=323.13^o#.

#arctan(5/12)=x# means #tanx=(5/12)#.

As #tanx=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#

Let #A = arc sin (-3/5)# and #B=arc tan (5/12)#.
Then, #sin A = -3/5, cos A = +-4/5#.

#cos 2A = 1-2 sin^2A=7/25#

#sin 2A=2 sin A cos A=-24/25 and 24/25#, respectively..

Also, #tan B=5/12, (sin B =5/13, cos B=12/13) and (sin B =-5/13, cos B=-12/13) #

The given expression is

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

= #+-(84+-120)/375#.

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