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

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Answer 1 · 2016-05-30T04:35:23

#=+-24/325 and +-144/325#.

Let #a = arc sin (-3/5)#. Then, #sin a = -3/5<0#.

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

So, #cos a = +-4/5#

Let # b = arc tan (5/12)#. Then, #tan b = 5/12>0#.

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

So, #sin b = +-(5/13) and cos b = +- (12/13)#.

Now, the given expression

# = cos (2a-b)=cos 2a cos b + sin 2a sin b#

#=(1-2 sin^2 a)cos b + (2 sin a cos a) sin b#

#=(1-2(-3/5)^2)(+-12/13)+2 (-3/5)(+-4/5)(+-5/13)#

#=+-(7/25)(12/13)+-(12/65)#

#=+-84/325+-12/65#

#=+-144/325 and +-24/325.#