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

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

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

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

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

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

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

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