How do you evaluate #sin (tan^-1(1/4) + cos^-1(-2/3))#?

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Answer 1 · 2016-05-14T14:42:03

#= 2/(3 sqrt 17)(+-1 +- 2 sqrt 5)#

Let #a = tan^(-1)(1/4). tan a =1/4 > 0.#.

So, the angle a is either in the 1st quadrant or in the 3rd.

Accordingly,

sin a = +-1/sqrt 17 and cos a = +-4/sqrt 17#.

Let #b = cos^(-1)(-2/3). cos b =-2/3 < 0.#.

So, the angle b is either in the 2nd quadrant or in the 3rd.

Accordingly, #sin b = +-sqrt 5/3#..

Now, the given expression is

sin (a + b ) = sin a cos b + cos a sin b

#=+-(1/sqrt 17)(-2/3) +- (4/sqrt 17)( sqrt 5 / 3)#

#= 2/(3 sqrt 17)(+-1 +- 2 sqrt 5)#