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

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Answer 1 · 2016-05-15T05:28:38

#= 1/(4 sqrt 10)(( +-1 +- 3 sqrt 15 )#.

Let #a = sin^(-1)(1/4)#. Then, #sin a = 1/4 > 0#.

So, a is in 1st or 2nd quadrant.

Accordingly, #cos a = +-sqrt 15/4#

Let #b = tan^(-1)(-3)#. Then, #tan b = -3 < 0#.

So, b is in 2nd or 4th quadrant.

Accordingly, #sin b = +-3/sqrt 10 and cos b = +- 1/sqrt 10#.

Now, the given expression = sin ( a + b ) = sin a cos b + cos a sin b

#= (1/4)(+-1/sqrt 10) + (+-sqrt 15 / 4 )( +-3/sqrt 10 )#

#= 1/(4 sqrt 10)(( +-1 +- 3 sqrt 15 )#.