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

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

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

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