Question

How do you simplify `\sin [ \sin ^ { - 1} 0+ \cos ^ { - 1} ( - 1) ]`?

Answer 1

`sin[sin^-1 0+cos^-1 (-1)]=0`

Explanation:

We know that ,

`diamondsin^-1 0=arc sin0=0`

`diamondcos^-1(-1)=arc cos(-1)=pi`

So,

`sin[sin^-1 0+cos^-1 (-1)]=sin[0+pi]=sinpi=0`