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Evaluate : sin[pi/3-sin^(-1)(-1/2)]...

Evaluate : `sin[pi/3-sin^(-1)(-1/2)]`

A

0

B

1

C

`-1`

D

`(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
We have `sin[(pi)/(3)-"sin"^(-1)(-(1)/(2))]`
`=sin[(pi)/(3)+"sin"^(-1)((1)/(2))][becausesin^(-1)(-theta)=-sin^(-1)theta]`
`=sin[(pi)/(3)+(pi)/(6)] " "[because"sin"(pi)/(6)=(1)/(2)]`
`=sin[(2pi+pi)/(6)]="sin"(3pi)/(6)="sin"(pi)/(2)=1`
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