Question

If `sinx=1/6`, find `cos(x-pi/2)`?

Answer 1

Given `sinx=1/6`

Now `cos(x-pi/2)`

`=>cos(-(pi/2-x))`

as `cos(-theta)=costheta`

`=>cos(pi/2-x)=sinx=1/6`

Answer 2

`cos(x-pi/2)=1/6`

Explanation:

As `cos(-theta)=costheta`,

we have `cos(x-pi/2)=cos(-(pi/2-x))=cos(pi/2-x)`

But, `cos(pi/2-x)=sinx=1/6`

Hence, `cos(x-pi/2)=1/6`