Question

Question #76702

Answer 1

Let `arcsin(3/5)=theta`

So `sintheta=3/5`

Now we know

`cos^2theta+sin^2theta=1`

`=>cos^2theta=1-sin^2theta`

`=>cos^2theta=1-(3/5)^2=16/25`

`=>costheta=sqrt(16/25)=4/5`

`=>cos(arcsin(3/5))=4/5`