Question

How do you evaluate `sin[arcsin(3/5)-arccos(3/5) ]`?

Answer 1

`-7/25`

Explanation:

Let `theta = arc sin(3/5). sin theta = 3/5`.
Restricting to the principal value,
`arc cos(3/5)=pi/2-arc sin(3/5)= pi/2-theta`.

#sin(arcsin (3/5)-arc cos (3/5)) =sin(theta-(pi/2-theta))=-sin(pi/2-2theta)=-cos 2theta=-(1-2 sin^2theta)=-(1-2(3/5)^2)=-7/25#