Question

How do you prove that `1 - cos(5theta)cos(3theta) - sin(5theta)sin(3theta) = 2sin^2theta`?

Answer 1

Factor.

`1 - (cos5thetacos3theta + sin5thetasin3theta) = 2sin^2theta`

Note that `cosAcosB + sinAsinB = cos(A - B)`.

`1 - (cos(5theta - 3theta)) = 2sin^2theta`

`1 - cos(2theta) = 2sin^2theta`

Now use `cos(A + B) = cosAcosB - sinAsinB` (because `cos2theta = cos(theta + theta)`) to find an expansion for `cos(2theta)`.

`1 - (cos^2theta - sin^2theta) = 2sin^2theta`

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

Now apply `sin^2x + cos^2x = 1`. This implies that `sin^2x = 1 - cos^2x`.

`sin^2theta + sin^2theta = 2sin^2theta`

`2sin^2theta = 2sin^2theta`

`LHS = RHS`

Since both sides are equal for all values of `theta`, this identity has been proved.

Hopefully this helps!