Question

Question #ed504

Answer 1

We start by squaring both sides to get rid of the square root.

`(costheta- sin theta)^2 = (sqrt(cos(pi/4 + theta))^2`

`cos^2theta - 2costhetasintheta + sin^2theta = cos(pi/4 + theta)`

We now use the identities `cos^2beta + sin^2beta = 1` and `cos(A + B) = cosAcosB - sinAsinB` to simplify.

`1 - 2costhetasintheta = cos(pi/4)costheta - sin(pi/4)sin theta`

`1 - 2costhetasintheta = costheta/sqrt(2) - sintheta/sqrt2`

As you can see, the two sides aren't equal, so the identity is false.

Hopefully this helps!

Answer 2

I think the question should be-
Prove that

`costheta-sintheta=sqrt2cos(pi/4+theta)`

LHS

`=costheta-sintheta`

`=sqrt2(1/sqrt2costheta-1/sqrt2sintheta)`

`=sqrt2(cos(pi/4)costheta-sin(pi/4)sintheta)`

`=sqrt2cos(pi/4+theta)=RHS`

Proved