Question

(sin Θ − cos Θ)^2 − (sin Θ + cos Θ)^2

Answer 1

More generally, we have

`(a-b)^2-(a+b)^2=(a^2-2ab+b^2)-(a^2+2ab+b^2)=-4ab`

so that

`(\sin\theta-\cos\theta)^2-(\sin\theta+\cos\theta)^2=-4\sin\theta\cos\theta=-2\sin2\theta`