Question

(asin θ-bcos θ)²=?​​

Answer 1

`(a\sin(\theta) - b\cos(\theta))^2\\= a^2\sin^2(\theta) - 2ab\sin(\theta)\cos(\theta) + b^2\cos^2(\theta)\\= a^2\sin^2(\theta) - ab\sin(2\theta) + b^2\cos^2(\theta)`