Prove that (i) (cos α – cos β)^2 + (sin α – sin β)^2 = 4 sin^2 (α - β)/2

Question

Prove that

(i) (cos α – cos β)² + (sin α – sin β)² = 4 sin²(\(\frac{\alpha-\beta}{2}\)) 

(ii) sin A sin(60° + A) sin(60° – A) = sin 3A

Answer 1

(i) LHS = (cos α – cos β)² + (sin α – sin β)²

(ii) LHS = 4 sin A sin (60° + A).sin (60° – A)

= 4 sin A {sin (60° + A).sin (60° – A)}

= 4 sin A {sin² 60° – sin² A}

= 4 sin A {\(\frac{3}{4}\) – sin² A}

= 3 sin A – 4 sin³ A

= sin 3A

= RHS