Question

Prove that Sin²A = Cos²(A-B) + Cos²B - 2Cos(A-B)CosACosB​

Answer 1

We have to prove that sin^2 a= cos^2(a-b) + cos^2 b - 2cos(a-b)cos a.cos b

R.H.S= cos^2(a-b)-2cos(a-b)cos a.cos b+cos^2 b

=cos(a-b){cos(a-b)-2cos a.cos b} + cos^2 b

=cos(a-b){cosa.cosb+sina.sinb-2cosa.cosb} + cos^2 b

=cos(a-b){sina.sinb - cosa.cosb} + cos^2 b

= cos (b-a){-cos(a+b)} + cos^2 b. (Since cos(-A)= cosA)

= -{cos(b+a).cos(b-a)} + cos^2 b

-{cos^2 b - sin^2 a)+ cos^2b. (Since cos(a+b).cos(a-b)= cos^2 a - sin^2 b)

= -cos^2 b + sin^2 a + cos^2 b = sin^2 a=LHS