Question

prove that cos(60_theta)cos(60+theta)=1÷4(4cos^2theta_3)​

Answer 1

Step-by-step explanation:

We know that ; cos(A-B) =cosA.cosB + sinA.sinB

We know that ; cos(A-B) =cosA.cosB + sinA.sinB& cos(A+B) =cosA.cosB - sinA.sinB

Cos(60-theta) cos(60+theta)= (cos60. Cos theta + sin60. Sin theta) (cos60. Cos theta - sin60. Sin theta)

= (1÷2cos theta + √3÷2sin theta) ( 1÷2cos theta - √3÷2sin theta)

= (1÷2cos theta)² - (√3÷2sin theta)²

= cos² theta ÷ 4 - 3÷4 sin² theta

= 1÷4 {cos² theta - 3(1 - cos² theta)}

= 1÷4 (cos² theta - 3+3 cos² theta)

=1÷4 (4 cos² theta - 3) [proove]