Prove that,cosθ cos(60°-θ) cos(60°+θ)= $\frac{1}{4}$ cos 3θ
Prove that,cosθ cos(60°-θ) cos(60°+θ)= $\frac{1}{4}$ cos 3θ Solution: cos θ cos(60°– θ) cos(60°+ θ) = $\frac{1}{2}$ cos θ {2 cos(60°- θ) …
Prove that,cosθ cos(60°-θ) cos(60°+θ)= $\frac{1}{4}$ cos 3θ Solution: cos θ cos(60°– θ) cos(60°+ θ) = $\frac{1}{2}$ cos θ {2 cos(60°- θ) …