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) cos(60°- θ)
= $\frac{1}{2}$ cos θ {cos(60°- θ +60°+ θ)+cos(60°- θ -60°- θ)}
= $\frac{1}{2}$ cos θ {cos(120°)+cos(-2θ)}
= $\frac{1}{2}$ cos θ (cos120°+cos2θ)
= $\frac{1}{2}$ cos θ (-$\frac{1}{2}$+ cos2θ)
=-$\frac{1}{4}$ cos θ + $\frac{1}{2}$ cosθ cos2θ
=-$\frac{1}{4}$ cos θ +$\frac{1}{4}$ (2 cos θ cos 2θ )
=- $\frac{1}{4}$ cos θ + $\frac{1}{4}$ {cos(θ+2θ) + cos(θ-2θ)}
=- $\frac{1}{4}$ cos θ +$\frac{1}{4}${cos3 θ +cos(-θ)}
=- $\frac{1}{4}$ cos θ + $\frac{1}{4}$ cos3θ +$\frac{1}{4}$ cos θ
=$\frac{1}{4}$ cos3θ [Proved]