Question -
Answer -
Let us consider LHS:
4 cos x cos (π/3 + x) cos (π/3 – x) = 2 cos x (2 cos(π/3 + x) cos (π/3 – x))
By using the formula,
2 cos A cos B = cos (A + B) + cos (A – B)
2 cos x (2 cos (π/3+x) cos (π/3 – x)) = 2 cos x (cos(π/3+x + π/3-x) + cos (π/3+x – π/3+ x))
= 2 cos x (cos (2π/3) + cos (2x))
= 2 cos x {cos 120° + cos 2x}
= 2 cos x {cos (180° – 60°) + cos 2x}
= 2 cos x (cos 2x – cos 60°) (since, {cos (180° – A) =– cos A})
= 2 cos 2x cos x – 2 cos x cos 60°
= (cos (x + 2x) + cos (2x – x)) – (2cos x)/2
= cos 3x + cos x – cos x
= cos 3x
= RHS
Hence Proved.