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