Question -
Answer -
Let us consider RHS:
4 (cos6 x – sin6 x)
Upon expansion we get,
4 (cos6 x – sin6 x) =4 [(cos2 x)3 – (sin2 x)3]
= 4 (cos2 x – sin2 x)(cos4 x + sin4 x + cos2 x sin2 x)
By using the formula,
a3 – b3 = (a-b) (a2 +b2 + ab)
= 4 cos 2x (cos4 x + sin4 x+ cos2 x sin2 x + cos2 x sin2 x– cos2 x sin
We know, cos 2x = cos2 x – sin2 x
So,
= 4 cos 2x (cos4 x + sin4 x+ 2 cos2 x sin2 x – cos2 x sin2 x)
= 4 cos 2x [(cos2 x)2 +(sin2 x)2 + 2 cos2 x sin2 x– cos2 x sin2 x]
We know, a2 + b2 + 2ab= (a + b)2
= 4 cos 2x [(1)2 – 1/4 (4 cos2 xsin2 x)]
= 4 cos 2x [(1)2 – 1/4 (2 cos x sin x)2]
We know, sin 2x = 2sin x cos x
= 4 cos 2x [(12) – 1/4 (sin 2x)2]
= 4 cos 2x (1 – 1/4 sin2 2x)
We know, sin2 x = 1 – cos2 x
= 4 cos 2x [1 – 1/4 (1 – cos2 2x)]
= 4 cos 2x [1 – 1/4 + 1/4 cos2 2x]
= 4 cos 2x [3/4 + 1/4 cos2 2x]
= 4 (3/4 cos 2x + 1/4 cos3 2x)
= 3 cos 2x + cos3 2x
= cos3 2x + 3 cos 2x
= LHS
Hence proved.