Prove :64 {Cos^8(x) + Sin^8(x)} = cos8 x + 28cos 4x + 35

we know (cos^2 x + sin ^2 x) = 1

square both sides


cos^4 x + sin ^4 x + 2 cos^2 x sin ^2 x = 1

or cos^4 x + sin ^4 x = 1- 2 cos^2 x sin ^2 x

multiply by 2 to get

2(cos^4 x + sin ^4 x) = 2- 4 cos^2 x sin ^2 x = 2 – ( 2 sin x cos x)^2 = 2 – sin ^2 2x

multiply by 2 again to get

4(cos^4 x + sin ^4 x) = 4 – 2 sin ^2 2x = 3 + cos 4x

again square both sides to get

16(cos^8x + sin ^ 8x+ 2 cos ^4 x sin ^4 x) = 9 + cos^2 4x + 6 cos 4x

or 16(cos ^ 8x + sin ^8x ) = 9 + cos^2 4x + 6 cos 4x – 32 cos ^4 x sin ^4 x
= 9 + cos^2 4x + 6 cos 4x – 2* (2 cos x sin x)^4
= 9 + cos^2 4x + 6 cos 4x – 2* (sin 2x)^4
multiply by 4 on both sides

64(cos ^ 8x + sin ^8x) = 36 + 4 cos^2 4x + 24 cos 4x – 2* (2sin^2 2x)^2
= 36 + 4 cos^2 4x + 24 cos 4x – 2* (1- cos 4x)^2
= 36 + 4 cos^2 4x + 24 cos 4x -2(1- 2cos 4x + cos^2 4x)
= 34 + 2 cos^2 4x + 28 cos 4x
= 34 + (1+ cos 8x) + 28 cos 4x
= 35 + cos 8x + 28 cos 4x