Q.8. Trigonometric Identity

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

Answer 8:
64(cos^8 x + sin^8 x)
= 4[(2cos^2 x)^4 + (2sin^2 x)^4]
= 4[(1 + cos2x)^4 + (1 - cos2x)^4]
= 4[(1 + 4cos2x + 6cos^2 2x + 4cos^3 2x + cos^4 2x) + (1 - 4cos2x + 6cos^2 2x - 4cos^3 2x + cos^4 2x)]
= 8(1 + 6cos^2 2x + cos^4 2x)
= 8 + 24(1 + cos4x) + 2(2cos^2 2x)^2
= 32 + 24cos4x + 2(1 + cos4x)^2
= 32 + 24cos4x + 2(1 + 2cos4x + cos^2 4x)
= 32 + 24cos4x + 2 + 4cos4x + 1 + cos8x
= 35 + 28cos4x + cos8x
= cos8x + 28cos4x + 35.

This answer of mine was not selected as the best. However, I selected it as I liked the question and my answer to it. Selection of BA is the asker's prerogative and has to be respected.

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