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Prove:


\cos {}^(6) ( \alpha ) + \sin {}^(6) ( \alpha ) = (1)/(4) (1 + 3 \cos {}^(2) (2 \alpha ) = (1)/(8) (5 + 3 \cos(4 \alpha ) )

1 Answer

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Use the double angle identity:


\cos^2\alpha=\frac{1+\cos2\alpha}2


\sin^2\alpha=\frac{1-\cos2\alpha}2

This lets us write


\cos^6\alpha+\sin^6\alpha=\left(\frac{1+\cos2\alpha}2\right)^3+\left(\frac{1-\cos2\alpha}2\right)^3


=\frac{1+3\cos2\alpha+3\cos^22\alpha+\cos^32\alpha}8+\frac{1-3\cos2\alpha+3\cos^22\alpha-\cos^32\alpha}8


=\frac{2+6\cos^22\alpha}8=\frac{1+3\cos^22\alpha}4

Use the identity again to write


=\frac{1+3\frac{1+\cos4\alpha}2}4


=\frac{5+3\cos4\alpha}8
User Sudarshan Tanwar
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