<s>[INST] How should I go about proving this identity? Found this on a math worksheet but wasn't able to prove it.
$$4(\cos^6x+\sin^6x)=1+3\cos^2(2x)$$
 [/INST] Looking at the LHS of this equation, we can lower the exponent as follow:
$4(cos^6( x) +sin^6( x ) )= 4(cos^{2\cdot 3}(x) + sin^{2\cdot 3}(x) )= 4[(\dfrac{1+cos(2x)}{2})^3+(\dfrac{1-cos(2x)}{2})^3] = 4 \cdot \dfrac{2 + 2 cos^2(2x) + 4 cos^2(2x)}{2^3}= 4 \cdot \dfrac{2 + 6cos^2(2x)}{8} = \dfrac{2 + 6 cos^2(2x)}{2} = 1 + 3cos^2(2x)$
So we get $4(cos^6( x) +sin^6( x ) )= 1 + 3cos^2(2x)$ as wanted.
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