How should I go about proving this identity?

Question

Found this on a math worksheet but wasn't able to prove it.

$$4(\cos^6x+\sin^6x)=1+3\cos^2(2x)$$

Answer 1

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.

Answer 2

You can set $u=e^{ix}$

$4(\cos^6 x+\sin^6 x)=\frac 4{64}\Big((u+\frac 1u)^6-(u-\frac 1u)^6\Big)=\frac 1{16}\times 2\times(6u^4+20+\frac 6{u^4})$

In the binomial dev $1,6,15,20,15,6,1$ you get to keep and double only $6,20,6$ the other terms are cancelling each other due to sign.

$=\frac 18\Big(6(u^2+\frac 1{u^2})^2+8\Big)=1+\frac34(u^2+\frac 1{u^2})^2=1+3\cos(2x)^2$