How does cos(4x) = (sinx)^4+(cosx)^4-6(sinx)^2(cosx)^2

I am completely stumped. The hint is that you use the complex exponential function.

I hope you know that e^(i*x) = Cos(x) + i*Sin(x)

Now raise both sides to the power of 4:
[e^(i*x)]^4 = [Cos(x) + i*Sin(x)] ^4

or e^(i*4x) = [Cos(x) + i*Sin(x)] ^4

But, e^(i*4x) = Cos(4x) + i*Sin(4x)

So, Cos(4x) + i*Sin(4x) = [Cos(x) + i*Sin(x)] ^4
Just expand the RHS of the above equation (a bit tedious) and separate out the real and imaginary parts. The real part is equal to Cos(4x) and the imaginary part is equal to Sin(4x)...(as you can see from the LHS)

i) By multiple angle identity, cos(2A) = 1 - 2sin²A and sin(2A) = 2sin(A)cos(A)

==> cos(4x) = 1 - 2sin²(2x) = 1 - 2[2sin(x)cos(x)]²

ii) ==> cos(4x) = 1 - 8sin²x*cos²x

= (sin²x + cos²x)² - 8sin²x*cos²x [Since sin²x + cos²x = 1 and 1² = 1]

= (sinx)^4 + (cosx)^4 + 2sin²x*cos²x - 8sin²x*cos²x

= (sinx)^4 + (cosx)^4 - 6sin²x*cos²x [Proved]