How does (cosx+isinx)^4 equate to 1-8 cos^2(x)+8 cos^4(x)-4 i cos(x) sin(x)+8 i cos^3(x) sin(x)

Question

I can't figure out how (cosx+isinx)^4 expands to 1-8 cos^2(x)+8 cos^4(x)-4 i cos(x) sin(x)+8 i cos^3(x) sin(x)

I got it equal to sin^4(x)+cos^4(x)+i (4 sin(x) cos^3(x)-4 sin^3(x) cos(x))-6 sin^2(x) cos^2(x)

How do I make it equal to 1-8 cos^2(x)+8 cos^4(x)-4 i cos(x) sin(x)+8 i cos^3(x) sin(x)

Answer 1

expanding we get
$(\cos(x)+i\sin(x))^4=\cos(x)^4-6\cos(x)^2\sin(x)^2+\sin(x)^4+i(4\cos(x)^3\sin(x)-4\cos(x)\sin(x)^3)$ note that $\cos(4x)=8\, \left( \cos \left( x \right) \right) ^{4}-8\, \left( \cos \left( x \right) \right) ^{2}+1 $ and $\sin(4x)=8\, \left( \cos \left( x \right) \right) ^{3}\sin \left( x \right) -4 \,\cos \left( x \right) \sin \left( x \right) $

Answer 2

Hint:

$$(\cos x + i \sin x)^4 = \left(e^{xi}\right)^4 = e^{4xi} = \cos 4x + i \sin 4x.$$

Now use $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$ to express it in terms of $\sin x, \cos x$.