Obtain another expression for $(\cos θ + i \sin θ)^4$ by direct multiplication (i.e., expand the bracket). Use the two expressions to show $$ \cos 4\theta = 8 \cos^4 \theta − 8 \cos^2 \theta + 1,\\ \sin 4\theta = 8\cos^3\theta \sin\theta − 4 \cos \theta\sin \theta. $$ You may use the well-known identity: $\sin^2 \theta + \cos^2 \theta = 1$, but do not use any multiple angle formula.
I got using DMT that $(\cos \theta + i\sin \theta)^4 = (cos 4\theta + i\sin 4\theta)$ and using direct multiplication, I got $\cos^4\theta + \sin^4\theta - 4\cos^3\theta*\sin\theta*i - 4\cos\theta \sin^3\theta i - 6\cos^2\theta\sin^2\theta$
By De Moivre’s theorem, $$ (\cos\theta+i\sin\theta)^4=\cos 4\theta+i\sin 4\theta. $$ On the other hand $$ (a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4 $$ Expand the binomial and equate the real and imaginary parts. Where you find $\sin^2\theta$, substitute $1-\cos^2\theta$.
You have, almost correctly, $$ \cos4\theta+i\sin4\theta= \cos^4\theta+4i\cos^3\theta\sin\theta+6i^2\cos^2\theta\sin^2\theta +4i^3\cos\theta\sin^3\theta+i^4\sin^4\theta $$ Now $i^2=-1$, $i^3=-i$ and $i^4=1$, so, by equating the real and imaginary parts, we get $$ \cos4\theta=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\ \sin4\theta=4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta $$ Now it's just a matter of substituting $\sin^2\theta=1-\cos^2\theta$, so, for example, we have \begin{align} \sin4\theta&=4\sin\theta(\cos^3\theta-\cos\theta\sin^2\theta)\\ &=4\sin\theta(\cos^3\theta-\cos\theta+\cos^3\theta)\\ &=8\cos^3\theta\sin\theta-4\cos\theta\sin\theta. \end{align} Do similarly for $\cos4\theta$.