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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 $\cos^4\theta + \sin^4\theta – 4\cos^3\theta*\sin\theta*i – 4\cos\theta \sin^3\theta i – 6\cos^2\theta\sin^2\theta$

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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 $\cos4\theta$.

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