Using de Moivre's Theorem, show that: $\cos(5\theta)=\cos^5\theta-10 \cos^3\theta \sin^2\theta+5 \cos\theta \sin^4\theta$

Question

I started stating that: $$(\cos\theta+i \sin\theta)^5=\sum_{k=0}^5 \binom{5}{k} (\cos\theta)^{5-k}\cdot (i \sin\theta)^k$$ Then, I apply the binomial theorem and develop, so I ended up with this, but arranged:

$$=\cos^5(\theta)-10 \cos^3(\theta) \sin^2(\theta)+5 \cos(\theta) \sin^4(\theta)+5\cos^4(\theta)-i \sin(\theta)\\-10 \cos^2(\theta) i \sin^3(\theta)+i \sin^5(\theta)$$

So you can see I get what I wanted to obtain, but I can't or dont know how to get rid of this part: $$5 \cos^4(\theta)-i \sin(\theta)-10 \cos^2(\theta) i \sin^3(\theta)+i \sin^5(\theta)$$ Until now, I got: $$i \sin(\theta)[\sin^2(\theta)\cos^2(\theta)]$$ I really don't know how to proceed, any help is appreciated.

Answer 1

On the left hand side, use de Moivre's theorem. On the right hand side, make sure you expand carefully.

Then use the fact that two complex numbers are equal if and only if their real and imaginary parts are equal.

Answer 2

There seems to be a mistake in your expansion; your term which reads $5\cos^4\theta - i\sin\theta$, should in fact be $5i\cos^4\theta\sin\theta$.

By De Moivre’s theorem, $(\cos\theta + i\sin\theta)^5=\cos 5\theta + i\sin 5\theta$. You should then equate real and imaginary parts; since the real part of the expansion is $\cos 5\theta$, you can ignore all the imaginary terms you have in the expansion.

Answer 3

$$(\cos\theta+i \sin\theta)^5=\sum_{k=0}^5 \binom{5}{k} (\cos\theta)^{5-k}\cdot (i \sin\theta)^k$$

The idea is to let the "real part of LHS" equals the "real part of RHS". (surely, "imaginary part of LHS" equals the "imaginary part of RHS"). But for this problem, we only need to show the "real part = real part".

For LHS, the real part is:

$$\Re(\cos\theta+i \sin\theta)^5=\Re e^{i5\theta}=\Re(\cos(5\theta)+i\sin(5\theta))=\cos(5\theta)$$

For RHS, the real part doesn't contain the factor $i$, and this happens only for those terms when $k$ takes even integers, namely, $k=0, 2, 4$.

For RHS, the real part is:

$$\Re \sum_{k=0}^5 \binom{5}{k} (\cos\theta)^{5-k}\cdot (i \sin\theta)^k=\cos^5(\theta)-10 \cos^3(\theta) \sin^2(\theta)+5 \cos(\theta) \sin^4(\theta)$$

Therefore, we get:

$$\cos(5\theta)=\cos^5(\theta)-10 \cos^3(\theta) \sin^2(\theta)+5 \cos(\theta) \sin^4(\theta)$$

The proof is completed.