# Complex numbers and trig identities: $\cos(3\theta) + i \sin(3\theta)$

Using the equally rule $a + bi = c + di$ and trigonometric identities how do I make...

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

Note that $$(\cos(t)+i\sin(t))^n=(\cos(nt)+i\sin(nt)),~~n\in\mathbb Z$$ and $(a+b)^3=a^3+3a^2b+3ab^2+b^3,~~~(a-b)^3=a^3-3a^2b+3ab^2-b^3$.

• $\color{blue}{\bf +1}\;$ for my friend $\quad\color{red}{\bf \ddot \smile}\quad \color{blue}{\bf \checkmark}\;$ – Namaste Mar 23 '13 at 13:31
• @amWhy: Thanks Angel. – mrs Mar 25 '13 at 7:01

Use the fact that

$$\cos{(n x)} + i \sin{( n x)} = (\cos{x} + i \sin{x})^n$$

You are interested in the case $n=3$. Expand the binomial and equate real and imaginary parts.

From the wording, it appears that you are asked to solve the problem without using DeMoivre's Theorem!

Then you need information about $\cos 3x$ and $\sin 3x$ from another source. Luckily, you probably have all the tools needed.

For $\cos 3x$, write it as $\cos(2x+x)$, and use the usual cosine of a sum rule. We get $\cos 2x\cos x-\sin 2x\sin x$. Using the double-angle identities, you then get $(\cos^2 x-\sin^2 x)\cos x-2\sin^2 x\cos x$. I used the rule $\cos 2x=\cos^2 x-\sin^2 x$ because it fits in well with what you are asked to prove.

We conclude that $\cos 3x=\cos^3 x-3\sin^2 x\cos x$.

You will need similar information about $\sin 3x$. Be guided by the ultimate result you are looking for. But start with $\sin 3x=\sin(2x+x)=\sin 2x \cos x+\cos 2x\sin x$. Of course you will then use $\sin 2x=2\sin x\cos x$, and an appropriate identity for $\cos 2x$.

$(\cos x+i\sin x)^3=(\cos 3x+i\sin 3x)$(by De Moivre's theorem)

But $(\cos x+i\sin x)^3=\cos^3x+i^3\sin^3x+3\cos x i \sin x(\cos x+i\sin x)=\cos^3 x-i\sin^3x+3i\cos^2 x\sin x-3\cos x\sin^2x=\cos^3x+i^3\sin^3x+3\cos x i \sin x(\cos x+i\sin x)=\cos^3 x-i\sin^3x+3i(1-\sin^2x) \sin x-3\cos x(1-\cos^2 x)=4\cos ^3x-3\cos x+i(3\sin x-4\sin^3x)$

Now by equating we have, $\cos 3x=4\cos ^3x-3\cos x$ and $\sin 3x=3\sin x-4\sin^3x$