Question: Prove $$\cos(3x) = \cos^3(x) - 3\cos(x)\sin^2(x) $$ by using De'Moivres Theorem

So far (learning complex numbers at the moment) that De Moivre's theorem states that

if $z$ $=$ $r\text{cis}(\theta)$ then $z^n = r^n\text{cis}(n\theta)$

so with this question I was thinking if

$$ z = \cos(3\theta) + i\sin(3\theta) $$


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

and then expanding and comparing the real part? Is that the right way to go for this question?

  • $\begingroup$ I would say yes. Do you in fact get the identity above? $\endgroup$ Jan 26, 2016 at 6:17
  • $\begingroup$ To obtain $\sin x$, $\cos x$, $\tan x$, $\csc x$, $\sec x$, and $\cot x$, type \sin x, \cos x, \tan x, \csc x, \sec x, and \cot x, respectively, when you are in math mode. $\endgroup$ Jan 26, 2016 at 18:05

2 Answers 2


The solution can be completed in this manner:

We know, by De-Moivre's Theorem, $$(\cos x + i \sin x)^3=\cos 3x + i \sin 3x$$

Therefore, we can write $$\cos^3 x + 3i\cos^2x\sin x + 3i^2\cos x\sin^2 x + i^3 \sin^3 x=\cos 3x + i \sin 3x$$ or, $$\cos^3 x + 3i\cos^2x\sin x - 3\cos x\sin^2 x - i \sin^3 x=\cos 3x + i \sin 3x$$ or, $$(\cos^3 x- 3\cos x\sin^2 x) + i(3\cos^2x\sin x - \sin^3 x)=\cos 3x + i \sin 3x$$

As far as your problem is concerned, just compare the real parts of this equation.

Hope this helps you.


From Euler's Formula, we have

$$e^{i\theta}=\cos(\theta)+i\sin(\theta) \tag 1$$

Letting $\theta \to n\theta$ in $(1)$ reveals that

$$e^{in\theta}=\cos(n\theta)+i\sin(n\theta) \tag 2$$

Since $e^{in\theta}=\left(e^{i\theta} \right)^n$, then we have from $(1)$ and $(2)$

$$\left(\cos(\theta)+i\sin(\theta)\right)^n=\cos(n\theta)+i\sin(n\theta) \tag 3$$

which is De Moivre's Fomula.

Finally, letting $n=3$ in $(3)$ and taking the real part reveals


And we are done!


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