Is there a nice geometric, intuitive or picture proof as to why the easily algebraically provable identity $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$ is true?

Note I'm not looking for a computational proof like the one linked to, more a proof without words or intuitive style proof, thanks.

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5 Answers5


Enhancing my diagram for the angle-sum formula (currently featured in Wikipedia) to use three angles will get you pretty close ...

enter image description here


$$\begin{align} \cos(\alpha+\beta+\gamma) &= \cos\alpha \cos\beta \cos\gamma - \sin\alpha \sin\beta \cos\gamma - \sin\alpha \cos\beta\sin\gamma - \cos\alpha \sin\beta\sin\gamma \\ \sin(\alpha+\beta+\gamma) &= \sin\alpha \cos\beta \cos\gamma + \cos\alpha \sin\beta \cos\gamma + \cos\alpha \cos\beta \sin\gamma - \sin\alpha \sin\beta \sin\gamma \end{align}$$

With $\alpha = \beta = \gamma = \theta$, these become ... $$\begin{align} \cos 3\theta &= \cos^3\theta - 3 \sin^2\theta \cos\theta \\ \sin 3\theta &= 3\cos^2\theta \sin\theta - \sin^3\theta \end{align}$$ ... which the Pythagorean identity helps us rewrite as ... $$\begin{align} \cos 3\theta &= \cos^3\theta - 3 (1-\cos^2\theta) \cos\theta = 4\cos^3\theta - 3 \cos\theta \\ \sin 3\theta &= 3(1-\sin^2\theta) \sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta \end{align}$$

Off-hand, I don't know of a diagram that goes directly from $\cos 3\theta$ to $4\cos^3\theta-3\cos\theta$.

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    That is absolutely amazing. – bobby Jun 30 '14 at 13:29
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    Very colorful! =) – ARNIE BEBITA-DRIS Sep 04 '14 at 12:36
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    I just love it when someone has the time and effort to make a truly helpful and fascinating visual explanation. Sometimes when the numbers and letters just swim around, a picture can be very helpful – Asimov Sep 04 '14 at 14:17
  • How do you get something featured on Wikipedia? Do you submit it somewhere and then wait for someone to place it on an article? – étale-cohomology Aug 09 '17 at 22:31
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    @étale-cohomology: The "wiki" in "Wikipedia" indicates that the content is user-editable. Anybody can go in and make improvements. As for my diagrams: I'd posted them [here](https://math.stackexchange.com/a/1342/409), and then someone posted "similar" versions to Wikipedia (without credit or notice); someone from here at Math.SE edited the post so that I'd get credit. (Thanks again @ChrisSherlock!) I, myself, recently went in and edited the article's description of those diagrams and provided a link to my new website for such things in the article's list of references. – Blue Aug 09 '17 at 23:00

Assemble four congruent right triangles with hypotenuse $1$ and angle $\theta$ as shown:


In the diagram, the marked angles are all $\theta$. Since $OQ = \cos\theta$, we have $OU = \cos^2\theta$, so $OS = 2\cos^2\theta$, and so $$ OS' = 2\cos^3\theta $$ On the other hand, $$ OS' = OP' + P'Q' + Q'S' = \cos(3\theta) + P'Q' + \cos(\theta) $$ and $$ OS' = OR' - Q'R' + Q'S' = \cos(\theta) - Q'R' + \cos(\theta) $$ Since $P'Q' = Q'R'$, adding these together yields $$ \cos(3\theta) + 3\cos(\theta) = 2OS' = 4\cos^3\theta $$

I hope the diagram is reasonably self-explanatory, that it's clear what needs to be proved to justify the steps of the argument, and that it's easy to prove those things once you've identified them. But if I've misjudged this, let me know and I'll fill in some more details.


I don't think it is a very good idea to try an find a geometric picture for every trig identity because that would take exceedingly long. I will show you that your formula comes from the idea that a rotation by $n\theta$ is equal to $n$ rotations by $\theta$.

Denote $\langle a,b\rangle$ as the vector to the ordered pair $(a,b)$. We will discuss rotations of these vectors. First, consider that the rotation of the vector $\langle 1,0\rangle$ by an angle of $\theta$ is given by $\langle \cos\theta,\sin\theta\rangle$. Similarly, rotation of the vector $\langle 1,0\rangle$ by $3\theta$ is given by $\langle \cos(3\theta),\sin(3\theta)\rangle$.

I will now define a new type of multiplication between vectors. Let $$ \langle a,b\rangle\ltimes\langle c,d \rangle = \langle ac-bd,ac+bc\rangle$$

It so happens that if the vector $\langle a,b\rangle$ makes an angle $\phi_1$ with the horizontal and $\langle c,d \rangle$ makes an angle $\phi_2$ then $\langle a,b\rangle\ltimes\langle c,d \rangle$ makes an angle $\phi_1 + \phi_2\!$. Using this, there is another way to find the rotation of a vector by $3\theta$. We can rotate it three times by $\theta$ rather than rotating it once by $3\theta$. In other words, we will find $\left(\langle \cos\theta,\sin\theta \rangle \ltimes \cos\theta,\sin\theta \rangle\right) \ltimes \cos\theta,\sin\theta \rangle.$ Using this, one could suspect that $$\begin{align}\langle \cos(3\theta),\sin(3\theta)\rangle &= \left(\langle \cos\theta,\sin\theta \rangle \ltimes \cos\theta,\sin\theta \rangle\right) \ltimes \cos\theta,\sin\theta \rangle \\ &= \langle \cos^2\!\theta-\sin^2\!\theta,2\sin\theta\cos\theta\rangle \ltimes \langle \cos\theta, \sin\theta \rangle \\ &= \langle \cos^{3}\!\theta-\sin^2\!\theta\cos\theta-2\!\sin^2\!\theta\!\cos\theta, 2\!\sin\theta\!\cos^2\!\theta+\sin\theta\cos^2\!\theta-\sin^3\!\theta\rangle \\ &=\langle 4\cos^3\!\theta-3\cos\theta,3\sin\theta-4\sin^3\!\theta\rangle\end{align}$$

Where the last step follows from the Pythagorean identities. If we then equate the first elements and the second elements we find the familiar results

$$\begin{align} \cos(3\theta) &= 4\cos^3\!\theta-3\cos\theta \\ \sin(3\theta) &= 3\sin\theta-4\sin^3\!\theta\end{align}$$

This is essentially a result of complex numbers and is called De moivre's formula. You can see it being used to prove your identity here. Our type of multiplication mirrors complex multiplication which is inherently rotational in nature. You can see why here. De moivre's formula says that a rotation by $n\theta$ is equal to $n$ rotations by $\theta$.

The best geometric proof of your formula comes from this basic idea which is so simple that no picture is needed to represent it.

This is advantageous because it can be used to show other identites such as $$\cos(5\theta) = 16 \cos^5\! \theta - 20 \cos^3 \!\theta + 5 \cos \theta$$

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  • This is neither geometric nor intuitive and doesn't address the question in a natural way at all. – beep-boop Jul 02 '14 at 22:14
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    @alexqwx: most of the proofs, including those using De Moivre's, are geometrically based. However, it is hard to find any that I would call intuitive. – robjohn Jul 07 '14 at 00:18

This is not a direct proof but a link to the Fourier series: $4\cos^3\theta=3\cos\theta+\cos3\theta$.

enter image description here

Similarly, for the fifth power: $16\cos^5\theta=10\cos\theta+5\cos3\theta+\cos5\theta$.

enter image description here

The main lesson is that to represent the $k^{th}$ power, the odd harmonics from $1$ to $k$ suffice.

The explanation is not so difficult: the Fourier coefficients are computed from the integrals $$\int_0^{2\pi}\cos^k\theta\ e^{in\theta}d\theta=\int_0^{2\pi}\left(\frac{e^{i\theta}-e^{-i\theta}}2\right)^ke^{in\theta}d\theta.$$ When developing, the lowest exponent of $e^{i\theta}$ is $n-k$. If $n>k$, the integrand is oscillatory and the integral vanishes.

If you admit that, then $\cos^3\theta=a\cos\theta+b\cos3\theta$.

From the figure, at $\theta=0$ the values add up to $1$ and at $\theta=\pi/2$ the slopes cancel out, so that $$a+b=1\\a-3b=0,$$ $$\cos^3\theta=\frac34\cos\theta+\frac14\cos3\theta.$$

Similarly, $\cos^5\theta=a\cos\theta+b\cos3\theta+c\cos5\theta$, at $\theta=0$ the values add up to $1$ and at $\theta=\pi/2$ the first and third derivatives cancel out, so that $$a+b+c=1\\ a-3b+5c=0\\ a-27b+125c=0,$$ $$\cos^5\theta=\frac{10}{16}\cos\theta+\frac{5}{16}\cos3\theta+\frac{1}{16}\cos5\theta.$$


yes.you can write: $ cos(3\theta) = cos(2\theta +\theta) $ =$$cos(2\theta)cos(\theta) - sin(2\theta)sin\theta$$ =$$2cos^3(\theta)-cos \theta -sin(2\theta) sin(\theta) $$=$$2cos^3\theta -cos\theta -2sin^2\theta cos\theta$$=$$2cos^3\theta - cos \theta - 2(cos\theta -cos^3 \theta)$$ so we have : $$4cos^3\theta -3cos\theta$$

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