Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How would I solve the following double angle identity.

$$\cos 3x =4\cos^3x-3\cos x $$

I know $\,\cos 3x = \cos(2x+x)$

So know I have $\,\cos 2x +\cos x \,$ , Which is $\,(2\cos^2x-1)\cos x$

But I am not sure what to do next.

share|cite|improve this question
Note $\cos(a+b) \neq \cos(a) + \cos(b),$ and $\cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b).$ – user2468 Jul 27 '12 at 15:37
I think triple angle identity would fit better, no? – Raskolnikov Jul 27 '12 at 15:39
Oh I see now so it would (cos2x)cosx-sin2x(sinx) I think I understand how to solve the problem. Thanks for your help. – Fernando Martinez Jul 27 '12 at 15:40
This is a triple-angle identity, not a double-angle identity. And one may solve equations or prove identities, but to speak of "solving" an identity is somewhat confused. – Michael Hardy Jul 27 '12 at 16:17
up vote 16 down vote accepted

$$e^{ix}=\cos x+i\sin x\Rightarrow e^{3ix}=(\cos x+i\sin x)^3\Rightarrow\cos 3x+i\sin 3x=(\cos x+i\sin x)^3$$

Now expand the cube and equate the real and imaginary parts of both sides to get the answer.

share|cite|improve this answer
+1 Beautiful, though it might be over the OP's level... – DonAntonio Jul 27 '12 at 15:52
But that's the real proof. And by the same proof you get formulas for $\cos(nx)$, $\sin(nx)$ for any integers $n \ge 2$. And furthermore you can copy the formulas right off of Pascal's triangle. – Lee Mosher Jul 27 '12 at 23:26
@LeeMosher - why is this not a real proof ? looks OK to me – Belgi Jul 28 '12 at 0:20

\begin{eqnarray} \cos(3x) &=& \cos(2x+x)\\ &=& \cos(2x)\cos x - \sin(2x)\sin x\\ &=& (\cos^2 x - \sin^2 x)\cos x - 2\sin^2 x\cos x\\ &=& \cos^3 x -(1-\cos^2 x)\cos x - 2 (1 -\cos^2 x)\cos x\\ &=& 2 \cos^3 x -\cos x + \cos^3 x -2 \cos x\\ &=& 4 \cos^3 x - 3 \cos x \end{eqnarray}

share|cite|improve this answer

I might as well: there is in fact a useful recurrence (Chebyshev) that you can use for expressing $\cos\,nx$ entirely in terms of powers of $\cos\,x$:


In this case, you can start with $\cos\,x$ and $\cos\,2x=2\cos^2 x-1$:

$$\begin{align*} \cos\,3x&=(2\cos\,x)(2\cos^2 x-1)-\cos\,x\\ &=4\cos^3 x-2\cos\,x-\cos\,x=4\cos^3 x-3\cos\,x \end{align*}$$

share|cite|improve this answer

I will attempt to answer my own question now.

$$\begin{eqnarray} \cos(2x)(\cos x)-\sin(2x)(\sin x)& =& (2\cos^2 x-1)(\cos x)-2\sin x\cdot\cos x\cdot\sin x\\ &=&2\cos^3x-\cos x-2\sin^2 x\cos x\\ &=&2\cos^3x-\cos x-2(1-\cos^2 x)(\cos x)\\ &=&2\cos^3x-\cos x-2\cos x+2\cos^3 x\\ &=&4\cos^3 x-3\cos x \end{eqnarray}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.