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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.

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    $\begingroup$ Note $\cos(a+b) \neq \cos(a) + \cos(b),$ and $\cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b).$ $\endgroup$
    – user2468
    Jul 27, 2012 at 15:37
  • $\begingroup$ I think triple angle identity would fit better, no? $\endgroup$ Jul 27, 2012 at 15:39
  • $\begingroup$ Oh I see now so it would (cos2x)cosx-sin2x(sinx) I think I understand how to solve the problem. Thanks for your help. $\endgroup$ Jul 27, 2012 at 15:40
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    $\begingroup$ 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. $\endgroup$ Jul 27, 2012 at 16:17

5 Answers 5

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

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

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    $\begingroup$ +1 Beautiful, though it might be over the OP's level... $\endgroup$
    – DonAntonio
    Jul 27, 2012 at 15:52
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    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Jul 27, 2012 at 23:26
  • $\begingroup$ @LeeMosher - why is this not a real proof ? looks OK to me $\endgroup$
    – Belgi
    Jul 28, 2012 at 0:20
  • $\begingroup$ Beautiful. Thank you. $\endgroup$
    – Kashmiri
    Nov 12, 2020 at 17:09
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\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}

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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 $\cos\,x$:

$$\cos((n+1)x)=2\cos(x)\cos(nx)-\cos((n-1)x)$$

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

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

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  • $\begingroup$ Thank you. This will help me a lot. $\endgroup$
    – Kashmiri
    Nov 12, 2020 at 17:14
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I will attempt to answer my own question now.

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

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In general, a product of sin and cos functions can be easily solved via Fourier series expansion and convolution.

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