I was asked to find a formula for $\sin(5x)$ in terms of $\sin(x)$ and $\cos(x)$.
I thought about using euler formula which gives:
$$\sin(5x) = e^{i\sin(5x)} = \cos(\sin(5x))+i\cdot sin(\sin(5x))$$

Do you think that's was the intention of this excerise?

Another solution might be:
$$\sin(5x) = \sin(4x + x) = \sin(4x)\cos(x) + \cos(4x)\sin(x)$$

  • 3
    $\begingroup$ Your first equality is wrong: $ \sin 5x = \operatorname{Im}(e^{i5x}) = \operatorname{Im}((e^{ix})^5) $. Then, use for instance a binomial expansion. $\endgroup$ – Clement C. May 5 '14 at 12:04
  • $\begingroup$ Have you checked the De Moivre's formula? $\endgroup$ – ajotatxe May 5 '14 at 12:06

$\cos 5x + i\sin 5x = e^{5ix} = (e^{ix})^5 = (\cos x + i\sin x)^5$

Now just expand the right-hand-side, and equate real and imaginary parts.


$\sin5x=\sin(3x+2x)=\sin3x\cos2x+\cos3x\sin2x=(3\sin x-4\sin^3x)(2\cos^2x-1)+(4\cos^3x-3\cos x)(2\sin x\cos x)$

Now, I think you can take it from here. It is just calculation part you are left with.

The formula that were used are:

$\sin(a+b)=\sin a\cos b+\cos a\sin b$

$\cos 2x=2\cos^2x-1$

$\sin 2x=2\sin x\cos x$

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

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

I expect you are familiar with these formula. Actually most of these can be derived from the first one itself.


$$\begin{align} \sin(5x) &= \frac{e^{i5x} - e^{-i5x}}{2i}\\ &= \frac{\left(\cos(x) + i\sin(x)\right)^5 - \left(\cos(-x) + i\sin(-x)\right)^5}{2i}\\ &= \frac{\left(\cos(x) + i\sin(x)\right)^5 - \left(\cos(x) - i\sin(x)\right)^5}{2i}\\ &\text{at this point the question is answered, but if you want to use binomial expansion...}\\ &= \frac{\sum_{k=0}^5{5 \choose k}\cos(x)^{5 - k} i^k\sin(x)^k - \sum_{k=0}^5{5 \choose k}\cos(x)^{5 - k}(-1)^k i^k\sin(x)^k}{2i}\\ &= \sin(x)^5 - 10\cos(x)^2 \sin(x)^3 + 5\cos(x)^4\sin(x) \end{align}$$

Euler's formula let's you never memorize another trig formula again.

  • $\begingroup$ Thanks, but one thing. How did you initialy got that $\sin(5x) = {{e^{i5x} - e^{-i5x}} \over 2i}$? $\endgroup$ – AnnieOK May 5 '14 at 13:31
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    $\begingroup$ Well Euler's formula is $e^{ix} = \cos(x) + i\sin(x)$, if you also put in $-x$ you get $e^{-ix} = \cos(x) - i\sin(x)$, then you can just subtract the second formula from the first. $\endgroup$ – DanielV May 5 '14 at 13:32
  • $\begingroup$ NP, actually I think TonyK approach is more straightforward than this one. $\endgroup$ – DanielV May 5 '14 at 13:34
  • $\begingroup$ OK, I sohuld approve his answer then. $\endgroup$ – AnnieOK May 5 '14 at 13:37
  • $\begingroup$ @AnnieOK You should choose whichever one helps you the most. $\endgroup$ – DanielV May 5 '14 at 13:43

See Chebyshev Polynomials of the Second Kind, specifically equations (4) and (22).


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