Solving $\sin(5x) = \sin(x)$

Question

If I have an equation:

$$\sin(5x) = \sin(x)$$

In what case can I equate $$5x = x$$

Is it only when there is a multiply of $2\pi n$ on either side, where n is any integer so

$$ 5x = x+2\pi n$$

Also with this method can I get every possible solution or does that not work?

I know I can use the sine addition formula but I want to see others I way I can solve this.

The easiest way would be to write that $\sin(x)=\sin(y) \iff x=y+2\pi n \vee x=\pi-y+2\pi n$. — Galc127, Mar 15 '16 at 07:44
No, $\cos(x)=\cos(y) \iff x=y+2\pi n \vee x=-y+2\pi n$. It comes from the following identities - $\sin(\pi-y)=\sin(y),\cos(-y)=\cos(y)$. — Galc127, Mar 15 '16 at 07:46
Related (and very recent) question: "Finding roots of $\sin(x)=\sin(ax)$ without resorting to complex analysis". — Blue, Mar 15 '16 at 07:55
@Galc127 thank you very much, not sure why they don't teach this to us at school.. — bigfocalchord, Mar 15 '16 at 07:56
@Galc127 I was wondering is there is a proof to sin(x)=sin(y) thing? — bigfocalchord, Mar 15 '16 at 07:57

Eric S. · Accepted Answer · 2016-03-15T09:08:27.247

5

You might find this useful:

$$ \sin(A)=\sin(B) \iff A=B +n2\pi\ \mbox{ or }\ A=\pi-B +n2\pi, $$ with $n\in\mathbb{Z}$. $$ \cos(A)=\cos(B) \iff A=B +n2\pi\ \mbox{ or }\ A=-B +n2\pi, $$ with $n\in\mathbb{Z}$.

So you need to solve two equations separately: $$5x=x+n2\pi$$ and $$5x=\pi-x+n2\pi.$$ Good luck :)

Edit: One way to prove $$ \sin(x)=\sin(y) \iff x=y +n2\pi\ \mbox{ or }\ x=\pi-y +n2\pi, $$ is to just verify the solutions work, using a consequence of Euler's formula: $$\sin(x)=\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}.$$ So, here we go: $$ \sin(x)=\sin(y) \iff \frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}=\frac{\mathrm{e}^{iy}-\mathrm{e}^{-iy}}{2i} \iff \mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}. $$

Now note that $x=y+n2\pi$ is indeed a solution: $$\mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{i(y+n2\pi)}-\mathrm{e}^{-i(y+n2\pi)}=\mathrm{e}^{iy}\mathrm{e}^{in2\pi}-\mathrm{e}^{-iy}\mathrm{e}^{-in2\pi}=\mathrm{e}^{iy}-\mathrm{e}^{-iy} $$
and that $x=\pi-y+n2\pi$ is indeed a solution: $$\mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{i(\pi-y+n2\pi)}-\mathrm{e}^{-i(\pi-y+n2\pi)}=\mathrm{e}^{-iy}\mathrm{e}^{i(2n+1)\pi}-\mathrm{e}^{iy}\mathrm{e}^{-i(2n+1)\pi}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}.$$

The last equalities use $\mathrm{e}^{in\pi}$ is equal to $1$ if $n$ is even and equal to $-1$ if $n$ is odd.

edited Mar 15 '16 at 09:08

answered Mar 15 '16 at 08:39

Eric S.

2,800

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What is the proof for why this method works? – bigfocalchord Mar 15 '16 at 08:40
For me it is intuitively clear when you look at the graph of such functions. For example, if you draw $\sin(x)$ yourself for $x\in[0,\pi]$, you can see that for every $x$ and $\pi-x$ the graph has the same value. I hope this is understandable, I don't have time to draw the picture right now. Maybe someone else can? Or I'll do it when I have more time :) – Eric S. Mar 15 '16 at 08:43
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btw, are these formulas not in your text book / learned to you at school? (judging from an earlier comment you made.) That's a shame :( – Eric S. Mar 15 '16 at 08:53
nope! :( , I even asked my teacher if sin(xyz) = sin(xyz), in what cases are they equal to each other and he had no clue.. he just said to use the sine addition formula.. – bigfocalchord Mar 15 '16 at 08:55
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That's a shame. Anyway, I've added a proof for $\sin(x)=\sin(y)$, but it uses complex exponents, don't know if you're familiar with them. If there are any questions, please let me know! – Eric S. Mar 15 '16 at 09:10
Thank you for the "complex" way of proofing this. Can you also explain to me with elementary trig functions instead of complex? Sorry if I'm bothering :) – bigfocalchord Mar 15 '16 at 09:18
Let us continue this discussion in chat. – bigfocalchord Mar 15 '16 at 09:33
I can't think of a such a proof, and I couldn't find it on the internet either. I guess most text books either just claim the result, or prove it using complex exponents, depending on the level. - But again, drawing the graph gives you the right intuition. – Eric S. Mar 15 '16 at 09:34
Ah okay, thank you so much for the extended help! :) – bigfocalchord Mar 15 '16 at 09:34

egreg · Answer 2 · 2016-03-15T08:56:21.267

Suppose $\sin x=\sin y$. Then for the cosine there are just two possibilities, since $\cos^2x=1-\sin^2x=1-\sin^2y=\cos^2y$: either $\cos x=\cos y$ or $\cos x=-\cos y$.

In the first case, the angles $x$ and $y$ must differ by an integer multiple of $2\pi$; in the second case, if we write $z=\pi-y$, we have $\sin z=\sin y$ and $\cos z=-\cos y$, so $\sin x=\sin z$ and $\cos x=\cos z$, so we're in the above case and we conclude $x$ differs from $z=\pi-y$ by an integer multiple of $2\pi$.

Therefore the equation $\sin x=\sin y$ is solved by the two cases

$x=y+2k\pi$
$x=\pi-y+2k\pi$

In your particular case the first one becomes $5x=x+2k\pi$ that can also be written $$ x=k\frac{\pi}{2} $$ and the second case becomes $5x=\pi-x+2k\pi$, that is, $$ x=\frac{\pi}{6}+k\frac{\pi}{3} $$

Some solutions, for instance $\pi/2$ appear in both sets.

With a similar reasoning you can solve $\cos x=\cos y$ with the two sets

$x=y+2k\pi$
$x=-y+2k\pi$

If you have $\sin x=\cos y$, you reduce it to the previous cases by noting it is the same as $\sin x=\sin(\pi/2-y)$ (or $\cos(\pi/2-x)=\cos y$).

score 2 · Answer 3 · answered Mar 15 '16 at 09:20

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You may also use this identity: \begin{align} \sin(5x)-\sin(x)&=2\sin(\frac{5x-x}{2})\cos(\frac{5x+x}{2})\\ &=2\sin(2x)\cos(3x)\\ \end{align}

answered Mar 15 '16 at 09:20

Math-fun

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score 1 · Answer 4 · answered Mar 15 '16 at 08:25

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$$\sin5x=\sin x\Rightarrow 5x=x+n.2\pi$$ or $$5x=\pi-x+n.2\pi$$ which leads to solutions of the form $$x=n.\frac{\pi}{2}\cup x=(2n+1)\frac{\pi}{6}$$

answered Mar 15 '16 at 08:25

David Quinn

34,121

Sil · Answer 5 · 2016-03-15T09:26:31.013

Apart from trivial solution ($x=y+2n\pi$), you can see others geometrically from looking at the Unit circle:

For angle $x$ in $1$st quadrant you have $y$ from $2$nd quadrant such $y = \pi - x$ which gives you $\sin x =\sin y$. Taking periodicity into account, you have

$$ y = \pi - x + 2n\pi $$

Similarly for $x$ being from $3$rd quadrant you see that it will have same $\sin$ value as $y$ from $4$th quadrant satisfying $2\pi - y = x - \pi$, which is just the same equality as above in disguise (put $n=1$).

So overall you have $$ y = x + 2n\pi $$ $$ y = \pi - x + 2n\pi $$

for $n \in \mathbb{Z}$. Now just plug in your $x$ and $5x$ for $x$ and $y$ to get your solution.

Martín-Blas Pérez Pinilla · Answer 6 · 2016-03-15T08:51:20.840

-1

Alternative solution: expand the LHS: $$16\sin^5 x - 20\sin^3 x + 5\sin x = \sin(5x) = \sin x,$$ and $s = \sin x$ is a solution of $$(16s^4 - 20s^2 + 4)s = 16s^5 - 20s^3 + 4s = 0.$$

edited Mar 15 '16 at 08:51

answered Mar 15 '16 at 07:53

Martín-Blas Pérez Pinilla

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You're joking, aren't you? – egreg Mar 15 '16 at 08:02
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"better"? Are you sure? – 5xum Mar 15 '16 at 08:11
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Just casually solve the quintic :/ – K.Power Mar 15 '16 at 08:14
@egreg, is really so difficult to solving a biquadratic? – Martín-Blas Pérez Pinilla Mar 15 '16 at 08:39
@K.Power, the quintic has a linear factor and the quartic factor is biquadratic. – Martín-Blas Pérez Pinilla Mar 15 '16 at 08:40
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@Martín-BlasPérezPinilla This is the worst method. Please, try the same with $\sin97x=\sin33x$ and we'll see. Then compare with the “$x=y$ or $x=\pi-y$” method. – egreg Mar 15 '16 at 08:42
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I don't really have a problem with saying that this is an alternative solution, but it is objectivelly not better. – 5xum Mar 15 '16 at 08:45
@egreg, this method is surely bad for $\sin97x = \sin33x$, but not for this case. – Martín-Blas Pérez Pinilla Mar 15 '16 at 08:51
@5xum, I've changed the word. – Martín-Blas Pérez Pinilla Mar 15 '16 at 08:52
@Martín-BlasPérezPinilla So I've retracted the downvote. – egreg Mar 15 '16 at 08:53
I've also retracted the close vote. Still, a warning to the OP: This alternative solution works in this particular case, because the quintic polynomial is still easy to solve. It can get really ugly really fast if you have bigger coefficients instead of $5$. – 5xum Mar 15 '16 at 09:02
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I have not retracted the downvote, as I think this answer only blurs thing for the OP. As he's still getting familiar with the $"x=y$ or $x=\pi-y"$ stuff, I see no need to complicate things for him by throwing in an otherwise useless approach... – Eric S. Mar 15 '16 at 09:13
Just to follow your approach, say you do solve the quintic for $s$ and you arrive at $\sin(x)=a$, with $a$ some (possibly beastly) expression, then what? Surely you can't rewrite to $\sin(x)=\sin(...)$ and use the identity you tried to avoid... So, how do you continue when you have $s$? – Eric S. Mar 15 '16 at 10:30
@EricS., $s = 0,\pm 1/2,\pm 1,\implies x = \arcsin s = 0,\pm\pi/6,\pm\pi/2$ in the interval $[-\pi,\pi]$ and now use the periodicity of both functions. – Martín-Blas Pérez Pinilla Mar 15 '16 at 11:47

Solving $\sin(5x) = \sin(x)$

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