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If you were given an equation $\sin(x)=\sin(ax)$ (say $a$ is a natural number), how would you go about finding all the roots on $[0,2\pi)$ without delving into complex numbers?

From a simple geometric analysis it is obvious that solving for $\pi-x=ax$ would yield 4 solutions, and $x=0$ and $x=\pi$ are another 2 obvious solutions. From complex analysis though we know that there could be many roots in this interval depending on the $a$. Is there any way to find all these roots using only techniques from real analysis?

K.Power
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3 Answers3

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$$\sin(x) - \sin(ax) = 2\sin\left(\frac{x-ax}{2}\right)\cos\left(\frac{x+ax}{2}\right) = 0 $$ This identity is not difficult to prove.

Solving $\frac{x-ax}{2} = {\pi}n$ and $\frac{x+ax}{2} = \frac {\pi}{2} + \pi{n}$ so that

$$x=\frac{2\pi{n}}{1-a} , \frac{2\pi{n} + \pi}{1+a}$$ for $a \neq 1, -1$. The case $a=1$ is trivial, and $a=-1$ yields solutions whenever $\sin$ vanishes, namely $\pi{n}$ which is clear since $\sin$ is an odd function.

MathematicsStudent1122
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$\sin(x) = \sin(ax)\iff x=ax[2\pi]\; \text{or}\; x=\pi-ax[2\pi]\iff \exists k,k'\in \mathbb Z, x=\frac{2k\pi}{1-a} \; \text{or}\; x=\frac{(2k'+1)\pi}{1+a}$

Gabriel Romon
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Your problem is equivalent to finding roots of $T_a(x)-x$ and $T_a(x)+x$ where $T_a$ is the $a$th order Chebychev polynomial.

Square each side (gaining candidate solutions) to get

$$1-\cos^2(x) = 1-\cos^2(ax),$$ which is equivalent to $\cos^2(x)=T_a(\cos(x))^2$ since $cos(ax)=T_a(\cos(x))$ for integer $a$. This means solutions to $$T_a(\cos(x))\pm \cos(x) = 0$$ are candidate solutions.

Chebychev polynomials can be defined iteratively. To compute real solutions we can use Newton's method.

abnry
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