Ask Uncle Colin: Matching sines

Dear Uncle Colin,

How do I solve $\sin (3x)=\sin (5x)$ for $0\le x<360$?

Seems I Need Extra Smarts

Hi, SINES, and thanks for your message!

This involves a formula I always have to look up, or work out from scratch. Today I am in a working out from scratch sort of mood.

The formula we need

The key thing is to rewrite what we’re trying to solve based around the mean of $3x$ and $5x$ - which is, of course, $4x$. Moving it all to one side:

$\sin (4x-x)-\sin (4x+x)=0$

We can expand those:

$\sin (4x)\cos (x)-\cos (4x)\sin (x)-(\sin (4x)\cos (x)+\cos (4x)\sin (x))=0$

The $\sin (4x)\cos (x)$ terms cancel out and we’re left with:

$-2\cos (4x)\sin (x)=0$

Now we’re cooking!

Two factors, separately

Either $\sin (x)=0$, which gives simple answers of 0 and 180 degrees, or $\cos (4x)=0$, which is somewhat trickier.

Since $0\le x<360$, we have $0\le 4x<1440$, and in that interval, $\cos (4x)=0$ at $4x=90,270,450,630,810,990,1170$ and $1330$ degrees.

Mapping back to $x$, we get $x=22.5,67.5,112.5,157.5,202.5,247.5,292.5,$ and $337.5$ degrees.

(A quick sanity check: $\cos (x)=0$ has two solutions over the circle, so $\cos (4x)=0$ should have eight. We’re good.)

That gives us a total of ten solutions.

Hope that helps!

- Uncle Colin