Answer:
[tex]-\frac12(sin 10\theta + sin 9\theta)[/tex]
Step-by-step explanation:
Mumble. There are formulas that do that, which you either commit to memory, or you quickly derive from the sine of a sum or a sine of a difference. I refuse to the day to memorize, so let's rewrite them down.
[tex]sin (x+y) =sin x\ cos y+ cosx\ siny\\sin (x-y) = sinx\ cos y - cos x \sin y[/tex]
Let's subtract them.
[tex]sin (x+y) - sin (x-y) = -2 cos x\ sin y \rightarro\\cos x \sin y =-\frac12[ sin (x+y) - sin (x-y)][/tex]
Ok, we got the formula. If you remember it, good. If not, know how to get if needed. At this point we can start crunching number. The and difference of the two angles is [tex]x+y= \frac{20\theta}2 = 10\theta\\ x-y =-\frac{18\theta}2 = -9\theta[/tex]
Careful with the order, y is the one inside the sine, so the difference will end negative. We can finally replace and we get:
[tex]-\frac12[sin 10\theta - sin (-9\theta)] = -\frac12(sin 10\theta + sin 9\theta)[/tex]
Note in the last passage you took the minus sign out of the sine since it's an odd function; if it was even you would have just ignored it.
Feel free to distribute the -1/2 if needed, i would prefer with it collected.
