EDIT:
I misattributed the solution to factoring. The teacher in fact used a the trig identity:
$\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$
I apologise for the time wasted on my expediant attempt at factoring trig. I am now more educated on trig identities.
Here is the original post
In a proof by induction question I am doing at the moment, the solutions seem to think that:
$$\cos(k\alpha)~\cos(\alpha) = \cos((k+1)\alpha)$$ and that $$-\sin(k\alpha)~sin(\alpha) = -sin((k+1)\alpha)$$ Does this hold true for all trig functions? Why? Would you be able to do similar factoring with addition? Thanks
Facts checking:
$$\cos\left(2\frac\pi6\right)\cos\left(\frac\pi6\right)=\frac{\sqrt3}4$$
vs.
$$\cos\left(3\frac\pi6\right)=0.$$
If we combine your first equation with the well-known addition formula, we get that
$$\sin(k\alpha)\sin(\alpha)=0$$ for all $k$ and all $\alpha$ !?