-4

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

Thomas J.
  • 19
  • 4
    Both are wrong. – Kavi Rama Murthy Jun 16 '20 at 07:56
  • 1
    Those equations aren't true in general. You should edit your question to instead show the problem and its solution, and then we can explain what it did. (Your question currently has this difficulty.) My guess is you've seen an equation of the form $a+c=b+d$, not known why the author thought it too obvious to explain, and concluded $a=b$ and $c=d$ are claimed, when really it uses $a-b=d-c$. – J.G. Jun 16 '20 at 07:56
  • 1
    Take $k=1$ and $x=\frac{\pi}{2}$. What do you get? – Matti P. Jun 16 '20 at 07:58
  • The equations are not so far from being true. $$\exp(k\alpha)\exp(\alpha)=\exp((k+1)\alpha).$$ –  Jun 16 '20 at 08:05
  • 2
    In fact$$\cos(k+1)\alpha=\cos k\alpha\cos\alpha-\sin k\alpha\sin\alpha,,\sin(k+1)\alpha=\sin k\alpha\cos\alpha+\cos k\alpha\sin\alpha.$$ – J.G. Jun 16 '20 at 08:07
  • Yes, I have now come to that realization. I did not take the two expressions into account as a group, but rather individually. You are the only commenter to mention this, so kudos. – Thomas J. Jun 16 '20 at 08:26

1 Answers1

1

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 $\alpha$ !?