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I don't understand how equating/comparing coefficients of trigonometric functions and identities works. I will use this question as an example. The question is as follows.

Use the basic definition of periodicity to show algebraically that the period of $f(x) = \sin(nx)$ is $\frac{2\pi}{n}$ for all $n > 0$.

The answer I got is as follows:

The period of a function $f(x)$ is the smallest $p > 0$ such that $f(x+p) = f(x)$ \begin{gather*} \therefore \sin[n(x+p)] = \sin(nx) \\ \therefore \sin(nx+np)=\sin(nx) \\ \therefore \sin(nx) \cos(np) + \cos(nx)\sin(np) = \sin(nx) \end{gather*} Here is the part where I need help with. The answer in the book continues as follows:

Equating coefficients of $\sin(nx)$ and $\cos(nx)$, \begin{gather*} \cos(np) = 1,\sin(np) = 0 \\ \therefore np = 2k\pi, k \subset \Bbb R \\ \therefore p = \frac{2k\pi}{n}, k \subset \Bbb R \end{gather*}

I just don't understand how they came to the conclusion that $\cos(np) = 1$ and that $\sin(np) = 0$.Can someone explain?

Maths explorer
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  • You have a typo after $\therefore$. – Bernard Oct 02 '20 at 13:54
  • @Bernard To be more specific, after the third $\therefore$. At least, that's where I saw a typo. – Arthur Oct 02 '20 at 13:55
  • The logical structure of your argument makes no sense. The "therefore's" don't belong where you've written them – Adam Rubinson Oct 02 '20 at 13:57
  • @Arthur: Yes. I focused on this specific place, and did't even see the others… – Bernard Oct 02 '20 at 13:59
  • What the proof is using is that if $A\sin \theta + B \cos \theta = C \sin \theta + D \cos \theta$ for all $\theta$, then $A = C$ and $B = D$. This is sort of intuitive by just looking at them. The proper reason is linear independence, but you could also argue as follows: since the equality holds for all $x$, substitute in $x = 0$ and $x = \pi/2n$ to get your desired result. – Izaak van Dongen Oct 02 '20 at 14:58

2 Answers2

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Very simple: in $$\sin nx\cos np+\cos nx\sin np,$$ the functions $\sin nx$ and $\cos nx$ (as functions of $x$) are linearly independent.

Bernard
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Your 3rd line of working is:

$ \therefore \sin(nx) \cos(nx) + \cos(nx)\sin(np) = \sin(nx) $

However, you've made a "silly mistake" here. It should say,

$ \therefore \sin(nx) \cos(np) + \cos(nx)\sin(np) = \sin(nx) $

Now we've got that sorted, see Bernard's answer.

For a proof of Bernard's answer, i.e. to show that that $\sin(u(x))$ and $\cos(u(x))$ are independent, see: Linear independence of $\sin(x)$ and $\cos(x)$

Adam Rubinson
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