Find the zeros of sin and cosine

Prove that $\sin x = 0$ if and only if $x = n \pi$ for $n \in \mathbb{Z}$ .
Find all $x \in \mathbb{R}$ such that $\cos x = 0$ .

Proof. First, since $\sin x = - \sin x$ implies that if $\sin x = 0$ then $\sin (-x) =0$ , it is sufficient to show the statement holds for all $n \in \mathbb{Z}_{>0}$ (and noting that $\sin 0 = 0$ ).
From Theorem 2.3 (Apostol, Section 2.5) we know $\sin 0 = \sin \pi = 0$ . Hence, the statement holds for the case $n =0$ and $n =1$ . Now, we use induction twice, first for the even integers, and then for the odd integers.

Assume the statement is true for some even $m \in \mathbb{Z}_{\geq 0}$ , i.e., $\sin (m \pi) = 0$ . Then, using the periodicity of the sine function

$0 = \sin (m \pi) = \sin (m \pi + 2 \pi) = \sin ((m+2)\pi).$

Hence, the statement is true for all even $n \in \mathbb{Z}_{\geq 0}$ .

Next, assume it is true for some odd $m \in \mathbb{Z}_{\geq 0}$ . Then,

$0 = \sin m \pi = \sin (m\pi + 2 \pi) = \sin ((m+2)\pi).$

Hence, it is true for all odd $n \in \mathbb{Z}_{\geq 0}$ . Therefore, it is true for all nonnegative integers $n$ , and hence, for all $n \in \mathbb{Z}$ .

Conversely, we must show that these are the only real values for which sine is 0. By the periodicity of sine, it is sufficient to show it true over any $2 \pi$ interval. We choose the interval $(-\pi, \pi)$ and show $\sin x = 0 \iff x = 0$ for all $x \in (-\pi, \pi)$ . From above, we know $\sin 0 =0$ . Then, from the fundamental properties of the sine and cosine, we have the inequalities,

$0 < \cos x < \frac{\sin x }{x} < \frac{1}{\cos x} \qquad x \in \left(0, \frac{\pi}{2} \right).$

Hence, both $\sin x$ and $\cos x$ are positive on $\left(0, \frac{\pi}{2} \right)$ . But, from the co-relation identities, we know

$\sin \left( \frac{\pi}{2} + x \right) = \cos x.$

Thus, for $x \in \left( \frac{\pi}{2} , \pi)$ $\sin x \neq 0$ (since $\cos x \neq 0$ for $x \in \left( 0, \frac{\pi}{2} \right)$ .) But, we also know $\sin \frac{\pi}{2} = 1$ . Hence, for $x \in (0, \pi)$ we have $\sin x \neq 0$ . Since $\sin$ is an odd function,

$\sin (-x) = - \sin x \quad \implies \quad \sin (-x) \neq 0 \quad \text{for } x \in (0, \pi).$

Therefore, $\sin x \neq 0$ for $x \in (-\pi, 0)$ . Therefore, we have $\sin x = 0 \implies x = 0$ for $x \in (-\pi, \pi). \qquad \blacksquare$
Claim: $\cos x = 0$ if and only if $x = \frac{\pi}{2} + n \pi$ .
Proof. Since $\cos x = 0 \implies \sin \left( x + \frac{\pi}{2} \right) =0$ we apply part (a) to conclude

$x + \frac{\pi}{2} = n \pi \quad \implies \quad x = \frac{\pi}{2} + n \pi.$

(where we’ve just pulled an extra factor of $\pi$ out of $n \pi$ to make this addition so that our answer looks like the one in the book) $. \qquad \blacksquare$

3 comments

May 13, 2020 at 9:08 am

Anonymous says:

\sin x = – \sin x
I think it is missing a minus sign on the RHS.

February 8, 2016 at 10:33 am

Anonymous says:

in part a) proof for even values of $n$ , i think you forgot an $n$ after the $m$ in the $\sin(m + 2 \pi)$ .

- February 8, 2016 at 10:47 am
  
  RoRi says:
  
  Hi, I don’t think I forgot an $n$ , I think I forgot a $\pi$ . I added a few words also to hopefully clear up that we are using the periodicity of sine so that the induction hypothesis is $\sin (m \pi) = 0$ and then from periodicity we know $\sin (m \pi + 2 \pi) = 0$ and so we get the next even number since $\sin (m \pi + 2 \pi) = \sin ((m+2) \pi) = 0$ . I also fixed a few other typos throughout the post. Hopefully it’s correct now.
  
  (Also, I hope you don’t mind I added LaTeX to your comment).

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Find the zeros of sin and cosine

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