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Find the zeros of sin and cosine

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

  1. 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

  2. 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

2 comments

    • 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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