- Prove that if and only if for .
- Find all such that .
- Proof. First, since implies that if then , it is sufficient to show the statement holds for all (and noting that ).
From Theorem 2.3 (Apostol, Section 2.5) we know . Hence, the statement holds for the case and . Now, we use induction twice, first for the even integers, and then for the odd integers.
Assume the statement is true for some even , i.e., . Then, using the periodicity of the sine function
Hence, the statement is true for all even .
Next, assume it is true for some odd . Then,
Hence, it is true for all odd . Therefore, it is true for all nonnegative integers , and hence, for all .
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 interval. We choose the interval and show for all . From above, we know . Then, from the fundamental properties of the sine and cosine, we have the inequalities,
Hence, both and are positive on . But, from the co-relation identities, we know
Thus, for (since for .) But, we also know . Hence, for we have . Since is an odd function,
Therefore, for . Therefore, we have for
- Claim: if and only if .
Proof. Since we apply part (a) to conclude
(where we’ve just pulled an extra factor of out of to make this addition so that our answer looks like the one in the book)
\sin x = – \sin x
I think it is missing a minus sign on the RHS.
in part a) proof for even values of , i think you forgot an after the in the .
Hi, I don’t think I forgot an , I think I forgot a . I added a few words also to hopefully clear up that we are using the periodicity of sine so that the induction hypothesis is and then from periodicity we know and so we get the next even number since . 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).