Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I prove that the sine function on the domain $[-1/2, 1/2]$ is injective?

I completely understand the concept but I'm having trouble writing a proof for this. Thanks in advance.

share|cite|improve this question

If a continuous function $f(x)$ is strictly increasing or decreasing over an interval $N$, then it is injective over $N$.

The derivative of $\sin(x)$ is strictly positive over $[-\frac{1}{2},\frac{1}{2}]$, therefore it is strictly increasing over $[-\frac{1}{2},\frac{1}{2}]$.

share|cite|improve this answer

If ZettaSuro's assumption may not be used (or you can't be bothered to prove it) try using this: Generally, when proving injectivity, show that for $f(a_1) = f(a_2)$ we must have $a_1 = a_2$ . So choose any such such values in the image, and show that they may only be equal if $a_1 = a_2$. I like to think of this as proving no horizontal line crosses the graph of the function $f:[\frac{-1}{2},\frac{1}{2}] \rightarrow \mathbb{R} \ \ \ \ f(x)=\sin(x)$ twice. You must of course decide the level of rigour required yourself.

share|cite|improve this answer

The defining series for $\sin u $ and $\cos u$ are alternating for small $u>0$. It follows that $$\eqalign{\sin u&\geq u-{u^3\over 6}=u\left(1-{u^2\over6}\right)\geq{23 u\over24}\qquad\bigl(0<u\leq{1\over2}\bigr),\cr \cos u&\geq1-{u^2\over2}\geq {3\over8}\qquad\bigl(0<|u|\leq{1\over2}\bigr)\ .\cr}$$ Assume now that $-{1\over2}\leq x<y\leq{1\over2}$. Then $0<{y-x\over2}\leq{1\over2}$ and $\left|{x+y\over2}\right|\leq{1\over2}$. Therefore $$\sin y-\sin x=2\sin{y-x\over2}\cos{x+y\over2}\geq 2\ {23\over24}{y-x\over2}\ \cdot{3\over8}>0\ .$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.