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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.

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3 Answers

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 $[-\frac{1}{2},\frac{1}{2}]$.

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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.

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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 $$\sin y-\sin x=2\sin{y-x\over2}\cos{x+y\over2}\geq 2\ {23\over24}{y-x\over2}\ \cdot{3\over8}>0\ .$$

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