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How prove that $\sin (x) $ is strictly increasing on $(\frac {-\pi}{2},\frac {\pi}{2}) $?

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    Are you allowed derivatives? If not, why not use the unit circle? – 2017-01-23
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    I think there is a typo, otherwise it is true vacuously :) – 2017-01-23
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    How do you define $\sin$? – 2017-01-23

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Let $I$ be your interval, and take $x,y\in I$ with $x0\\ \frac{\sin{y}-\sin{x}}{y-x}&>0\\ \sin{y}-\sin{x}&>0\\ \sin{y}&>\sin{x} \end{align*} This shows that $\sin$ is an increasing function on $I$.

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    If derivatives are allowed, $\sin'x>0$ is simpler. – 2017-01-23
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we have by the first derivative $$(\sin(x))'=\cos(x)$$ and this function is positive in the gieven interval

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    how do i know that $\cos(x)$ is positive for $x\in(-\pi/2,\pi/2)$? – 2017-01-23
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$$\sin a-\sin b=2\sin\frac{a-b}2\cos\frac{a+b}2.$$

For $a>b$, and $a,b\in(-\frac\pi2,\frac\pi2)$, both factors are positive (first argument in the first quadrant, second argument in the first or fourth quadrant).