Hey i need to prove that if the tangent of $y=sinx$ creating an angle $\alpha$ with the positive direction of the $X$ axis ($\alpha$ is in radians) so $\frac{3}{4}\pi\leq \alpha\leq \pi$ or $0\leq \alpha\leq \frac{\pi}{4}$.
So i now that $y'=cosx$ and then $cosx=tg\alpha$ and $-1\leq cos\alpha\leq1$ so also $-1\leq tg\alpha\leq 1$ but i don't know why and how to get to $0$ and $\pi$ in the inequalities.
Thanks.
If you have the tangent at the point $(x_0,\sin{x_0})$ then the slope is $\cos{x_0}$. Therefore as you said: $$-1\leq\cos{x_0}\leq 1 \implies -1\leq\tan{\alpha}\leq 1$$
The angle formed by the tangent and the position direction of the axis $x$ will be between $0$ and $\pi$. Then the result is just knowing the behavior of $\tan$ in that interval.
Between $0$ and $\frac{\pi}{2}$ $\tan$ is continous and increasing in $0$ is $0$ and in $\frac{\pi}{4}$ is $1$ therefore for any $\alpha$ great than $\frac{\pi}{4}$ and less than $\frac{\pi}{2}$ the condition is not accomplished.
Between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ is continous and increasing again in $\frac{3\pi}{4}$ is $-1$ and $\pi$ is $0$ (the rest of the interval is not in the problem).