Use differentiation to show that
$$y = \sin x$$ has a maximum turning point at $\left(\frac{\pi}2, 1\right)$ and a minimum turning point of $\left(\frac{3\pi}2, -1\right)$.
I know that the turning points occur when $\frac{dy}{dx} = 0$. So …
$$\begin{align}\frac{dy}{dx} &= \cos x \\[6pt]
\implies\quad \cos x &= 0 \\[6pt]
\implies\quad \cos^{-1}(0) &= 90^\circ
\end{align}$$
$\cos x $ is positive in the 1st and 4th quadrants, so $x = 90^\circ$ or $x = 270^\circ$
I am not sure how to progress from here or if indeed this is the correct approach.
Best Answer
You're correct. Just change your measures from degrees to radians and you will get $$x=90^\circ = \frac{\pi}{2} \\ x=270^\circ = \frac{3 \pi}{2}$$ If you plug these back in the original equation you will get $$ \sin\left(\frac{\pi}{2}\right)=1 \\ \sin\left(\frac{3\pi}{2}\right)=-1$$ Which gives you your points $\left(\frac{\pi}{2},1\right),\left(\frac{3\pi}{2},-1\right)$ as extrema (maximum and minimum respectively) to the function.