Finding the max/min turning points of $y=\sin x$ 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 ...

$$\begin{align}\frac{dy}{dx} &amp;= \cos x \\[6pt]
\implies\quad \cos x &amp;= 0 \\[6pt]
\implies\quad \cos^{-1}(0) &amp;= 90^\circ
\end{align}$$

$\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.
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.
Let $f(x) = \sin x$, then $f'(x)=\cos x = 0 \iff x = 90^{\circ}$ or $x = 270^{\circ}$, and $f''(x) = - \sin x$ has $f''(90^{\circ}) = -1&lt; 0$, and $f''(270^{\circ}) = 1 &gt; 0$. Thus the max occurs at $x = 90^{\circ}$, and min happens at $x = 270^{\circ}$.