Trig trig trig

Differentiating gives

 

$$f^{\prime }(x)=\sin (\sin (\cos x))\cdot \cos (\cos (x))\cdot \sin (x)$$

 

This is zero if and only if

 

$$\sin (\sin (\cos x))=0\text{ or }\cos (\cos (x))=0\text{ or }\sin (x)=0$$

 

Consider the first of these three conditions:

 

$$\sin (\sin (\cos x))=0\Rightarrow \sin (\cos x)=n\pi ,n\in \mathbb{Z}$$

 

Since $|\sin(X)|\leq 1$ for any real $X$ and $\pi > 1$ we must choose $n=0$ in the previous equation.

 

Thus,

$$\sin (\sin (\cos x))=0\Rightarrow \sin (\cos x)=0\Rightarrow \cos x=m\pi ,m\in \mathbb{Z}$$

Similarly, we must choose $m=0$ in this expression. We can thus conclude that

$$\sin (\sin (\cos x))=0\iff x=(r+\frac{1}{2})\pi ,r\in \mathbb{Z}$$

 

Consider the second of the three conditions:

 

$$\cos (\cos (x))=0\iff \cos (x)=(r+\frac{1}{2})\pi ,r\in \mathbb{Z}$$

Since $\frac{1}{2}\pi> 1$ there are no real solutions to this condition.

 

Consider the third of the three conditions:

 

$$\sin (x)=0\iff x=n\pi ,n\in \mathbb{Z}$$

 

 

Combining all three conditions gives us the locations of the turning points:

 $$f^{\prime }(x)=0\iff x=\frac{N\pi }{2},N\in \mathbb{Z}$$

We now need to consider whether they are maxima, minima or something else. We could look at the second derivative, but this will be complicated and the boundedness of $\sin(x)$ and $\cos(x)$ allows us to make shortcuts as follows:

 

Notice that $f(x) = 1$ when $x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{5\pi}{2}, \dots$. Since $f(x)$ is continuous and differentiable and $|f(x)|\leq 1$ these points must be maxima. The even multiples of $\frac{\pi}{2}$ must therefore be minima, at which the function takes the values $f(x) = \cos(\sin 1) \approx 0.666$.

A plot of the graph confirms this calculation:

  

Image