Geometric interpretation of the fact that the extrema of $\sin x(\sin x+2\cos x)$ are the golden ratio and its conjugate I stumbled upon the function $$f(x)=\sin(x)(\sin(x)+2\cos(x)).$$
Now I noticed this function has maximum and minimum values $$\frac{1\pm\sqrt5}{2}$$
which are exactly the golden ratio and its conjugate.
Computationally, this can be verified, but I wondered if there is also an intuitive (geometric) explanation of the golden ratio appearing here.
 A: You mean in fact that the maximal or minimal values taken by the function are the golden ratio $\Phi$ or its conjugate $1-\Phi$.
Here is an explanation. The derivative of
$$f(x)=\sin(x)(\sin(x)+2\cos(x)) \ \ (*)$$
is
$$f'(x)=\cos x (\sin x + 2 \cos x)+\sin x(\cos x -2 sin x).$$
which can be written 
$$f'(x)=2 (\cos^2 x - \sin^2 x) + 2 sin x \cos x=2 \cos 2x +\sin 2 x$$ 
Thus, $f'(x)$ is equal to zero if and only if $\tan 2x=-2$. Knowing relationship $\tan 2x = (2 \tan x)/(1- (\tan x)^2 )$, we have now to solve $2T/(1-T^2)=-2$ (by setting $T=\tan x$) which amounts to: 
$$T^2=T+1 \ \ \ (1)$$
Thus $T$ is either $T_1=1.618....$ (golden ratio), or $T_2=-0.618...$ (its conjugate).
Now, we have to go back to (*) because it is the extremal values of $f$ that we are interest in. We can write the expression of $f(x)$ under a form that only involves $\tan x$:
$$f(x)=\sin^2 x + 2 \sin \cos x=\dfrac{\tan^2x}{1+\tan^2 x}+\dfrac{2 \tan x}{1+\tan^2 x} \ \ (2)$$
The extreme values of $f(x)$ are obtained for one of the $T_k$s above.
Let us denote by $T$ any of these two (that both verify (1)).
Equation (2) becomes : $\dfrac{T^2+2T}{1+T^2}$ which is equal... to $T$.
Indeed, $T^2+2T=(1+T^2)T$ is immediately brought back to (1).

