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.

  • $\begingroup$ ummmm maybe the golden ratio times $\pi$ $\endgroup$ – Will Jagy Mar 10 '16 at 23:17
  • $\begingroup$ perhaps you mean the value of $f$ give me a minute $\endgroup$ – Will Jagy Mar 10 '16 at 23:18
  • $\begingroup$ $$ f(x) = \frac{1}{2} - \frac{1}{2} \cos 2x + \sin 2x $$ $\endgroup$ – Will Jagy Mar 10 '16 at 23:25
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    $\begingroup$ @WillJagy Why stop there? $$f(x) = \frac{1}{2}+\frac{\sqrt{5}}{2}\sin(2x-\tan^{-1}\frac12)$$ $\endgroup$ – Semiclassical Mar 10 '16 at 23:34

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

$$f(x)=\sin(x)(\sin(x)+2\cos(x)) \ \ (*)$$


$$f'(x)=\cos x (\sin x + 2 \cos x)+\sin x(\cos x -2 sin x).$$

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).

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    $\begingroup$ Is it possible to know why you have called this function: "goniometric" ? $\endgroup$ – Jean Marie Mar 11 '16 at 0:37
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    $\begingroup$ See books.google.com/… (I hope the link works. Ping me if it doesn't and I'll try again.) $\endgroup$ – Barry Cipra Mar 11 '16 at 0:58
  • $\begingroup$ @Barry Cipra Thanks! $\endgroup$ – Jean Marie Mar 11 '16 at 1:02

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