# An Intriguing Identity: $\cos(2x) \overset{?}{=} \log_{\cos(1)}\frac{\cos(\cos(x))}{\cos(\sin(x))}$

I was playing around with Desmos and noticed that: $$\cos(2x)=\log_{\cos(1)}\frac{\cos(\cos(x))}{\cos(\sin(x))}$$ I derived this from: $$\int_{\cos(x)}^{\sin(x)} \tan(u)du$$ when I noticed it looked similar to a sine wave, $\ln(\cos(1))\cos(2x)$, did some algebra and came up with the identity. It seems to check out on a graphing calculator relatively well, except I have no idea why this works. This seems very intriguing to me because it includes nestled trig functions and the cosine of 1, which is very odd to equal a simple cosine function. If anybody can prove this, it could be interesting to see. I have a feeling this is slightly off, because of how much the complexities of the sides differ, so if anybody can disprove it, that would also be great. Thank you.

• All aboard the up vote train... Choo choo! (+1 interesting question) – mrnovice Apr 25 '17 at 1:39
• One way to prove that they are the same is the calculate that they have the same derivative, and then check that they agree at some point. The later bit would be the hard part. I'm a little busy now, but if this is unanswered in an hour or two I'll take a stab at it. – Stella Biderman Apr 25 '17 at 1:42
• These seem to be close but not exactly the same. For $x=1$ I get LHS $= -0.416146836547$, RHS $= -0.409733922774$. – Jair Taylor Apr 25 '17 at 1:46
• Ok, so I calculated the derivative $$\frac {\cos(\cos(x))\cos(x)\sin(\sin(x))+\cos(\sin(x))\sin(\cos(x))\sin(x)} {\ln(\cos(1))\cos(\cos(x))\cos(\sin(x))}=2\cos(2x)$$ – Jacob Claassen Apr 25 '17 at 1:47
• At $x=\pi/6$ the LHS is a rational number and the RHS is a (most surely) transcendental number, mmmh. – Richard Apr 25 '17 at 1:49

The answers above already showed that this is not a true identity so I just want to show why this looks like a good approximation. Let us define

$$h(x) := \frac{\cos(\cos(x))}{\cos(\sin(x))}, f(x) := \log_{\cos(1)} h(x).$$

There are a few key properties that are satisfied by $h(x)$ (and $f(x)$) which make it look like the trigonometric function $\cos(2x)$:

1. Just like $\cos(2x)$, the function $h(x)$ (and hence $f(x)$) is $\pi$-periodic because $$h(x + \pi) = \frac{\cos(\cos(x + \pi))}{\cos(\sin(x + \pi))} = \frac{\cos(-\cos(x))}{\cos(-\sin(x))} = \frac{\cos(\cos(x))}{\cos(\sin(x))} = h(x).$$ If $f$ weren't periodic, by plotting $f$ and $\cos(2x)$ you would immediately see that they can't be equal.
2. In addition, the function $f(x)$ shares another symmetry with $\cos(2x)$. Namely, $$h \left( x + \frac{\pi}{2} \right) = \frac{\cos \left( \cos \left( x + \frac{\pi}{2} \right) \right)}{\cos \left( \sin \left( x + \frac{\pi}{2} \right) \right)} = \frac{\cos(-\sin(x))}{\cos(\cos(x))} = \frac{\cos(\sin(x))}{\cos(\cos(x))} = \frac{1}{h(x)} \implies \\ f \left( x + \frac{\pi}{2} \right) = \log_{\cos(1)} \left( \frac{1}{h(x)} \right) = - \log_{\cos(1)}(h(x)) = -f(x)$$ while $$\cos \left( 2 \left(x + \frac{\pi}{2} \right) \right) = \cos(2x + \pi) = - \cos(2x).$$ Hence, the approximation on the whole of $\mathbb{R}$ looks "as good" as the approximation on $\left[ 0, \frac{\pi}{2} \right]$ and again, without this symmetry it would be more clear graphically that the functions cannot be identical.
3. The function $f$ agrees with $\cos(2x)$ on five different points on $[0,\pi]$ (and on infinitely many points on $\mathbb{R}$ because of the periodicity). Namely, $$f(0) = \log_{\cos(1)} \frac{\cos(\cos(0)}{\cos(\sin(0))} = \log_{\cos(1)} \frac{\cos(1)}{\cos(0)} = \log_{\cos(1)} \cos(1) = 1 = \cos(2 \cdot 0), \\ f \left( \frac{\pi}{4} \right) = \log_{\cos(1)} \frac{\cos \left( \cos \left( \frac{\pi}{4} \right) \right)}{\cos \left( \sin \left( \frac{\pi}{4} \right) \right)} = \log_{\cos(1)} 1 = 0 = \cos \left( 2 \cdot \frac{\pi}{4} \right), \\ f \left( \frac{\pi}{2} \right) = -f(0) = -1 = \cos \left( 2 \frac{\pi}{2} \right), \\ f \left( \frac{3\pi}{4} \right) = -f \left( \frac{\pi}{4} \right) = 0 = \cos \left( 2 \cdot \frac{3\pi}{4} \right), \\ f(\pi) = f(0) = 1 = \cos(2\pi).$$
4. The derivatives of $f$ and $\cos(2x)$ agree on three of the five points above (at $x = 0, \frac{\pi}{2}$ and $x = \pi$).

Given all the properties above it is not surprising that the functions looks quite the same without zooming in. For example, if instead of $f$ we would take an Hermite interpolation polynomial which agrees with $\cos(2x)$ at the points $x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$ and whose derivatives agree with the derivatives of $\cos(2x)$ at $x = 0, \frac{\pi}{2}, \pi$, we would get a polynomial approximation given by

Plotting it against $\cos(2x)$ on $[0,4]$ we would get

You can see that without zooming in, it is impossible to see the difference between $\cos(2x)$ and this interpolation polynomial on $[0,\pi]$. The only reason we see the difference is that the polynomial is not periodic so after being close to $\cos(2x)$ on $[0,\pi]$ it starts shooting up. For $f$, we do not have this phenomenon so the interpolation looks to our eyes even better.

• This is an excellent response and deserves a bounty. – mweiss Apr 27 '17 at 20:14
• @mweiss: Thanks! I actually enjoyed "reverse engineering" this function to see why it is such a decent fit :) – levap Apr 27 '17 at 20:26

It does not seem to be true, as can be seen by plotting the difference of the left and right sides of the "equation":

However, the differences are quite small, and it would still be interesting to see if someone can provide a goodreason why they are approximately equal.

No: they have different second derivatives at $x=0$. As we all know, the second derivative of $\cos{2x}$ at zero is $-4$. On the other hand, the right-hand side is $1/\log{(\cos{1})}$ times $$\log{(\cos{\cos{x}})}-\log{(\cos{\sin{x}})}.$$ Differentiating once gives $$\sin{x} \tan{(\cos{x})} + \cos{x} \tan{(\sin{x})},$$ and differentiating one more time, $$\cos{x} \tan{(\cos{x})} - \sin^2{x} \sec^2{(\cos{x})} -\sin{x}\tan{(\sin{x})} +\cos^2{x} \sec^2{(\sin{x})}.$$ Putting $x=0$ gives $1+\tan{1}$, so the second derivative is $$\frac{1+\tan{1}}{\log{(\cos{1})}},$$ which it is easy to check is not $-4$ (it's about $-4.15$).

Perhaps a more telling way to see why the identity fails is by comparing the second derivatives of each term:

In red is the second derivative of the LHS and in blue that of the RHS. This way, the relative error is large enough to rule out the possibility of floating point error.

I know this is a bit late, but I wanted to give one more perspective with the Fourier transform. Let $$f(x) = \log_{\cos(1)}\frac{\cos(\cos(x))}{\cos(\sin(x))}.$$ We can use MATLAB to take the discrete Fourier transform of $f(x)$ with the following code:

N=2^5;  %Number of grid points
L = 2*pi;   %Length of interval
dx = L/N;   %Spatial step size
k = [0:N/2-1 0 -N/2+1:-1]*(2*pi/L); %Wavenumbers in order
x = (0:(N-1))*dx;   %Grid points
y = log(cos(cos(x))./cos(sin(x)))/log(cos(1));  %f(x)
yt = fft(y);    %Discrete Fourier transform of f(x)


Looking at the the real part of the spectrum gives us the following.

The large spikes correspond to $k = 2$, and the smaller bumps correspond to $k = 6$. $k = 10$ also experiences slight activation. With this level of resolution none of the other nodes are active within $10^{-4}.$ In particular this gives us some insight as to why $f(x) \approx \cos(2x).$ Using the output from MATLAB at this resolution we can approximate $f(x)$ as

$$f(x) \approx \frac{1}{16}\left(15.9245\cos \left(2x\right)+0.0748\cos \left(6x\right)+0.0007\cos \left(10x\right)\right).$$