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I expected to always find cos and sin functions to be identical to each other with the only exception being that their phases will differ by π/2

So what I am trying to say here is that we should always expect to find:

f(sin(θ)) = f(cos(θ - π/2))

However, discounting the phase shift, you should always expect the shapes of the curves to be identical. Surely any cos function should be found to be identical in form (shape) to its sibling sinusoidal equivalent, only shifted along by π/2 radians, right?

However, to my astonishment, I discovered a scenario when this was not the case! What I expect to be a universal law of logic, was breached.

Consider the following curves:

$$\frac{10\sin\left(x\right)}{x}$$

and

$$\frac{10\cos\left(x\right)}{x}$$

Graphing these functions (for example, on https://www.desmos.com/calculator), you will find the two curves differ in appearance!! They are not a phase shifted mirror of one another!!

I find this quite unexpected and surprising and am quite amazed by this discovery!

Dissecting the cos equation, I can see where it diverges, but it is unclear to me how I was wrong to assume sine and cosines operations would always yield the same shape only phase shifted? It seems counterintuitive that this assumption is not universally always true. Your thoughts please.

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  • $\begingroup$ That's because your function is not solely of the trigonometric functions, but here you divide by $x$ as well. Why would you expect these complete functions to be the same? $\endgroup$ – David G. Stork Aug 5 '18 at 0:35
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    $\begingroup$ What is the $f$ you're imagining such that $f(\sin x)=\frac{10 \sin x}{x}$? $\endgroup$ – hmakholm left over Monica Aug 5 '18 at 2:59
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You are right, $\cos$ and $\sin$ are similar when we shift the argument $x$ with a proper phase. When considering $\frac{\sin(x)}{x}$ and "shifting the phase", actually all you do is a translation of $x$ by $\frac{\pi}{2}$. So it should become $$\frac{cos(x)}{x+\pi/2}$$ whose graph now has the same shape as the previous one.

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  • $\begingroup$ Maybe $\frac{\cos x}{x+\pi /2}$ is better? $\endgroup$ – user529760 Aug 5 '18 at 0:39
  • $\begingroup$ Oh, indeed! Thank you very much. @user529760 $\endgroup$ – Suzet Aug 5 '18 at 0:40
  • $\begingroup$ You're welcome! $\endgroup$ – user529760 Aug 5 '18 at 0:42
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Your confusion is about assuming that you have $$f(\sin x)=\frac {10\sin x} {x}$$ and $$f(\cos x)=\frac {10\cos x}{x}$$ for some function f.

These two functions , namely, $\frac {10\sin x}{x}$ and $\frac {10\cos x}{x}$ are not $f(\sin x)$ and $f(\cos x)$ at all.

What is the function $f(x)$ where we have $$f(\sin x)=\frac {10\sin x}{x}$$ and $$f(\cos x)=\frac {10\cos x}{x}$$

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