What is $\sin(\cos(x))$?

Question

As the title asks, what is $\sin(\cos(x))$. I want it to be reduced more, if possible. I don't know where to start. I could manipulate the expression in many ways, but none allow me to remove either the $\sin$ or $\cos$. For example:$$\sin(\cos(x))=\sqrt{1-(\cos(\cos(x))^2}$$

Then I'd need to know what $\cos(\cos(x))$ is. Anyways, I want to know if this is solvable such that I don't have a trig function inside another trig function, and a solution if one such exists.

normally in engineering, sin or cos in another sin or cos has no practical application. But theoretically you can use it, however forget the conceptual view anymore. — ar2015, Nov 26 '15 at 02:34
Well, for all practical purposes the answer is "no, it's not possible to make that not have nested trig function" unless you're willing to trade that for infinite series. — Milo Brandt, Nov 26 '15 at 02:34
$\frac12 i e^{-\frac12 i (e^{-i x}+e^{i x})}-\frac12 i e^{\frac12 i (e^{-i x}+e^{i x})}$ — Rocket Man, Nov 26 '15 at 02:36
It appears to be very close to $(\sin(1)) \cdot \cos x$, but not exactly — Rob, Nov 26 '15 at 02:36
I don't think there's any (good) way to express this without embedded trig functions. I think the most meaningful thing you can say is that, for a given angle, $\cos$ turns it into the adjacent length in the relevant triangle, and then we put that length into $\sin$ as if it were an angle. But that's pretty meaningless. — Addem, Nov 26 '15 at 02:41
When you say a solution, do you mean solving $\sin (\cos x)=0$? If so this would be equivalent to solving $\cos x = n\pi$. — Addem, Nov 26 '15 at 02:42
You could expand it in a power series. That would give you a good way to approximate it to whatever accuracy you like. A power series around $x=0$ has a bunch of $\sin(1)$'s and $\cos(1)$'s which are hard to evaluate but at $x=\pi/2$ the power series expansion is $-(x-\pi/2)+(1/3)(x-\pi/2)^3-(1/10)(x-\pi/2)^5 +\cdots$. — user293378, Nov 26 '15 at 02:49
The graph of sin(cos x) looks very much like an ordinary sine wave... is that an illusion? — Yan King Yin, Sep 15 '23 at 07:53

score 4 · Accepted Answer · answered Nov 26 '15 at 03:16

As already said in comments and answers, you cannot do much with embedded trigonometirc functions except developing as Taylor series.

Built around $x=0$ (remember that, for any $x$, $|\cos(x)| \leq 1$), you should get

$$\sin(\cos(x))=\sin (1)-\frac{1}{2} x^2 \cos (1)+x^4 \left(\frac{\cos (1)}{24}-\frac{\sin (1)}{8}\right)+x^6 \left(\frac{\sin (1)}{48}+\frac{7 \cos (1)}{360}\right)+x^8 \left(\frac{\sin (1)}{960}-\frac{209 \cos (1)}{40320}\right)+x^{10} \left(\frac{1259 \cos (1)}{3628800}-\frac{193 \sin (1)}{241920}\right)+O\left(x^{11}\right)$$

$$\cos(\cos(x))=\cos (1)+\frac{1}{2} x^2 \sin (1)-x^4 \left(\frac{\sin (1)}{24}+\frac{\cos (1)}{8}\right)+x^6 \left(\frac{\cos (1)}{48}-\frac{7 \sin (1)}{360}\right)+x^8 \left(\frac{209 \sin (1)}{40320}+\frac{\cos (1)}{960}\right)-x^{10} \left(\frac{1259 \sin (1)}{3628800}+\frac{193 \cos (1)}{241920}\right)+O\left(x^{11}\right)$$

One might also be inclined to take a Fourier series, given that $\sin(\cos(x))$ is periodic. I just suggest this since that's a little more natural for a periodic function than Taylor expansion (though the coefficients are harder to compute; Mathematica insists they have a closed form in terms of the Bessel function though) — Milo Brandt, Nov 26 '15 at 04:14

score 2 · Answer 2 · answered Nov 26 '15 at 04:18

2

Rationalizing $\cos (x)$ in function of $\tan (x/2)=t$ you have $\cos (x)=\frac {1-t^2}{1+t^2}$ hence $\sin(\cos (x))=\sin(\frac{1-t^2}{1+t^2})$ and you can if you want to developpe as Taylor series this last expression.

answered Nov 26 '15 at 04:18

Piquito

29,594

This is a good idea, indeed ! $+1$ – Claude Leibovici Nov 26 '15 at 15:24
You are very welcome. This leads to nice Taylor series $$\sin(\cos (x))=\sin (1)-2 t^2 \cos (1)+t^4 (2 \cos (1)-2 \sin (1))+t^6 \left(4 \sin (1)-\frac{2 \cos (1)}{3}\right)+O\left(t^7\right)$$ $$\cos(\cos (x))=\cos (1)+2 t^2 \sin (1)-2 t^4 (\sin (1)+\cos (1))+t^6 \left(\frac{2 \sin (1)}{3}+4 \cos (1)\right)+O\left(t^7\right)$$ – Claude Leibovici Nov 26 '15 at 16:52

score 1 · Answer 3 · edited Apr 13 '17 at 12:20

I want it to be reduced more, if possible.

Not possible.

I want to know if this is solvable such that I don't have a trig function inside another trig function, and a solution if one such exists.

Not “solvable”, and no such simplified solution exists.

The only remotely interesting thing I can come up with concerning this topic is that $$\int_0^\tfrac\pi2\sin(\sin x)~dx~=~\int_0^\tfrac\pi2\sin(\cos x)~dx~=~\frac\pi2~H_0(1),$$ and $$\int_0^\tfrac\pi2\cos(\cos x)~dx~=~\int_0^\tfrac\pi2\cos(\sin x)~dx~=~\frac\pi2~J_0(1).$$ See Bessel and Struve functions for more information.

What is $\sin(\cos(x))$?

3 Answers3