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