True Statement

For $0 < x < \frac \pi 2$ , $\cos x$ decreases as $x$ increases. Since $\sin x = x - \dfrac {x^3}{3!} + \dfrac {x^5}{5!} - \cdots$ , $\implies \sin x < x$ . Therefore $\cos (\sin x) > \cos x$ .

Now consider $f(x) = \cos x$ and $g(x) = \sin (\cos x)$ . We note that $f(x) = g(x)$ , when $x = \dfrac \pi 2$ that is out the range of $0 < x < \dfrac \pi 2$ , and $f'(x) = - \sin x < 0$ and $g'(x) = -\sin x \cos(\cos x) < 0$ for $0 < x < \dfrac \pi 2$ . They are both monotonically decreasing functions. Since $f(0) = 1$ and $g(0) = \sin 1 < f(0)$ , and $f(x)$ and $g(x)$ do not intersect between $0 < x < \dfrac \pi 2$ , $f(x) > g(x)$ between the interval.

Therefore we have $\boxed{\textbf C}: \sin (\cos x) < \cos x < \cos (\sin x)$ .

In the folllowing range we can,say that sinx<x......(1) ..so,cos(sinx)>cosx.....(2) Put...x=cosx in equation 1 So,cosx>sin(cosx)......(3) Therefore, Cos(sinx)>cosx>sin(cosx).........

$0<x<\dfrac{π}{2}\implies \dfrac{π}{2}>1>\sin x>0,\dfrac{π}{2}>1>\cos x>0$

For these values of $\sin x$ and $\cos x, \sin (\cos x)<\cos x$

$\sin x<x\implies \cos (\sin x) >\cos x$ .

Therefore $\boxed {\sin (\cos x) <\cos x<\cos (\sin x) }$