is $\cos x < \cos (\sin x)$ in $(0,\pi/2)$? i need to quickly verify if $\cos x < \cos(\sin x)$ in $(0,\pi/2)$
i dont have any solid ground to prove it except by some trivil way?can anyone show a concrete way to check it?
 A: True: $x > \sin x \to \cos (x) < \cos (\sin x)$ 
A: we can use the following two facts:(a) $\cos(x)$ is decreasing on $[0. \pi/2],$
(b) $ \sin x < x \text{ for } 0 < x \le \pi/2$ this means $$\cos x  < \cos (\sin x), 0 < x \le \pi/2.$$
p.s. in fact the inequality is true for all $x$ except for integer multiples of $2\pi$ where the equality holds. 
A: Yes. 
Let $f(x) = x - \sin(x)$. We'll just need it on $]0,\frac{\pi}{2}[$
$f'(x) = 1 - \cos(x)$, but we know that $\cos(x) < 1$ holds on $]0,\frac{\pi}{2}[$. Hence, $f'(x) > 0$ on $]0,\frac{\pi}{2}[$. So, $f$ is increasing strictly on $]0,\frac{\pi}{2}[$. But $f(0) = 0$. Thus $f(x) > 0$ on $]0,\frac{\pi}{2}[$. In other words, $x > \sin(x)$ on $]0,\frac{\pi}{2}[$.
However, $\cos$ is decreasing strictly on $]0,\frac{\pi}{2}[$; hence $\cos(x) < \cos(\sin(x))$ on $]0,\frac{\pi}{2}[$
A: $ \cos x $ decreases in the interval $ 0 < x < \pi/2 $ with derivative sign -\sin x <0,
and $ \sin x $ increases in the interval $ 0 < x < \pi/2, $ having derivative signs > 0. 
For the given combined function derivative is $ -\sin (\sin x) \cos x $ which has
reduced magnitude with a negative sign, reduces faster than $ \cos x. $ 
So given inequality holds good.
A: Use the facts:-
(1)cosine is a decreasing function
(2) sin(x)=x=0 at x=0
(3)cos(x)<1 in (0,π/2)
Since cos x and 1 are derivative of sin x and x respectively,it implies that in interval [0,π/2], x grows faster than sin(x) and is hence always greater than sin(x) which means cos(sin(x))>cos(x).
A: Observe, 
For all $x_0, x_1 \in (0, \pi /2 ) $ 
$$
x_0 < x_1\implies cos(x_0) > cos(x_1)
$$
Next note that for any $x \in (0, \pi /2) $
$$ x < sin(x)  \implies cos(x) > cos(sin(x)) $$
So   $cos(x) > cos(sin(x)) $, while $x \in (0, \pi /2)$. 
