How to show: $\sin [tx+(1-t)y]\leq t\sin x+(1-t)\sin y,$ where $0\leq t\leq 1,$ and $\pi\leq x, y\leq 2\pi$. 
How can I show that
  $$\sin [tx+(1-t)y]\leq t\sin x+(1-t)\sin y$$ where $0\leq t\leq 1$ and $\pi\leq x, y\leq 2\pi$? 

I know I have to use convex function property, but I am not able to prove this inequality.
 A: In order to avoid confusion,  we use the letters $a$ and $b$ instead of $x$ and $y$. The inequality obviously holds if $a=b$, so we assume that $a\ne b$.
Recall that if the second derivative of the function $f(x)$ is positive over an interval, then the curve $y=f(x)$ is what in elementary calculus courses is called "facing up" in that interval. That means in particular that the secant line joining $(a,f(a))$ to $(b,f(b))$ lies "above" the curve except at the endpoints $a$ and $b$.
As $t$ ranges from $0$ to $1$, the point $ta+(1-t)b$ ranges over all points in the interval whose endpoints are $a$ and $b$.  The secant line $\ell$ that joins $(a,f(a))$ to $(b,f(b))$ has equation
$$\frac{y-f(a)}{x-a}=\frac{f(b)-f(a)}{b-a}.\tag{$\ast$}$$
Put $x =at+(1-t)b$.  Then $x-a=at-a+(1-t)b=(b-a)(1-t)$. A little manipulation of $(\ast)$ yields
$$y-f(a)=(1-t)(f(b)-f(a)),$$
which quickly yields
$$y=tf(a)+(1-t)f(b).\tag{$\ast\ast$}$$
Thus the $y$-coordinate of the point on $\ell$ that has $x$-coordinate equal to $ta+(1-t)b$ is $tf(a)+(1-t)f(b)$.
If the second derivative of $f$ is positive between $a$ and $b$, then the secant line $\ell$ is above or on the curve between $a$ and $b$. That gives the inequality
$$tf(a)+(1-t)f(b) \ge f(at+(1-t)b).\tag{$\ast\ast\ast$}$$
Finally, let $f(x)=\sin x$. Then $f''(x)=-\sin x$, so $f''(x) \ge 0$ on the interval with endpoints $a$ and $b$, if $a$ and $b$ are in the interval $[\pi,2\pi]$. Thus the desired inequality follows from $(\ast\ast\ast)$.  
A: Here is a pictorial proof for the problem. $\sin(x)$ is convex in $(\pi, 2\pi)$. Hence, the center of mass of two points $(x,\sin(x))$ and $(y, \sin(y))$ will always lie above the curve.
