# Integral inequality for sin function

Let $$0 and $$t\geq 0$$ real numbers. Is it true that $$\int_t^{t+r} \sin(x)\, dx \leq \int_{\frac{\pi}{2}-\frac{r}{2}}^{\frac{\pi}{2}+\frac{r}{2}}\sin(x)\, dx \,?$$

I suspect that yes, since both integration intervals have length $$r$$ and $$\sin$$ has maximum in second one (RHS).

Hints are welcome.

Well, one way to do it (though unelegant) would be to write out the solutions to the integrals explicitly. So, you are asking us to prove that: $$\cos(t)-\cos(r+t)\leq \cos(\pi/2-r/2)-\cos(\pi/2+r/2)$$ This is true, since by one of the Prosthaphaeresis Formulas, namely: $$\cos(\alpha)-\cos(\beta)=-2\sin\left[\frac{1}{2}(\alpha+\beta)\right]\sin\left[\frac{1}{2}(\alpha-\beta)\right]$$ We have that: $$\cos(t)-\cos(r+t)=2\sin(r/2)\sin(r/2+t)$$ And: $$\cos(\pi/2-r/2)-\cos(\pi/2+r/2)=2\sin(r/2)$$ And fortunately, since $$0, we have that $$\sin(r/2)> 0$$ and since $$r/2+t$$ is a real number, $$\sin(r/2+t)\leq 1$$.
• Fortunately, $\sin\leq 1$!!! – Sigur Nov 27 '18 at 21:40