0
$\begingroup$

Let $0<r<1$ 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.

$\endgroup$
1
$\begingroup$

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 $0<r<1$, we have that $\sin(r/2)> 0$ and since $r/2+t$ is a real number, $\sin(r/2+t)\leq 1$.

$\endgroup$
  • 1
    $\begingroup$ Fortunately, $\sin\leq 1$!!! $\endgroup$ – Sigur Nov 27 '18 at 21:40
  • $\begingroup$ Indeed. It would be nice however if someone could figure out a solution to this problem without having to explicitly solve the integrals. $\endgroup$ – projectilemotion Nov 27 '18 at 21:42
  • $\begingroup$ Sometimes, hard problems have simple solutions... and usually we don't look to the simple ones. Thanks. $\endgroup$ – Sigur Nov 27 '18 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.