Convolution of sine and cosine.

Question

So I came across this question while studying for the GRE Subject Exam, and I am not really sure how I am supposed to handle it. Let $$ f(x) = \int _0 ^{\pi} \sin t \cos (x+t) dt $$ I am to find where $f$ gets its minimum and maximum values when $x \in [0,2\pi]$.

Since it is a GRE-question, I believe that the answer should not be that difficult. Could someone please explain?

Answer 1

easiest way: calculate integral $$\int _0 ^{\pi} \sin t \cos (x+t) dt=\frac{1}{2}\int _0 ^{\pi} (\sin (2t+x) -\sin (x) )dt=\\ \frac{1}{2}(-\frac{1}{2}\cos (2t+x)-t\sin (x))|^{\pi}_{0}=-\frac{\pi}{2}\sin (x)$$

Answer 2

Using the formula $\cos(x + t) = \cos x \cos t - \sin x \sin t$, you can split the integral and just calculate it.

\begin{align*}\int_0^\pi \sin t\cos(x + t)\ dt &= \int_0^\pi \sin t(\cos x \cos t - \sin x \sin t)\ dt\\ &=\cos x \underbrace{\int_0^\pi \sin t \cos t\ dt}_{=\,0} - \sin x\underbrace{\int_0^\pi \sin^2 t\ dt}_{=\, \pi/2} \\ &=-\frac{\pi}{2}\sin(x), \end{align*}

which is maximized at $x = 3\pi/2$.

I'm sure there's a better perspective or way to do things, but this isn't that bad.

Answer 3

By the Cauchy-Schwarz inequality, we have $$\begin{aligned} \left|\int_0^{\pi}\sin(t)\cos(x+t)dt\right| &\leq \left(\int_{0}^{\pi}\sin^2(t) dt\right)^{1/2} \left(\int_0^{\pi}\cos^2(x+t)dt\right)^{1/2} \\ &= \frac{\pi}{2} \\ \end{aligned}$$ The right hand side is independent of $x$ because both integrands have period $\pi$.

Equality is achieved if and only if $\sin(t)$ and $\cos(x+t)$ are scalar multiples of each other. But this happens if and only if $x = \pi/2$ and $x=3\pi/2$. Specifically, $$\cos(t+3\pi/2) = \cos(t-\pi/2) = \sin(t) \quad\text{and}\quad\cos(t+\pi/2) = -\sin(t)$$ so the minimum is achieved at $x=\pi/2$, and the maximum is achieved at $x = 3\pi/2$.