Integral proof using comparison theorem

I came across the following question and I have no idea how to solve it. I would really appreciate if anyone can help me with this.

Show that:

$$\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \cos(t) \,dt \ge \frac{\pi\sqrt{2}}{4}$$

But I do not follow the steps.

$\displaystyle\int_{-\pi/4}^{\pi/4}\cos(t)dt=\sin(t)\Big|_{-\pi/4}^{\pi/4}=\sin(\pi/4)-\sin(-\pi/4)=\sin(\pi/4)+\sin(\pi/4)=2\sin(\pi/4)=\dfrac{2}{\sqrt{2}}=\dfrac{2\sqrt{2}}{2}=\dfrac{4\sqrt{2}}{4}\geq\dfrac{\pi\sqrt{2}}{4}$
where $\sin(-\pi/4)=-\sin(\pi/4)$ since $\sin$ is an odd function.
For $t \in [-\pi/4, \pi/4]$ we have $\cos(t) \ge 1/\sqrt{2}$. You should probably make a plot of the function $\cos(t)$ and see what it looks like on the interval $[-\pi/4, \pi/4]$.
So, $$\int_{-\pi/4}^{\pi/4} \cos(t) \, dt \ge \int_{-\pi/4}^{\pi/4} \frac{1}{\sqrt{2}} \, dt = \cdots$$
• @boniface316 You could use the geometric definition of cosine (ratio of adjacent side to hypotenuse in a right triangle) to justify why $\cos(t) \ge 1/\sqrt{2}$ holds on $[-\pi/4, \pi/4]$. – angryavian Aug 7 '18 at 3:37