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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:

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

I found the answer here: http://www.slader.com/textbook/9781938168024-calculus-volume-1/550/exercises/109/

But I do not follow the steps.

Thanks in advance

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$\displaystyle\int_{-\pi/4}^{\pi/4}\cos(t)dt=\sin(t)\Big|_{-\pi/4}^{\pi/4}=\sin(\pi/4)-\sin(-\pi/4)=\sin(\pi/4)+\sin(\pi/4)=2\sin(\pi/4)=\dfrac{2}{\sqrt{2}}=\dfrac{2\sqrt{2}}{2}=\dfrac{4\sqrt{2}}{4}\geq\dfrac{\pi\sqrt{2}}{4}$

where $\sin(-\pi/4)=-\sin(\pi/4)$ since $\sin$ is an odd function.

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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, $$\int_{-\pi/4}^{\pi/4} \cos(t) \, dt \ge \int_{-\pi/4}^{\pi/4} \frac{1}{\sqrt{2}} \, dt = \cdots$$

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  • $\begingroup$ I should have clarified this in my question, could you prove this without a plot? $\endgroup$ – boniface316 Aug 7 '18 at 3:24
  • $\begingroup$ @boniface316: It was proved without a plot? $\endgroup$ – copper.hat Aug 7 '18 at 3:37
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    $\begingroup$ @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 $[-\pi/4, \pi/4]$. $\endgroup$ – angryavian Aug 7 '18 at 3:37

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