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This is one of the questions I got in my math exam:

Evaluate without a calculator: $$\cos^{-1}\left(\,\sin\left(\pi^2\right)\,\right)$$

I just can't figure out how to evaluate $\sin(\pi^2)$ without using a calculator.

Thanks!

Blue
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hhalaweh
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    Recall $\cos^{-1}(\alpha)+\sin^{-1}(\alpha)=\frac{\pi}{2}$... – Aphelli Nov 05 '19 at 16:53
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    I wonder why you should be interested in evaluating $\sin(\pi^2)$ without a calculator: personally, evaluating $\arccos(\text{stuff I know})$ is not my strong suit. –  Nov 05 '19 at 16:56

2 Answers2

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Recall that $\arccos$ is a function $[-1,1]\to \left[0,\pi\right]$ and therefore for $\theta \in \left[0,\pi\right]$

$$\arccos(\cos \theta)=\theta$$

then note that by reflections, shifts, and periodicity

$$\pi^2\in\left(3\pi,3\pi +\frac \pi 2\right) \implies \sin \left(\pi^2\right)=-\sin(\pi^2-3\pi)=$$

$$=\sin(3\pi-\pi^2)=\cos \left(\frac \pi 2-(3\pi-\pi^2)\right)$$

user
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The angle $\pi^2=\pi\pi\approx3.14\pi$ is in the third quadrant, so its sine will be negative, which will place the inverse cosine in the second quadrant. For the reference triangle, $\cos^{-1}(\sin(\theta))=\frac\pi2-\theta$ since you're obviously looking for the complelementary angle. So the reference diagram looks like this:

enter image description here

So the inverse cosine in the second quadrant would be $\pi-((3.5-\pi)\pi)=(\pi-2.5)\pi\approx2.0156$. This matches my calculator's result.