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Normally I just look up or remember that $\cos(\pi/4)=\frac{\sqrt2}{2}$, or type "$\cos(\pi/4)$" into WolframAlpha to check the answer.

But what about the first time someone wanted to know what $\cos(\pi/4)$ was? How could they find out before trig tables and WolframAlpha?

How can you prove that $\cos(\pi/4)=\frac{\sqrt2}{2}$?

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    $\begingroup$ Use $\cos(2x) = 2\cos^2(x)-1$. On second thought. If this is for the first time then a geometrical explanations is probably best (see David's suggestion below) $\endgroup$
    – Winther
    Dec 13, 2014 at 15:49
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    $\begingroup$ Draw a $45^\circ$-$45^\circ$-$90^\circ$ triangle. Set the length of the hypotenuse to $1$. Solve for the other side lengths and calculate what you want. $\endgroup$
    – David Mitra
    Dec 13, 2014 at 15:50
  • $\begingroup$ is $\pi/4$ pronounced as "pi over four" or is it pronounced as "one quarter of half of a revolution"? $\endgroup$
    – John Joy
    Dec 14, 2014 at 2:19

3 Answers 3

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Use a $45^\circ-45^\circ -90^\circ $ triangle.

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  • $\begingroup$ What about $\cos(m\pi/n)$? $\endgroup$
    – john.abraham
    Dec 13, 2014 at 16:52
  • $\begingroup$ @john.abraham What are m and n there? $\endgroup$
    – Sawarnik
    Dec 13, 2014 at 18:33
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we consider a isosceles right angled triangel and we have by definition $\cos(\pi/4)=\frac{\frac{c}{2}}{a}$ and with $c=\sqrt{2}a$ we get $$\cos(\pi/4)=\frac{\sqrt{2}}{2}$$

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First of all $$\cos(x)=\sin(\frac{\pi}{2}-x)$$ therefore $$\cos(\frac{\pi}{4})=\sin(\frac{\pi}{2}-\frac{\pi}{4})=\sin(\frac{\pi}{4})$$ Secondly $$\sin^2(x)+\cos^2(x)=1$$ and thus $$\sin^2(\frac{\pi}{4})+\cos^2(\frac{\pi}{4})=1\Leftrightarrow 2\cos^2(\frac{\pi}{4})=1\Leftrightarrow \cos(\frac{\pi}{4})=|\frac{\sqrt{2}}{2}|$$ Since $\frac{\pi}{4}$ belongs to the first quadrant then $$\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$$

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