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I only know $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$, $180^\circ$, $270^\circ$, and $360^\circ$ as standard angles but how can I prove that $$\cos 18^\circ=\frac{1}{4}\sqrt{10+2\sqrt{5}}$$

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Please see here for help writing math with MathJax, and see here for help formatting posts with Markdown. – Zev Chonoles Jul 7 '13 at 19:50
up vote 12 down vote accepted

Consider the isosceles triangle pictured below:

enter image description here

By considering similar triangles, one has ${x\over 1}={1\over x-1}$. From this, it follows that $x={1+\sqrt 5\over 2}$. Drop a perpendicular from the top vertex to the base below. One sees that $\sin(18^\circ)={1\over 1+\sqrt 5}={\sqrt 5-1\over 4}$. Now use the Pythagorean Identity to find $\cos(18^\circ)$.

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Let $\theta = 18^\circ$.

Then $2\theta = 36^\circ$ and $3\theta = 54^\circ$.

Note that $90^\circ-3\theta = 2\theta$.

Thus, $\sin(90^\circ - 3\theta) = \sin (2\theta)$ or $\cos(3\theta) = \sin(2\theta)$

So, $4\cos^3\theta-3\cos\theta = 2\sin\theta\cos\theta$.

Since $\cos \theta$ can't be zero, we can divide by it to obtain,

$4\cos^2\theta-3 = 2\sin\theta$

Now use $\cos^2 \theta = 1 - \sin^2 \theta$ to obtain a quadratic in $\sin \theta$.

Solve it and then disregarding the negative root (since $\sin 18^\circ$ can't be negative), solve for $\cos 18^\circ$.

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