Can You Solve the Trigonometric Equation for \(\tan^2 9^\circ\)?

[anemone]

Here is this week's POTW:

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Prove that \(\displaystyle \tan^2 9^\circ=\sqrt{201+88\sqrt{5}}-\sqrt{200+88\sqrt{5}}\).

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[anemone]

Congratulations to the following members for their correct solution::)

1. greg1313
2. kaliprasad
3. lfdahl

Solution from greg1313:

Spoiler

All arguments are in degrees.

$$\cos72=\sin18=\sqrt{\frac{1-\cos36}{2}}$$

$$\begin{align*}\frac{1-\cos36}{2}&=(2\cos^236-1)^2 \\
&=4\cos^436-4\cos^236+1 \\
&\Rightarrow8\cos^436-8\cos^236+\cos36+1=0 \\
&\Rightarrow(2\cos36-1)(\cos36+1)(4\cos^236-2\cos36-1)=0 \\
&\Rightarrow\cos36=\frac{1+\sqrt5}{4}\end{align*}$$

$$\begin{align*}\cos36=\frac{1+\sqrt5}{4}&=2\cos^218-1 \\
&=2(2\cos^29-1)^2-1 \\
&=8\cos^49-8\cos^29+1 \\
&\Rightarrow32\cos^49-32\cos^29+3-\sqrt5=0 \\
&\Rightarrow\cos^29=\frac12+\sqrt{\frac{1}{32}(5+\sqrt5)}\end{align*}$$

$$\begin{align*}\tan^29&=\sec^29-1 \\
&=\frac{1}{\cos^29}-1 \\
&=\frac{1}{\frac12+\sqrt{\frac{1}{32}(5+\sqrt5)}}-1 \\
&=\frac{\frac12-\sqrt{\frac{1}{32}(5+\sqrt5)}}{\frac12+\sqrt{\frac{1}{32}(5+\sqrt5)}} \\
&=\frac{4-\sqrt{2(5+\sqrt5)}}{4+\sqrt{2(5+\sqrt5)}} \\
&=\frac{26+2\sqrt5-8\sqrt{2(5+\sqrt5)}}{6-2\sqrt5} \\
&=\frac{156+12\sqrt5-48\sqrt{2(5+\sqrt5)}+52\sqrt5+20-16\sqrt5\sqrt{2(5+\sqrt5)}}{16} \\
&=11+4\sqrt5-(3+\sqrt5)\sqrt{2(5+\sqrt5)} \\
&=\sqrt{(11+4\sqrt5)^2}-\sqrt{(3+\sqrt5)^2(10+2\sqrt5)} \\
&=\sqrt{201+88\sqrt5}-\sqrt{(14+6\sqrt5)(10+2\sqrt5)} \\
&=\sqrt{201+88\sqrt5}-\sqrt{200+88\sqrt5}\end{align*}$$