how to calculate the exact value of $\tan \frac{\pi}{10}$

Question

I have an extra homework: to calculate the exact value of $ \tan \frac{\pi}{10}$. From WolframAlpha calculator I know that it's $\sqrt{1-\frac{2}{\sqrt{5}}} $, but i have no idea how to calculate that.

Thank you in advance, Greg

Answer 1

look this How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$?

then you will get $\sin \frac{\pi}{10}$($\frac{2\pi }{5}+\frac{\pi}{10}=\frac{\pi}{2}$) ,then $\tan \frac{\pi}{10}$.

Answer 2

Your textbook probably has an example, where $\cos(\pi/5)$ (or $\sin(\pi/5)$) has been worked out. I betcha it also has formulas for $\sin(\alpha/2)$ and $\cos(\alpha/2)$ expressed in terms of $\cos\alpha$. Take it from there.

Answer 3

If $10x=\pi$ $\sin 2x=\cos 3x$ as $2x+3x=5x=\frac{\pi}{2}$ $\implies2\sin x \cos x=4\cos^3x-3\cos x$

$\implies 2\sin x=4\cos^2x-3$ as $\cos x≠0$

If $\sin x=t, 2t=4(1-t^2)-3\implies 4t^2+2t-1=0$

$$\implies t=\frac{-1±\sqrt{5}}{4}$$, but $\sin x>0$ as $0<x<\pi$

$$\sin \frac{\pi}{10}=\frac{\sqrt{5}-1}{4}$$

(1)So, $$\cos \frac{\pi}{10}=\sqrt{1-(\sin \frac{\pi}{10})^2}=\frac{\sqrt{10+2\sqrt5}}{4}$$

So, $$\tan \frac{\pi}{10}=\frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt5}}{4}}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt5}}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt5}}$$

$$=\sqrt{\frac{(\sqrt 5 -1)^2}{10+2\sqrt5}}=\sqrt{\frac{3-\sqrt 5}{\sqrt 5(\sqrt 5+1)}}=\sqrt{\frac{(3-\sqrt 5)(\sqrt 5 -1)}{\sqrt 5(\sqrt 5+1)(\sqrt 5 -1)}}=\sqrt{\frac{\sqrt 5-2}{\sqrt 5}}$$

Or(2) $$\cos \frac{\pi}{5}=1-2(\frac{\sqrt{5}-1}{4})^2=\frac{\sqrt 5 + 1}{4}$$

We know $$\cos2y=\frac{1-\tan^2y}{1+\tan^2y}\implies \tan^2y=\frac{1-\cos2y}{1+\cos2y}$$

So, $$\tan^2 \frac{\pi}{10}= \frac{1-\frac{\sqrt 5 + 1}{4}}{1+\frac{\sqrt 5 + 1}{4}}=\frac{3-\sqrt 5}{\sqrt 5(\sqrt 5+1)}$$ which we have already encountered in (1).

Answer 4

$$\tan\frac{3\pi}{10}=\tan(\frac{\pi}{2}-\frac{2\pi}{10})=\cot\frac{2\pi}{10}$$

$$\frac{3\tan\frac{\pi}{10}-\tan^3\frac{\pi}{10}}{1-3\tan^2\frac{\pi}{10}}=\frac{\cot^2\frac{\pi}{10}-1}{2\cot\frac{\pi}{10}}$$

$$(3\tan\frac{\pi}{10}-\tan^3\frac{\pi}{10})(2\cot\frac{\pi}{10})=(\cot^2\frac{\pi}{10}-1)(1-3\tan^2\frac{\pi}{10})$$

$$6-2\tan^2\frac{\pi}{10}=\cot^2\frac{\pi}{10}-4+3\tan^2\frac{\pi}{10}$$

$$5\tan^2\frac{\pi}{10}-10+\cot^2\frac{\pi}{10}=0$$

$$5\tan^4\frac{\pi}{10}-10\tan^2\frac{\pi}{10}+1=0$$

$$\tan^2\frac{\pi}{10}=\frac{10\pm\sqrt{100-20}}{10}=\frac{10\pm4\sqrt{5}}{10}=1+\frac{2}{\sqrt{5}}\;\textrm{or}\;1-\frac{2}{\sqrt{5}}(\textrm{rej.})$$

$$\tan\frac{\pi}{10}=\sqrt{1+\frac{2}{\sqrt{5}}}\;\textrm{or}\;-\sqrt{1+\frac{2}{\sqrt{5}}}(\textrm{rej.})$$

$$∴\tan\frac{\pi}{10}=\sqrt{1+\frac{2}{\sqrt{5}}}$$