Hi, I realise there has been a question already asked regarding this particular exact value, but this question requires for it to be done under different conditions, which is the part I require help in.
I have been having trouble with this for the past 30 minutes, I was able to do (i) and (ii) quite easily, but I am unsure how to actually get the exact value from the quadratic.
Please note that I only stumbled upon this question during self-study of harder trigonometry, and is not homework.
Any help would be greatly appreciated, thank you.
Best Answer
Consider an isosceles triangle $ABC$, having its vertex angle $\hat{C}$ equal to $\pi/5$, so the base angles $\hat{A}$ and $\hat{B}$ are $2\pi/5$. Draw the bisector of angle $\hat{B}$ meeting $AC$ in $D$.
Then the triangle $ADB$ is similar to $ABC$. Call $l$ the length of $AC$ and $x$ the length of $AB$. Then $BD$ and $DC$ also have length $x$ and $l\cos2\pi/5=x/2$.
By the similarity, we have $$ \frac{l}{x}=\frac{x}{l-x} $$ so $$ l^2-lx=x^2 $$ and, easily, $$ x=\frac{\sqrt{5}-1}{2}l $$ (the negative root must be discarded, of course). Thus $$ \cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4} $$ and it's easy to compute $\sin(\pi/5)$ and $\cos(\pi/5)$ from this.
One can use the outlined strategy, too. The key is using $\pi/10$ and not $x=\pi/5$. If $x=\pi/10$, then $$ \frac{\pi}{2}-3x=\frac{5\pi}{10}-\frac{3\pi}{10}=\frac{2\pi}{10}=2x $$ and therefore $$ \tan3x=\tan\left(\frac{\pi}{2}-2x\right)=\cot2x=\frac{1}{\tan2x} $$ Now we can apply the formulas and get $$ \frac{3\tan x-\tan^3x}{1-3\tan^2x}=\frac{1-\tan^2x}{2\tan x} $$ or, setting $t=\tan x$, $$ 6t^2-2t^4=1-3t^2-t^2+3t^4 $$ that reduces to $$ 5t^4-10t^2+1=0 $$ This gives $$ t^2=\frac{5+2\sqrt{5}}{5} $$ But $\tan(\pi/5)=\tan2x$ or $$ \tan\frac{\pi}{5}=\frac{2t}{1-t^2} $$ and just substituting will give the result.