question trig

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.


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 $AB$ 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.

  • $\begingroup$ Would you care to elaborate on why you were able to use tan(π/5) without any complications. As in, able to substitue π/5. Thanks. $\endgroup$ – missiledragon Sep 29 '14 at 3:36
  • $\begingroup$ I'm not sure of what you mean; since $\pi/5=2x$ (because $x$ is just a shorthand for $\pi/10$), we can apply the known formula $\tan(2x)=\frac{2\tan x}{1-\tan^2x}$. $\endgroup$ – egreg Sep 29 '14 at 9:16

Here is a different approach that more directly gives the correct polynomial.

Using De Moivre's Formula, we have $$ \begin{align} \color{#C00000}{\cos(5x)}+i\color{#5D60BD}{\sin(5x)} &=(\color{#C00000}{\cos^5(x)-10\cos^3(x)\sin^2(x)+5\cos(x)\sin^4(x)})\\ &+i(\color{#5D60BD}{5\cos^4(x)\sin(x)-10\cos^2(x)\sin^3(x)+\sin^5(x)})\tag{1} \end{align} $$ Since the smallest positive angle whose sine is $0$ is $\pi$, $\sin(\pi/5)\ne0$. Since the smallest positive angle whose cosine is $0$ is $\pi/2$, $\cos(\pi/5)\ne0$. Therefore, we can divide the imaginary part of $(1)$ by $\cos^4(x)\sin(x)$, with $x=\pi/5$, to get $$ 0=\frac{\sin(\pi)}{\cos^4(\pi/5)\sin(\pi/5)}=\tan^4(\pi/5)-10\tan^2(\pi/5)+5\tag{2} $$ $\tan(\pi/5)$ is the smallest positive root of $z^4-10z^2+5$, which is $$ \tan(\pi/5)=\sqrt{5-2\sqrt5}\tag{3} $$

De Moivre's Formula

Using the formulas for the sine and cosine of a sum, we get $$ \begin{align} &(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))\\ &=(\cos(x)\cos(y)-\sin(x)\sin(y))+i(\sin(x)\cos(y)+\cos(x)\sin(y))\\ &=\cos(x+y)+i\sin(x+y) \end{align} $$ from which, induction gives De Moivre's Formula: $$ (\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx) $$


For first part use

$\tan(a+b)= \dfrac{\tan a+\tan b}{1-\tan a \tan b}$

Put $a=x$ , $b=2x$

Further evaluate $\tan 2x$ using addition property described above by substituting $a=x$ and $b=x$

For second part use $\tan 3x$ and convert $\cot 2x$ into $\dfrac{1}{\tan 2x}$

Using the second result find $\tan 36$

by substituting square of $\tan x =p$

and solving quadratic equation

remember $\tan (0-45)$ is less than $1$ so reject $1$ of the solution of the quadratic equation

hope this helps.


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