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I need some help with this demonstration, please

I have tried with some identities but nothing.

I wanted to use this $$\sin(\pi/15)\cdot \sin(2\pi/15)\cdots\sin(7\pi/15)=\sqrt{15}$$

Bumblebee
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Carlos
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    I think you last identity is wrong. Because $\sin(\pi/15)\cdot \sin(2\pi/15)\cdots\sin(7\pi/15)\lt 1.$$ – Bumblebee Jun 10 '15 at 04:56
  • In fact, that product equals $\frac{\sqrt {15}}{128}$, which can be readily proved by considering the $15$th roots of unity. – Sahaj Jan 19 '24 at 18:47

3 Answers3

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We may prove: $$ \cos\frac{\pi}{15}-4\sin^2\frac{\pi}{15}=\sqrt{15}\sin\frac{\pi}{15} $$ by squaring both sides. By setting $\theta=\frac{\pi}{15}$, that leads to:

$$ \frac{13}{2}-2\cos(\theta)-\frac{15}{2}\cos(2\theta)+2\cos(3\theta)+2\cos(4\theta) = \frac{15}{2}-\frac{15}{2}\cos(2\theta)$$ or to: $$ -\cos(\theta)+\cos(3\theta)+\cos(4\theta) = \frac{1}{2} $$ so we just have to prove that $\cos(\theta)$ is a root of: $$ p(x) = 16x^4+8x^3-16x^2-8x+1.$$ That easily follows from: $$ \Phi_{30}(x) = x^8+x^7-x^5-x^4-x^3+x+1.$$

Jack D'Aurizio
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  • +1. Very nice solution. But I do not know what is $\Phi_{30}(x).$ Please can you explain. Thank you. – Bumblebee Jun 10 '15 at 04:54
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    @Nilan: it is the cyclotomic polynomial that is the minimal polynomial of $\exp\left(\frac{2\pi i}{30}\right)$. It is a palyndromic polynomial, hence by dividing it by $x^4$ you get a polynomial in $\left(x+\frac{1}{x}\right)$, then the minimal polynomial of $\cos\left(\frac{\pi}{15}\right).$ – Jack D'Aurizio Jun 10 '15 at 08:56
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Consider the equation $$\tan5\theta=\tan\dfrac{\pi}{3}$$ which has $5$ principle solutions $$\theta=\dfrac{n\pi}{5}+\dfrac{\pi}{15}\,\,\,\,\,\,\,\,n=0, 1, 2, 3, 4.$$ Therefore $\tan\dfrac{\pi}{15}$ is a root of the equation $$t^5-5\sqrt3t^4-10t^3+10\sqrt3t^2+5t-\sqrt3=0.$$ If we can find $\tan\dfrac{\pi}{15}$ using above equation we just have to show that $$\dfrac{1}{t}-\dfrac{4t}{\sqrt{t^2+1}}=\sqrt{15}.$$ After solve the above equation, I got that $$\color{Green}{\tan\dfrac{\pi}{15}=\sqrt{23-10\sqrt5-2\sqrt{255-114\sqrt5}}=\dfrac{3\sqrt3}{2}-\dfrac{\sqrt{15}}{2}-\dfrac{\sqrt{50-22\sqrt5}}{2}}.$$

Bumblebee
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If $\cot A-4\sin A=\pm\sqrt{15}$

Multiplying both sides by $\sin A$ which is clearly $\ne0,$

$\pm\sqrt{15}\sin A=\cos A-4\sin^2A=\cos A-4(1-\cos^2A)$

$\pm\sqrt{15}\sin A=4\cos^2A+\cos A-4$

Squaring both sides, $15(1-\cos^2A)=(4\cos^2A+\cos A-4)^2$

$\iff16\cos^4A+8\cos^3A-16\cos^2A-8\cos A+1=0\ \ \ \ (1)$

$\dfrac\pi{15}=12^\circ$ and as $\cos5\cdot\left(12^\circ\right)=\dfrac12$

and $\cos5y=16\cos^5y-20\cos^3y+5\cos y$

and $\cos5y=\dfrac12=\cos60^\circ$

$\implies5y=360^\circ m\pm60^\circ\iff y=72^\circ m\pm12^\circ$ where $m$ is any integer

So, the roots of $16\cos^5y-20\cos^3y+5\cos y=\dfrac12\iff32\cos^5y-40\cos^3y+10\cos y-1=0$

are $\cos y$ where $y=72^\circ m+12^\circ$ where $m\equiv0,1,2,3,4\pmod5$

But $\cos\left[72^\circ\cdot4+12^\circ\right]=\cdots=\dfrac12$

Now $\dfrac{32\cos^5y-40\cos^3y+10\cos y-1}{2\cos y-1}=16\cos^4y+8\cos^3y-16\cos^2y-8\cos y+1$

So, the roots of $16\cos^4y+8\cos^3y-16\cos^2y-8\cos y+1=0\ \ \ \ (2)$ (which is same as $(1)$)

are $\cos y$ where $y=72^\circ m+12^\circ$ where $m\equiv0,1,2,3\pmod5$

Now $\sin\left[72^\circ\cdot1+12^\circ\right]>0$ and $\cot\left[72^\circ\cdot1+12^\circ\right]<\cot60^\circ<\sqrt{15}$

$\sin\left[72^\circ\cdot2+12^\circ\right]=\cdots=\sin24^\circ$ and $\cot\left[72^\circ\cdot1+12^\circ\right]=\cdots=-\cot24^\circ<0$

$\implies\cot\left[72^\circ\cdot m+12^\circ\right]-4\sin\left[72^\circ\cdot m+12^\circ\right]=-\sqrt{15}$ for $m\equiv1,2$

$\sin\left[72^\circ\cdot3+12^\circ\right]=\cdots=-\sin48^\circ<0$ and $\cot\left[72^\circ\cdot3+12^\circ\right]=\cot48^\circ>0$

$\implies\cot\left[72^\circ\cdot m+12^\circ\right]-4\sin\left[72^\circ\cdot m+12^\circ\right]=\sqrt{15}$ for $m\equiv0,3$

lab bhattacharjee
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