Prove that $\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}=4\sqrt{3}$

Question

I'm trying to calculate the expression: $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}$$ and show that it is equal $4\sqrt{3}$.

I was trying to group the summands and calculate sums of $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}} \hspace{0.5cm}\text{and} \hspace{0.5cm} -\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}$$ where we get $$\frac{2\cos\frac{2\pi}{15}+1}{\sin\frac{2\pi}{15}}-\frac{2\cos\frac{4\pi}{15}-1}{\sin\frac{8\pi}{15}}$$ but unfortunately this sum is not simplified.
How to prove this equality?

Answer 1

Well, this is not a full solution (I will not do the calculations), but I guess this is one of the ways to reach the end ...

1. $\sin\left(\frac{\pi}{15}\right) = \sin\left(\frac{\pi}{6} - \frac{\pi}{10}\right)$
2. $\sin\left(\frac{\pi}{10}\right) = \sin\left(\frac{1}{2}\cdot\frac{\pi}{5}\right)$
3. for the $\frac{\pi}{5}$ use the identity: $\sin(5x) = 5\sin(x)-20\sin^3(x) + 16\sin^5(x)$. It has a straightforward solution

Rest of the angles are possible to calculate with the double angle formula.

99,9% not indended to be solved this way, but ...

Answer 2

$$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}\\ =\frac{1}{\sin12°}+\frac{1}{\sin24°}-\frac{1}{\sin48°}+\frac{1}{\sin96°} $$

We split them into two groups as shown:

$$\frac{1}{\sin12°}-\frac{1}{\sin48°} \hspace{0.5cm}\text{and}\hspace{0.5cm} \frac{1}{\sin24°}+\frac{1}{\sin96°}$$ And we see that (from sum-to-product and product-to-sum)

$\Large{\frac{1}{\sin12°}-\frac{1}{\sin48°}\\ = \frac{\sin48°-\sin12°}{\sin12°\sin48°}\\ = \frac{2\cos30°\sin18°}{{\frac12}(\cos36°-\cos60°)}\\=\frac{2(\frac{\sqrt3}{2})\sin18°}{{\frac12}(\cos36°-{\frac12})}\\=\frac{4\sqrt3\sin18°}{2\cos36°-1}}$

and

$\Large{\frac{1}{\sin24°}+\frac{1}{\sin96°}\\=\frac{\sin24°+\sin96°}{\sin24°\sin96°}\\=\frac{2\sin60°\cos36°}{{\frac12}(\cos72°-\cos120°)}\\=\frac{2({\frac{\sqrt3} 2})\cos36°}{{\frac12}(\sin18°+{\frac12})}\\=\frac{4\sqrt3\cos36°}{2\sin18°+1}}$

Hence it remains to find

$$\frac{4\sqrt3\sin18°}{2\cos36°-1} + \frac{4\sqrt3\cos36°}{2\sin18°+1}$$

From here we can determine $2\cos36°-1=2\sin18°\ $ and $\ 2\sin18°+1=2\cos36°$. By plugging in these into the denominators and simplifying we get $4\sqrt3$, which is what we want. $\ _\square$

Answer 3

I use degrees: $$\frac{1}{\sin 12^\circ}+\frac{1}{\sin 24^\circ}-\frac{1}{\sin48^\circ}+\frac{1}{\sin96^\circ}=\\ \frac{\sin 96^\circ+\sin 12^\circ}{\sin 12^\circ\sin 96^\circ}+\frac{\sin 48^\circ-\sin 24^\circ}{\sin 24^\circ\sin 48^\circ} =\\ \frac{2\sin 54^\circ\cos 42^\circ}{\sin 12^\circ\sin 96^\circ}+\frac{2\cos 36^\circ\sin 12^\circ}{\sin 24^\circ\sin 48^\circ} =\\ \frac{2\cos 36^\circ\require{cancel} \cancel{\sin 48^\circ}}{2\sin 12^\circ\cancel{\sin 48^\circ}\cos 48^\circ}+\frac{2\cos 36^\circ\cancel{\sin 12^\circ}}{2\cancel{\sin 12^\circ}\cos 12^\circ\sin 48^\circ} =\\ \cos 36^\circ\cdot \frac{\cos 12^\circ\sin 48^\circ+\sin 12^\circ\cos 48^\circ}{\sin 12^\circ\cos 12^\circ\sin 48^\circ\cos 48^\circ} =\\ \frac{4\cos 36^\circ\sin 60^\circ}{\sin 24^\circ\sin 96^\circ}=\\ \frac{4\cos 36^\circ\sin 60^\circ}{\frac12(\cos 72^\circ-\cos 120^\circ)}=\\ \frac{4\cos 36^\circ\sin 60^\circ}{\frac12((\cos 36^\circ-\frac12)+\frac12)}=4\sqrt{3}. $$ You can see see here: $\cos72^\circ = \cos36^\circ-\frac12$.

Answer 4

Hint:

Using $\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$,find $\alpha$,

$$\dfrac1{\sin12^\circ}+\dfrac1{\sin24^\circ}+\dfrac1{\sin48^\circ}+\dfrac1{\sin96^\circ}-\dfrac2{\sin48^\circ}$$

$$=\cot6^\circ-\cot96^\circ-\dfrac2{\sin48^\circ}$$

$\cot6^\circ-\cot96^\circ=\cot6^\circ+\tan 6^\circ=\dfrac2{\sin12^\circ}$

Now $\dfrac1{\sin12^\circ}-\dfrac1{\sin48^\circ}=\dfrac{\sin48^\circ-\sin12^\circ}{\sin48^\circ\sin12^\circ}=\dfrac{4\sin18^\circ\cos30^\circ}{\cos36^\circ-\cos60^\circ}$

Using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$, $\cos36^\circ-\cos60^\circ=\sin18^\circ$