Evaluating $\csc\left(\frac{\pi}{18}\right) + \csc\left(\frac{5\pi}{18}\right) - \csc\left(\frac{7\pi}{18}\right)$

Question

I need to evaluate $$\csc\left(\frac{\pi}{18}\right) + \csc\left(\frac{5\pi}{18}\right) - \csc\left(\frac{7\pi}{18}\right)$$

My work:

Taking $t = \frac{\pi}{18}$, the expression becomes:

$$\frac{2\cos 6t-\cos 12t + \cos 2u - \cos 4u - \cos 8u}{2\sin u\cdot \sin 5u\cdot \sin 7u}$$

According to WolframAlpha, this evaluates to $6$. But I am having trouble showing the same.

Then I tried complex exponentials:

Taking $u=e^{\frac{i\pi}{18}}$, the expression becomes:

$$2i\cdot \left(\frac{u}{u^2-1} + \frac{u^5}{u^{10}-1} + \frac{u^7}{1-u^{14}}\right)$$ but that too seems to be in vain.

Any help will be appreciated. I think the first approach might work out. If you write an answer, I would appreciate if you could provide a general approach to deal with such problems.

Thanks!

Answer 1

Noting that $$ \frac{1}{2}=\sin \left(3\left(\frac{\pi}{18}\right)\right)=3 \sin \left(\frac{\pi}{18}\right)-4 \sin ^3\left(\frac{\pi}{18}\right); $$$$ \frac{1}{2}=\sin \left(3\left(\frac{5 \pi}{18}\right)\right)=3 \sin \left(\frac{5 \pi}{18}\right)-4 \sin ^3\left(\frac{5 \pi}{18}\right) $$ and $$ \frac{1}{2}=\sin \left(-3\left(\frac{7 \pi}{18}\right)\right)=3 \sin \left(-\frac{7 \pi}{18}\right)-4 \sin ^3\left(-\frac{7 \pi}{18}\right), $$ The three roots of the cubic equation $8x^3-6x+1=0 \cdots (*)$ are $$\sin \left(\frac \pi{18}\right), \sin \left(\frac {5\pi}{18}\right) \textrm{ and } \sin \left(-\frac {7\pi}{18}\right) .$$ Transforming $(*)$ by $u=\frac 1x$ gives the equation $$ \frac{8}{u^3}-\frac{6}{u}+1=0 \Leftrightarrow u^3-6 u^2+8=0 \tag*{(**)} $$ whose roots are $$\csc \left(\frac \pi{18}\right), \csc \left(\frac {5\pi}{18}\right) \textrm{ and } \csc \left(-\frac {7\pi}{18}\right) .$$ Hence their sum is $$\csc \left(\frac \pi{18}\right)+ \csc \left(\frac {5\pi}{18}\right) + \csc \left(-\frac {7\pi}{18}\right)=\textrm{Sum of roots of }(**)= 6 $$ $$ \textrm{ i.e. } \csc \left(\frac \pi{18}\right)+ \csc \left(\frac {5\pi}{18}\right) - \csc \left(\frac {7\pi}{18}\right)= 6 $$

Answer 2

I'll switch over to degrees for my comfort : someone can rewrite all my work in terms of radians if necessary. The question is now : $\csc(10)+\csc(50) -\csc(70)= ?$

In expressions involving multiple angles, one can attempt to combine two angles whose sum and difference behave well with respect to the third angle i.e. they could lead to known angles such as $90,60,30,180$ etc., or to twice another angle, half another angle, maybe even equal to another angle etc. This helps considerably simplify computations.

For example, here we note that $10+(-70) = -60$, which is a desirable angle. The difference $10-(-70) = 80$ doesn't seem to immediately be related to $50$, but once we do combine those terms the connection actually becomes clear (which means, yes, if I'd failed then I'd have showed you that I failed and tried another combination. Turns out, it worked!)

$$ \csc(10)+\csc(-70) = \frac{1}{\sin(10)} + \frac{1}{\sin(-70)} = \frac{\sin(10)+\sin(-70)}{\sin(10)\sin(-70)} = \frac{4 \sin(-30)\cos(40)}{\cos(80) - \cos(-60)} = \frac{-2\cos(40)}{\cos(80) - \frac 12} = \frac{-4\cos(40)}{2\cos(80)-1} = \frac{-4\cos(40)}{2(2 \cos^2(40)-1)-1} = \frac{-4\cos(40)}{4 \cos^2 40-3} = \frac{4 \cos(40)}{3-4 \cos^2(40)}. $$

Now, remember that the leftover term is $\csc(50)$ which is just $\frac{1}{\sin(50)}$. However, $\cos(40)=\sin(50)$, so we just have $$ \csc(10)+\csc(-70)+\csc(50) = \frac{4 \sin(50)}{3-4 \sin^2(50)} + \frac{1}{\sin(50)} = \frac{3}{4 \sin^3(50) - 3 \sin(50)} = \frac{3}{-\sin(150)} =\frac{3}{-(-0.5)} = 6. $$

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Could other angles have been combined, though? Let's think about $50$ and $-70$ : their difference is $120$, which is desirable, and their sum is $-20$, which is twice of $-10$. So there is a possibility of simplification. Let's explore it.

$$ \csc(50)+ \csc(-70) = \frac{1}{\sin(50)} + \frac{1}{\sin(-70)} = \frac{\sin(-70)+\sin(50)}{\sin(-70)\sin(50)} = \frac{4 \sin(-10)\cos(60)}{\cos(120) - \cos(20)} = \frac{-2\sin(10)}{-\frac 12 - \cos(20)} = \frac{2\sin(10)}{\frac{1}{2}+\cos(20)} = \frac{4\sin(10)}{1+2\cos(20)} = \frac{4 \sin 10}{1+2(1-2\sin^2(10))} = \frac{4\sin 10}{3-4\sin^2(10)}. $$

and you can already see where this is going. Add $\frac{1}{\sin(10)}$ to this and simplify to obtain the answer. In other words, there were multiple combinations of angles that could have been used to solve this.

Answer 3

Observing that $\frac{10^\circ+50^\circ}2=30^\circ$ and $\frac{50^\circ-10^\circ}2=90^\circ-70^\circ$, one can proceed in the following manner:

$$\csc10^\circ+\csc50^\circ-\csc70^\circ$$ $$=\frac1{\sin10^\circ}+\frac1{\sin50^\circ}-\frac1{\sin70^\circ}$$ $$=\frac{2\sin30^\circ\cos20^\circ}{\sin10^\circ\sin50^\circ}-\frac1{\sin70^\circ}$$ $$=\frac{\sin70^\circ}{\sin10^\circ\sin50^\circ}-\frac1{\sin70^\circ}$$ $$=\frac{\sin^270^\circ-\sin(30^\circ+20^\circ)\sin(30^\circ-20^\circ)}{\sin10^\circ\sin50^\circ\sin70^\circ}$$

Using the identities $\sin(60^\circ-\theta)\sin\theta\sin(60^\circ+\theta)=\frac14\sin3\theta$ and $\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$, the above expression becomes:

$$\frac{\sin^270^\circ-\sin^230^\circ+\sin^220^\circ}{\frac14\sin30^\circ}$$ $$=\frac{1-\frac14}{\frac18}$$ $$=6$$

Answer 4

You can show:

1. $\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ} = \dfrac{1}{8}$

2. $\sin 50^{\circ} \sin 70^{\circ}+\sin 10^{\circ} \sin 70^{\circ}-\sin 50^{\circ} \sin 10^{\circ} = \dfrac{3}{4}$

For 1. We have the identity $\cos A \cos 2A \cos 4A = \dfrac{\sin 8A}{8\sin A}$ (multiply both sides by $8\sin A$).

Now, $\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ} = \cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ}$

$ = \dfrac{ \sin 160^{\circ}}{8\sin 20^{\circ}}=\dfrac{1}{8} $

2. $2 \left(\sin 50^{\circ} \sin 70^{\circ}+\sin 10^{\circ} \sin 70^{\circ}-\sin 50^{\circ} \sin 10^{\circ}\right)$ $=\cos 20^{\circ}-\cos 120^{\circ}+\cos 60^{\circ} - \cos 80^{\circ}+\cos 60^{\circ}- \cos 40^{\circ}$

$=\dfrac{3}{2}+ \left(\cos 20^{\circ}- \cos 40^{\circ}- \cos 80^{\circ}\right)$

$\cos 20^{\circ}- \left(\cos 40^{\circ}+ \cos 80^{\circ} \right)= \cos 20^{\circ}-\cos 20^{\circ}=0$