Find the exact value of $\DeclareMathOperator{\cosec}{cosec} \cosec(10^\circ) + \cosec(50^\circ) - \cosec(70^\circ)$

Question

Find the exact value of $\cosec(10^\circ) + \cosec(50^\circ) - \cosec(70^\circ)$.

The equation can be written as $$ \cosec(x) + \cosec(60^\circ-x) - \cosec(60^\circ+x), $$ where $x = 10^\circ$.

I also know that $$ \sin(x) + \sin(60^\circ-x) + \sin(60^\circ+x) = \frac{\sin(3x)}{4}. $$

This equation and identity have the same format, so I got the idea to use it.

Can I use this information to find the answer?

(This question is from a book called Advanced Problems in JEE, in case someone wants to know about it.)

Answer 1

I don't offhand know how to use your equation involving $\sin$. Instead, we have

$$\begin{equation}\begin{aligned} & \cosec(10^\circ) + \cosec(50^\circ) - \cosec(70^\circ) \\ & = \frac{1}{\sin(10^\circ)} + \frac{1}{\sin(50^\circ)} - \frac{1}{\sin(70^\circ)} \\ & = \frac{\sin(70^\circ)\sin(50^\circ) + \sin(70^\circ)\sin(10^\circ) - \sin(50^\circ)\sin(10^\circ)}{\sin(10^\circ)\sin(50^\circ)\sin(70^\circ)} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

As shown in multiple answers in Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$. (e.g., this one), with the denominator, we get

$$\sin(10^\circ)\sin(50^\circ)\sin(70^\circ) = \frac{1}{8} \tag{2}\label{eq2A}$$

The Product-to-sum identities section of Wikipedia's "List of Trigonometric Identities" article has

$$\sin \theta \,\sin \varphi ={\cos(\theta -\varphi )-\cos(\theta +\varphi ) \over 2} \tag{3}\label{eq3A}$$

Thus, the numerator of \eqref{eq1A} becomes

$$\begin{equation}\begin{aligned} & \frac{(\cos(20^{\circ}) - \cos(120^{\circ})) + (\cos(60^{\circ}) - \cos(80^{\circ})) - (\cos(40^{\circ}) - \cos(60^{\circ}))}{2} \\ & = \frac{(\cos(20^{\circ}) + \frac{1}{2}) + (\frac{1}{2} - \cos(80^{\circ})) - (\cos(40^{\circ}) - \frac{1}{2})}{2} \\ & = \frac{\frac{3}{2} + \cos(20^{\circ}) - \cos(80^{\circ}) - \cos(40^{\circ})}{2} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

From this Math SE answer, using the expansion of $\cos(80^{\circ}) = \cos(60^{\circ} + 20^{\circ})$ and $\cos(40^{\circ}) = \cos(60^{\circ} - 20^{\circ})$, we get $\cos(20^\circ) = \cos(80^\circ) + \cos(40^\circ) \;\to\; \cos(20^\circ) - \cos(80^\circ) - \cos(40^\circ) = 0$. Thus, \eqref{eq4A} simplifies to $\frac{3}{4}$. Using this and \eqref{eq2A} in \eqref{eq1A}, we end up with

$$\cosec(10^\circ) + \cosec(50^\circ) - \cosec(70^\circ) = \frac{\frac{3}{4}}{\frac{1}{8}} = 6 \tag{5}\label{eq5A}$$

Answer 2

Rewrite the first term using the triple angle identity with $x=30^\circ$:

$$\sin x = 3 \sin \frac x3 - 4 \sin^3 \frac x3$$

$$\begin{align*} \frac12 &= 3\sin10^\circ - 4\sin^310^\circ \tag{$*$} \\ \implies \csc 10^\circ &= 6 - 8\sin^2 10^\circ \end{align*}$$

Join the other terms with the sum identity,

$$\sin(x\pm y) = \sin x\cos y \pm \cos x \sin y$$

$$\begin{align*} \sin (60^\circ\pm t) &= \frac{\sqrt3}2 \cos t \pm \frac12 \sin t \\ \stackrel{t=10^\circ}\implies \csc 50^\circ - \csc 70^\circ &= \frac2{\sqrt3\cos10^\circ-\sin10^\circ} - \frac2{\sqrt3\cos10^\circ+\sin10^\circ} \\ &= \frac{4\sin10^\circ}{3\cos^210^\circ - \sin^210^\circ} \\ &= \frac{4\sin10^\circ}{3 - 4\sin^2 10^\circ} \end{align*}$$

By $(*)$, we end up with

$$\begin{align*} & 6 - 8\sin^2 10^\circ + \frac{4\sin10^\circ}{3-4\sin^210^\circ} \\ &= 6 + \frac{4\sin10^\circ - 24\sin^210^\circ + 32\sin^410^\circ}{3-4\sin^210^\circ} \\ &= \boxed{6} + \frac{4\sin10^\circ}{3-4\sin^210^\circ} \underbrace{\left(1 - 6\sin10^\circ + 8\sin^3 10^\circ\right)}_{=0} \end{align*}$$

Answer 3

Since you received good answers, a $1,400^+$ years old approximate answer

$$A=\csc (x)+\csc \left(\frac{\pi}{3}-x\right)-\csc \left(\frac{\pi}{3}+x\right)=\frac{3 \csc (x)}{2 \cos (2 x)+1}$$

For $x=\frac \pi {18}$, $\sin(x)\sim \frac{17}{97}$, $\cos(2x)\sim \frac{77}{82}$ $$A\sim \frac{11931}{2006}=6-\frac{105}{2006}$$ (relative error of $0.87$%)