Solve $\frac{\sin (10^\circ) \sin (30^\circ)}{\sin 40^\circ \sin (80^\circ-x^\circ)} = \frac{\sin 20^\circ}{\sin x}$

Question

The context to this is trivial I think, I was solving a geometry problem using the trigonometric version of Ceva, I got here and I was stuck, I tried using the sum-difference, product to sums, sums to products identities but my attempts failed and was rather tedious.

Answer 1

You can use \begin{equation} \sin a\sin b\sin c=\frac{1}{4}(\sin(a+b-c)+\sin(a-b+c)+\sin(-a+b+c)-\sin(a+b+c)) \end{equation} With this you get \begin{equation} \sin(40-x)+\sin(-20+x)+\sin(20+x)-\sin(40+x) \\ = \sin(-20+x)+\sin(60-x)+\sin(100-x)-\sin(140-x) \end{equation} Then you can use $\sin(140 - x) = \sin(180 - 140 + x) = \sin(40+x)$ to reduce this to \begin{equation} \sin(40-x)+\sin(20+x) =\sin(60-x)+\sin(100-x) \\ \sin(70-(30+x))+\sin(-10+(30+x)) - \sin(90-(30+x)) - \sin(130-(30+x)) = 0 \\ \end{equation} Now I use formula $\sin(a\pm(30+x))=\sin a\cos (30+x)\pm \cos a\sin (30+x)$. \begin{equation} \cos(30+x)(\sin 70 - \sin 10 - 1 -\sin130)+\sin(30+x)(-\cos 70 +\cos 10 + \cos 130) = 0 \\ -\cos(30+x)(\sin 250 + \sin 10 + 1 + \sin130)+\sin(30+x)(\cos 250 +\cos 10 + \cos 130) = 0 \end{equation} Now we can notice the following $(\cos 250 +\cos 10 + \cos 130) = \Re e^{i 10^\circ}(e^{i 0^\circ} + e^{i 120^\circ} + e^{i 240^\circ})$ and $(\sin 250 +\sin 10 + \sin 130) = \Im e^{i 10^\circ}(e^{i 0^\circ} + e^{i 120^\circ} + e^{i 240^\circ})$. But $e^{i 0^\circ} + e^{i 120^\circ} + e^{i 240^\circ} = 0$, which yields \begin{equation} -\cos(30+x) = 0 \end{equation} which has two solutions $x=60$ and $x=-120$.

Answer 2

Here is a completely different approach.

1. First part: existence of a unique solution:

The initial equation written under the form:

$$[\sin(20) \sin(40)]\frac{\sin(80-x)}{\sin(10)\sin(30)} = \sin(x)$$

(please note that I have dropped the "degree" sign everywhere)

can be transformed, using classical trigonometry formulas, into:

$$\tfrac12[\cos(20)-\cos(60)]\frac{\sin(80)\cos(x)-\sin(x)\cos(80)}{\sin(10)\tfrac12} = \sin(x)$$

Taking into account the fact that $\sin(10)=\cos(80)$:

$$[\cos(20)-\tfrac12]\left(\tan(80)\cos(x)-\sin(x)\right) = \sin(x)$$

$$\tan(x)=\tan(80)\dfrac{\cos(20)+\tfrac12}{\cos(20)-\tfrac12}\approx 1.7320508....\tag{3}$$

As the quantity in the RHS is positive and because $\tan$ is increasing on $[0,\pi/2)$, there exists a unique root $x_0$ on this interval. For the same reason, there is no root on the other interval $(\pi/2,\pi]$ (compatible with a triangle angular value).

The numerical value of the RHS of (3) appears to be $\sqrt{3}$, otherwise said $x_0=60$ degrees. Let us prove it in a rigorous way.

2. Second part: why is $x_0=60$ degrees exactly ?

Equivalently, we have to establish (by squaring the RHS of (3)) that:

$$3=\tan(80)^2\left(\dfrac{2\cos(20)+1}{2\cos(20)-1}\right)^2\tag{4}$$

Let us set $$c:=\cos(20).$$ $c$ is a root of (its minimal) polynomial $8x^3-6x-1$. For a proof, see here.

Otherwise said, $c$ is such that

$$\underbrace{8c^3-6c-1}_{P(c)}=0 \tag{5}$$

As we can write (4) under the form:

$$3=\left(\frac{1}{\cos^2(80)}-1 \right)\left(\dfrac{2\cos(20)+1}{2\cos(20)-1}\right)^2\tag{6}$$

As $\cos(4a)=8 \cos(a)^4 -8 \cos(a)^2 +1$,

(Chebyschev polynomial of order 4), taking $a=20$ degrees, (6) becomes a rational expression in "letter" $c$:

$$3=\left(\ \dfrac{1}{(8c^4 -8c^2 +1)^2}-1 \right)\left(\dfrac{2c+1}{2c-1}\right)^2=:\dfrac{N(c)}{D(c)}\tag{7}$$

It remains (using a Computing Algebra System) to obtain the euclidean division by polynomial $P$ (as defined in (5)) of the numerator and the denominator in (7) to get:

$$\require{cancel}\dfrac{N(c)}{D(c)}=\dfrac{Q(p)\xcancel{P(c)}+R(c)}{Q'(p)\xcancel{P(c)}+R'(c)}=\dfrac{R(c)}{R'(c)}$$

In this case, one obtains (trust me...) $R(c)=3/4$ and $R'(c)=1/4$ giving result $3$ as awaited.

Here is how I have obtained the $\color{red}{\text{rem}}$ainders $R(c)$ and $R'(c)$ with the C.A.S. available with Matlab:

[N,D]=numden((1/(8*c^4-8*c^2+1)^2-1)*((2*c-1)/(2*c+1))^2)
P=8*c^3-6*c-1;
rN=rem(N,P),rD=rem(D,P)
rN/rD