Prove that $$\frac{\sin 40^\circ}{\sin 80^\circ} +\frac{\sin 80^\circ}{\sin 20^\circ} -\frac{\sin 20^\circ}{\sin 40^\circ} =3.$$
I tried taking the least common multiple of the denominator and then simplifying the numerator but I do not get the value $3$.
Just for fun, a complicated proof by dissection:
The sine theorem gives that the length of the red segment is $\frac{\sin 40}{\sin 80}$, hence:
$$ 3+\frac{\sin 20}{\sin 40} = \frac{\sin 80}{\sin 20}+\frac{\sin 40}{\sin 80}$$
as wanted.
As $\sin(180^\circ-A)=\sin A,\sin(360^\circ-B)=-\sin B$
$\sin40^\circ=\cdots=-\sin320^\circ$ and $\sin20^\circ=\cdots=\sin160^\circ$
Now use $\sin2x=2\sin x\cos x$ twice to find $$\sin320^\circ=4\sin80^\circ\cos80^\circ\cos160^\circ$$
Similarly as $\cos(180^\circ-C)=-\cos C,$
$\sin80^\circ=4\sin20^\circ\cos20^\circ\cos40^\circ=-4\sin20^\circ\cos160^\circ\cos40^\circ$
and $\sin160^\circ=4\sin40^\circ\cos40^\circ\cos80^\circ$
So we need to establish $$4\cos40^\circ\cos80^\circ+4\cos80^\circ\cos160^\circ+4\cos160^\circ\cos40^\circ=-3$$
We shall establish something more generic: $$4\cos x\cos(120^\circ-x)+4\cos(120^\circ-x)\cos(120^\circ+x)+4\cos(120^\circ+x)\cos x=-3$$
Method $\#1:$
We have $\cos x\cos(120^\circ-x)+\cos(120^\circ+x)\cos x=\cos x(2\cos x\cos120^\circ)=-\cos^2x$
and using Werner Formula,
$2\cos(120^\circ-x)\cos(120^\circ+x)=\cos2x+\cos(2\cdot120^\circ)$ $=2\cos^2x-1+\cos(180^\circ+60^\circ)=2\cos^2x-1-\cos60^\circ=?$
Please complete the calculation.
Here we could utilize Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ as well.
Method $\#2:$ This is probably how the problem came into being.
If $\cos3x=\cos3A$
$3x=360^\circ m\pm3A$ where $m$ is any integer
$x=120^\circ m+ A$ where $m\equiv0,1,2\pmod3$
Now as $\cos3x=4\cos^3x-3\cos x,$
The roots of $4t^3-3t-\cos3A=0$ are
$t_1=\cos(120^\circ\cdot0+ x)=\cos x$
$t_2=\cos(120^\circ\cdot1+ x)=\cos(120^\circ+ x)$
$t_3=\cos(120^\circ\cdot2+ x)=\cos(240^\circ+ x)=\cos\{360^\circ-(120^\circ+x)\}=\cos(120^\circ-x)$
Now applying Vieta's formula,
$t_1+t_2+t_3=0$
$t_1\cdot t_2+t_2\cdot t_3+t_3\cdot t_1=-\dfrac34$
$t_1\cdot t_2\cdot t_3=\dfrac{\cos3A}4$
Since $$8\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}=\frac{8\sin20^{\circ}\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}}{\sin20^{\circ}}=\frac{\sin160^{\circ}}{\sin20^{\circ}}=1,$$ we obtain: $$\frac{\sin 40^\circ}{\sin 80^\circ} +\frac{\sin 80^\circ}{\sin 20^\circ} -\frac{\sin 20^\circ}{\sin 40^\circ}=$$ $$=\frac{1}{2\cos40^{\circ}}+4\cos20^{\circ}\cos40^{\circ}-\frac{1}{2\cos20^{\circ}}=$$ $$=\frac{\cos20^{\circ}-\cos40^{\circ}}{2\cos20^{\circ}\cos40^{\circ}}+2\cos60^{\circ}+2\cos20^{\circ}=$$ $$=\frac{\sin10^{\circ}}{2\cos20^{\circ}\cos40^{\circ}}+1+2\cos20^{\circ}=$$ $$=\frac{4\cos^280^{\circ}}{8\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}}+1+2\cos20^{\circ}=$$ $$=2+2\cos160^{\circ}+1+2\cos20^{\circ}=3.$$ Done!
