Trigonometric Equation $\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$

How can I solve this trigonometric equation? $$\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$$

Using this solution,

$$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$

Putting $x=12^\circ,$ $$\tan12^\circ\tan48^\circ\tan72^\circ=\tan36^\circ$$

$$\implies \tan 12^\circ \tan 48^\circ \tan 54^\circ \tan 72^\circ =\tan 54^\circ \tan36^\circ=\tan(90^\circ-36^\circ)\tan36^\circ=\cot36^\circ\tan36^\circ=1$$

Ok, there might be nicer ways to do this but here it goes. Let $\theta=\pi/15$ so you have

$$ \tan(\theta)\tan(4\theta)\tan(\frac{9}{2}\theta)\tan(6\theta) $$

then you use $\tan()=\frac{\sin()}{\cos()}$, and then you use the equivalences

$2\sin(\alpha)\sin(\beta)=\cos(\alpha-\beta)-\cos(\alpha+\beta),$

and

$2\cos(\alpha)\cos(\beta)=\cos(\alpha-\beta)+\cos(\alpha+\beta)$.

You should get

$\frac{\cos(3\theta)\cos(\frac{3}{2}\theta)-\cos(3\theta)\cos(\frac{21}{2}\theta)-\cos(5\theta)\cos(\frac{3}{2}\theta)+\cos(5\theta)\cos(\frac{21}{2}\theta)}{\cos(3\theta)\cos(\frac{3}{2}\theta)+\cos(3\theta)\cos(\frac{21}{2}\theta)+\cos(5\theta)\cos(\frac{3}{2}\theta)+\cos(5\theta)\cos(\frac{21}{2}\theta)} $

** I also used the fact that $\cos(-\alpha)=\cos(\alpha)$. Now, note that you have something of the form

$\frac{a-b-c+d}{a+b+c+d},$

which is equal to $1+\frac{-2b-2c}{a+b+c+d}$. Now, we shall show that $-2b-2c=0$. So we have

$ -2b-2c=-2\cos(3\theta)\cos(\frac{21}{2}\theta)-2\cos(5\theta)\cos(\frac{3}{2}\theta) $

using the formula above for the product of cosines we get

$-2\cos(3\theta)\cos(\frac{21}{2}\theta)-2\cos(5\theta)\cos(\frac{3}{2}\theta)=-\cos(\frac{15}{2}\theta)-\cos(\frac{27}{2}\theta)-\cos(\frac{7}{2}\theta)-\cos(\frac{13}{2}\theta).$

Now plug in the value of $\theta$, you get

$ -\cos(\pi/2)-\cos(\frac{27}{30}\pi)-\cos(\frac{7}{30}\pi)-\cos(\frac{13}{30}\pi) $

naturally the first element is $0$. Now, using the fact that $\cos(\alpha\pm\pi/2)=\mp\sin(\alpha)$ you can write

$ -\cos(\frac{27}{30}\pi)-\cos(\frac{7}{30}\pi)-\cos(\frac{13}{30}\pi)=\sin(6\pi/15)-\sin(\frac{4}{15}\pi)-\sin(\pi/15) $

Now you use $\sin(\alpha)\cos(\beta)=\frac{\sin(\alpha+\beta)+\cos(\alpha-\beta)}{2}$ looking for two numbers such that the equality holds. You should get then

$ \sin(6\pi/15)-\sin(\frac{4}{15}\pi)-\sin(\pi/15)=\sin(6\pi/15)-2\sin(\pi/6)\cos(\pi/10)=\sin(6\pi/15)-\cos(\pi/10) $

which it is easily seen to be zero.

Therefore

$ \tan(\theta)\tan(4\theta)\tan(\frac{9}{2}\theta)\tan(6\theta)=1, \qquad \theta=\pi/15 $

Put $z = \exp(i\pi/30)$ and use Euler's formula to get $$P = \tan\left(\frac{\pi}{15}\right) \times \tan\left(\frac{4\pi}{15}\right) \times \tan\left(\frac{3\pi}{10}\right) \times \tan\left(\frac{6\pi}{15}\right) \\ = \frac{1}{i^4} \frac{z^2-1/z^2}{z^2+1/z^2} \frac{z^8-1/z^8}{z^8+1/z^8} \frac{z^9-1/z^9}{z^9+1/z^9} \frac{z^{12}-1/z^{12}}{z^{12}+1/z^{12}} \\ = \frac{z^4-1}{z^4+1} \frac{z^{16}-1}{z^{16}+1} \frac{z^{18}-1}{z^{18}+1} \frac{z^{24}-1}{z^{24}+1}.$$ Now simplify moving from right to left, repeatedly applying the fact that $z^{30}=-1$. First we get $$ \frac{z^4-1}{z^4+1} \frac{z^{16}-1}{z^{16}+1} \frac{z^{42}-z^{18}-z^{24}+1}{z^{42}+z^{18}+z^{24}+1} = \frac{z^4-1}{z^4+1} \frac{z^{16}-1}{z^{16}+1} \frac{-z^{12}-z^{18}-z^{24}+1}{-z^{12}+z^{18}+z^{24}+1}$$ This is $$\frac{z^4-1}{z^4+1} \frac{-z^{28}-z^{34}-z^{40}+z^{16}+z^{12}+z^{18}+z^{24}-1} {-z^{28}+z^{34}+z^{40}+z^{16}-z^{12}+z^{18}+z^{24}+1}$$ which is in turn $$\frac{z^4-1}{z^4+1} \frac{-z^{28}+z^4+z^{10}+z^{16}+z^{12}+z^{18}+z^{24}-1} {-z^{28}-z^4-z^{10}+z^{16}-z^{12}+z^{18}+z^{24}+1}.$$ Call this fraction $f.$ We need to show that $f=1.$ The numerator is $$ -z^{32}+z^8+z^{14}+z^{20}+z^{16}+z^{22}+z^{28}-z^4\\ + z^{28}-z^4-z^{10}-z^{16}-z^{12}-z^{18}-z^{24}+1$$ The denominator is $$ -z^{32}-z^8-z^{14}+z^{20}-z^{16}+z^{22}+z^{28}+z^4\\ -z^{28}-z^4-z^{10}+z^{16}-z^{12}+z^{18}+z^{24}+1.$$ The difference between these two is $$2\times \left(z^{28}-z^{24}-z^{18}+z^{14}+z^8-z^4\right) = 2\times \left(z^{28}-z^{24}+1/z^{12}-1/z^{16}+z^8-z^4\right).$$ Now putting $u = z^4 = \exp(4 i\pi/30) = \exp(2i\pi/15)$ this becomes $$2\times \left(u^7-u^6+1/u^3-1/u^4+u^2-u\right).$$ But $u^5 = -1/2 + 1/2\sqrt{3}i$ and $u^{-5} = -1/2 - 1/2\sqrt{3}i$ so the inner term finally becomes $$\left(-\frac{1}{2} + \frac{1}{2}\sqrt{3}i\right) u^2 - \left(-\frac{1}{2} + \frac{1}{2}\sqrt{3}i\right) u + \left(-\frac{1}{2} - \frac{1}{2}\sqrt{3}i\right) u^2 - \left(-\frac{1}{2} - \frac{1}{2}\sqrt{3}i\right) u\\ + u^2 -u$$ which is $$u^2\left(-\frac{1}{2} + \frac{1}{2}\sqrt{3}i -\frac{1}{2} - \frac{1}{2}\sqrt{3}i +1\right)+ u\left(\frac{1}{2} - \frac{1}{2}\sqrt{3}i +\frac{1}{2} + \frac{1}{2}\sqrt{3}i -1\right) = 0.$$ This means that the difference between numerator and denominator of $f$ is zero and since both are nonzero we have $f=1$ as desired.

The motivation for this admittedly somewhat tedious calculation is that it demonstrates how a computer algebra system might carry out this sort of simplification without human intervention.

There is a more creative use of the above techniques at the following MSE link.