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?
$$\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$$
 A: Using this solution,
$$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$ 
Putting $x=12^\circ,$  $$\tan12^\circ\tan48^\circ\tan72^\circ=\tan36^\circ$$ 
$$\implies \tan 12^\circ \tan 48^\circ \tan 54^\circ \tan 72^\circ =\tan 54^\circ \tan36^\circ=\tan(90^\circ-36^\circ)\tan36^\circ=\cot36^\circ\tan36^\circ=1$$
A: 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
$
A: Let's use degree measure (for writing convenience):
we need to prove identity
$$
\tan 12^\circ \tan 48^\circ \tan 54^\circ \tan 72^\circ = 1.\tag{1}
$$
$$
\sin 12^\circ \sin 48^\circ \sin 54^\circ \sin 72^\circ =^?= \cos 12^\circ \cos 48^\circ \cos 54^\circ \cos 72^\circ;\tag{2}
$$
$$
\sin 12^\circ \sin 48^\circ \sin 54^\circ \sin 72^\circ =^?= \sin 78^\circ \sin 42^\circ \sin 36^\circ \sin 18^\circ;
$$
$$
(\sin 12^\circ \sin 72^\circ ) \cdot ( \sin 48^\circ \sin 54^\circ) =^?= (\sin 18^\circ \sin 78^\circ ) \cdot ( \sin 36^\circ \sin 42^\circ);
$$
$$
(\cos 60^\circ - \cos 84^\circ)(\cos 6^\circ - \cos 102^\circ) =^?=
(\cos 60^\circ - \cos 96^\circ)(\cos 6^\circ - \cos 78^\circ);
$$
$$
(\cos 60^\circ - \cos 84^\circ)(\cos 6^\circ + \cos 78^\circ) =^?=
(\cos 60^\circ + \cos 84^\circ)(\cos 6^\circ - \cos 78^\circ);
$$
$$
-\cos 84^\circ \cos 6^\circ + \cos 60^\circ \cos 78^\circ =^?=
\cos 84^\circ \cos 6^\circ - \cos 60^\circ \cos 78^\circ;\tag{3}
$$
$$
\cos 84^\circ \cos 6^\circ - \cos 60^\circ \cos 78^\circ=^?=0.\tag{4}
$$
Yes, $(4)$ is true  identity, because
$$
2\sin 6^\circ \cos 6^\circ = \sin 12^\circ,
$$
$$
\sin 6^\circ \cos 6^\circ = \frac{1}{2}\sin 12^\circ,
$$
$$
\cos 84^\circ \cos 6^\circ = \cos 60^\circ \cos 78^\circ.\tag{5}
$$
So, we get equation
$$
\sin (x) = 1. \tag{6}
$$
Solution is obvious...
