Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I solve this trigonometric equation?

$$\sin(x)=\tan(\frac{\pi}{15})\tan(\frac{4\pi}{15})\tan(\frac{3\pi}{10})\tan(\frac{6\pi}{15}).$$

share|improve this question
    
$\sin x ~=~ 1~$? –  Oleg567 Sep 24 '13 at 13:56
    
Why $\tan(\frac{\pi}{15})\tan(\frac{4\pi}{15})\tan(\frac{3\pi}{10})\tan(\frac{6\pi}{‌​15})=1$? –  user91500 Sep 24 '13 at 14:02
    
I explained it in answer now... –  Oleg567 Sep 24 '13 at 15:10

4 Answers 4

up vote 5 down vote accepted

Using this solution,

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

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

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

share|improve this answer
    
@Oleg567, please have a look into this solution? –  lab bhattacharjee Sep 28 '13 at 18:54
    
@Artin, how about this? –  lab bhattacharjee Sep 28 '13 at 18:55
    
Thanks very much. It's very well. –  user91500 Sep 29 '13 at 4:32
    
@Artin, my pleasure. I have prepared a list of each item containing four angles in degrees, the product of whose tangent ratios will be $1$ in the next comment. I think the formula mentioned should suffice –  lab bhattacharjee Sep 29 '13 at 18:49
1  
$$[1,59,61,87],[2,58,62,84],[3,29,31,89],[3,39,75,81],[3,57,63,81],[3,63,69,75],‌​[4,56,64,78],[5,55,65,75],[6,28,32,88],[6,42,66,78],[6,54,66,72],[7,53,67,69],[8,‌​52,66,68],[9,15,51,87],[9,27,33,87],[9,33,69,75],[9,51,63,69],[10,50,60,70],[11,4‌​9,57,71],[12,24,48,84],[12,26,34,86],[12,48,54,72],[13,47,51,73],[14,46,48,74],[1‌​5,21,27,87],[15,21,57,81],[15,25,35,85],[15,51,57,63],[16,42,44,76],[17,39,43,77]‌​,[18,24,36,84],[18,36,42,78],[19,33,41,79],[20,30,40,80],[21,23,37,83],[21,27,39,‌​81],[22,24,38,82],[27,33,39,75]$$ –  lab bhattacharjee Sep 29 '13 at 18:50

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...

share|improve this answer

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 $

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.