If $\alpha =2\pi/7$, find $\tan\alpha \tan 2\alpha + \tan 2\alpha \tan 4\alpha + \tan 4\alpha \tan \alpha$.
I tried by $7\alpha = 2\pi$
$4\alpha =2\pi-3\alpha$
$\sin 4\alpha = \sin 3\alpha$
But couldn't reach solution Please help
Asked
Active
Viewed 3,273 times
1
Em.
- 15,981
-
Please use MathJax and proper spelling to the best of your abilities. – Em. Jul 12 '17 at 02:47
-
@lab bhattacharjee I don't think that it's duplicate exactly because my solution is different than all solutions in your link. – Michael Rozenberg Jul 13 '17 at 03:53
2 Answers
2
Since $$\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}=\frac{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{\sin\frac{16\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{1}{8},$$ we obtain $$\tan\frac{2\pi}{7}\tan\frac{4\pi}{7}+\tan\frac{2\pi}{7}\tan\frac{8\pi}{7}+\tan\frac{4\pi}{7}\tan\frac{8\pi}{7}=$$ $$=8\left(\cos\frac{8\pi}{7}\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}\sin\frac{8\pi}{7}+\cos\frac{2\pi}{7}\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}\right)=$$ $$=4\left(\cos\frac{8\pi}{7}\left(\cos\frac{2\pi}{7}-\cos\frac{6\pi}{7}\right)+\cos\frac{4\pi}{7}\left(\cos\frac{6\pi}{7}-\cos\frac{4\pi}{7}\right)+\cos\frac{2\pi}{7}\left(\cos\frac{4\pi}{7}-\cos\frac{2\pi}{7}\right)\right)=$$ $$=2\left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}-\cos\frac{2\pi}{7}-\cos\frac{14\pi}{7}\right)+$$ $$+2\left(\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}-\cos\frac{6\pi}{7}-1\right)+$$ $$+2\left(\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}-\cos\frac{4\pi}{7}-1\right)=$$ $$=2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}-3\right)=$$ $$=2\left(\frac{2\sin\frac{\pi}{7}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)}{2\sin\frac{\pi}{7}}-3\right)=$$ $$=2\left(\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}-3\right)=2\left(-\frac{1}{2}-3\right)=-7.$$ Done!
Michael Rozenberg
- 194,933
1
$\displaystyle \tan \frac{k \pi}{7}; 1 \le k \le 6$ are roots of
$\tan 7 \theta = 0$ which gives the polynomial $$y^6 - \binom{7}{2} y^4 + \binom{7}{4} y^2- 7=0$$ with the substitution $ y = \tan \theta$
This also yields that $\displaystyle \tan^2 \frac{k \pi}{7}; k={2,4,8}$ are roots to the polynomial $$y^3 - \binom{7}{2} y^2 + \binom{7}{4} y - 7=0$$ so that from Vieta's formula $$\tan^2 \frac{2 \pi}{7}+\tan^2 \frac{4 \pi}{7}+\tan^2 \frac{8 \pi}{7} = 21$$
Further we have $$\tan \frac{2 \pi}{7}+\tan \frac{4 \pi}{7}+\tan \frac{8 \pi}{7} = \tan \frac{2 \pi}{7}\tan \frac{4 \pi}{7}\tan \frac{8 \pi}{7}$$
But from Vieta, $$tan^2 \frac{2 \pi}{7} \tan^2 \frac{4 \pi}{7} \tan^2 \frac{8 \pi}{7} = 7$$ and hence $$\tan \frac{2 \pi}{7}\tan \frac{4 \pi}{7}\tan \frac{8 \pi}{7} = -\sqrt 7$$
Using the identity 2$$\sum ab =(a+b+c)^2 -(a^2+b^2+c^2)$$ we obtain the required sum as $-7$
DXT
- 11,241
-
Do you mean $$ab+ac+bc = \frac{(a+b+c)^2 - (a^2 + b^2 + c^2)}{2}$$? – Zubin Mukerjee Jul 12 '17 at 03:02
-
@Durgesh, Have we not concurred : https://math.stackexchange.com/questions/823819/if-alpha-frac2-pi7-then-the-find-the-value-of-tan-alpha-tan2-alpha/824178#824178 – lab bhattacharjee Jul 12 '17 at 06:18