${1 \over \sqrt{3}} \sum\limits _{r=0}^4 \tan \left( \frac{\pi}{15}+\frac{r\pi}{5} \right)=$?

Question

$$\sum _{r=0}^4 \tan \left( \frac{\pi}{15}+\frac{r\pi}{5} \right)=k \sqrt{3}$$. Then evaluation of $k$

solution i try

$$\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(120^\circ)+\tan(156^\circ)$$

$$=\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(156^\circ)+\sqrt{3}$$

$$=\frac{\sin(60^\circ)}{\cos(12^\circ)\cos(48^\circ)}+\frac{\sin(240^\circ)}{\cos(84^\circ)\cos(156^\circ)}-\sqrt{3}.$$

plz help me how to simplify denominator.

See https://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series OR https://math.stackexchange.com/questions/823619/calculate-tan9-circ-tan27-circ-tan63-circ-tan81-circ/823655#823655 — lab bhattacharjee, May 18 '18 at 03:10
@DavidG.Stork Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see Guidelines for good use of $\LaTeX$ in question titles. (Having said that, thanks for taking your time to help with editing and improving posts on this site.) — Martin Sleziak, May 18 '18 at 05:23
thanks lab bhattacharjee and David. plez explain how do i calculate my sum( which i have left) — jacky, May 18 '18 at 10:57

score 1 · Accepted Answer · answered May 18 '18 at 18:31

$$S=\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(120^\circ)+\tan(156^\circ)$$

$$S=\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(156^\circ)+\sqrt{3}$$

$$S=\frac{\sin(60^\circ)}{\cos(12^\circ)\cos(48^\circ)}+\frac{\sin(240^\circ)}{\cos(84^\circ)\cos(156^\circ)}-\sqrt{3}.$$

$$S=\frac{\sqrt{3}}{\cos (60^\circ)+\cos(36^\circ)}-\frac{\sqrt{3}}{\cos(240^\circ)+\cos(72\circ)}-\sqrt{3}$$

use $\displaystyle \cos (36^\circ)=\frac{\sqrt{5}+1}{4}$ and $\displaystyle \cos(72^\circ)=\sin(18^\circ)=\frac{\sqrt{5}-1}{4}$

$$=4\sqrt{3}\bigg[\frac{6}{4}\bigg]-\sqrt{3}=5\sqrt{3}$$

1 Answers1