Tangents of sums

Geometry Level 3

Let e k e_k (for k = 0 , 1 , 2 , 3 , . . . k = 0, 1, 2, 3, ... ) be the k k th-degree elementary symmetric polynomial in the variables

x i = tan θ i x_i= \tan \theta_i

for i = 0 , 1 , 2 , i = 0, 1, 2, \ldots i.e.

e 0 = 1 e_0 = 1

e 1 = i tan θ i e_1 = \displaystyle \sum_i \tan \theta_i

e 2 = i < j tan θ i tan θ j e_2 =\displaystyle \sum_{i<j} \tan \theta_i \tan \theta_j

and so forth.

Find the value of

tan ( i θ i ) . \tan \left( \displaystyle \sum_i \theta_i \right) .

e 0 e 2 + e 4 e 1 + e 3 e 5 + \dfrac {e_0 - e_2 + e_4 - \ldots}{e_1 + e_3 - e_5 + \ldots} e 1 + e 3 e 5 + e 0 e 2 + e 4 \dfrac {e_1 + e_3 - e_5 + \ldots}{e_0 - e_2 + e_4 - \ldots} e 0 e 2 + e 4 e 1 e 3 + e 5 \dfrac {e_0 - e_2 + e_4 - \ldots}{e_1 - e_3 + e_5 - \ldots} e 1 e 3 + e 5 e 0 e 2 + e 4 \dfrac {e_1 - e_3 + e_5 - \ldots}{e_0 - e_2 + e_4 - \ldots}

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rindell Mabunga
Jul 4, 2014

I just let i = 3 i = 3 and found out that the answer is e 1 e 3 + e 5 . . . e 0 e 2 + e 4 . . . \frac{e_1 - e_3 + e_5 - ...}{e_0 - e_2 + e_4 - ...}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...