SOLUTION: Given d is not multiples of 2pi. Show that sinb+sin(b+d)+sin(b+2d)+…+sin(b+(n-1)d)=[sin(nd/2)sin(b+(n-1)d/2)]/sin(d/2).

Algebra ->  Trigonometry-basics -> SOLUTION: Given d is not multiples of 2pi. Show that sinb+sin(b+d)+sin(b+2d)+…+sin(b+(n-1)d)=[sin(nd/2)sin(b+(n-1)d/2)]/sin(d/2).      Log On


   

Question 1172411: Given d is not multiples of 2pi. Show that sinb+sin(b+d)+sin(b+2d)+…+sin(b+(n-1)d)=[sin(nd/2)sin(b+(n-1)d/2)]/sin(d/2).
Answer by ikleyn(52908) About Me  (Show Source):
You can put this solution on YOUR website!
.

Hello,  this problem is of the Math circle level;  so,  it is intended for those advanced school students

(advanced high school students)  who proudly solve such problems  on their own.


        It would be a  BAD  STYLE  from my side to provide a complete solution to you.


Nevertheless,  to give a  HINT  is not a bad style:  it is exactly the  BEST  POSSIBLE  help for such a student.


        THEREFORE,  I provide such hints here.


There are two ways to solve the problem.  First way is to use the method of mathematical induction.

It will require to manipulate with complicated trigonometric identities on the way.


        A better way is to use complex numbers and the de Moivre formula.


You introduce complex number  x = cos(b) + i*sin(b)  and   Z = cos(d) + i*sin(d).

You consider geometric progression   x + xZ + xZ%5E2 + . . . + xZ%5E%28n-1%29.

You sum it up using the standard formula for the first  n  terms of the  GP.
It gives you the right side of the formula you need to prove.

Next you use the de Moivre formula and the product formula for the separate terms of the  GP.

The last step is to separate and to equal the imaginary parts of both formulas.

It will give you the proof of your identity.