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Question

sinx+sin(2x)+...+sin(nx)=sin((n+1)*x/2)*sin(n*x/2)/sin(x/2)

Answer 1

Your problem got cut off. If you put spaces in it, Yahoo won't do that. I know, it's annoying.

Check this link, see if it's what you want:

http://www.gomath.com/Questions/question.php?question=14231

Answer 2

it can be found if u know a little bit of complex numbers.
we know that every complex number can be expressed in eulerian form e^(i*theta) = cos(theta) +isin(theta)
sinx +sin2x +sin3x ...... is thus the imaginary part of complex
e^(ix) + e^(i2x) +............. = [e^(ix){e^(inx)-1}]/(e^ix-1)....... (by applying the formula for the sum a g.p)

solving by rationalizing and all
we get e^(inx) -1 - e^1x(n+1) + e^ix ............1

now writing this in the form of cosx +isinx

now retourn to the first step where we assumed the given series to be the imaginary part of the complex number;-
cosx + cos2x +............... +i(sinx + sin2x+...........) ...............2

comparing the imaginary parts of the equation
we get this trigonometrical equation;-
[sin(nx) - sin(n+1)x + sinx]/2(1-cosx)

now solving this using the formula of sinA+sinB and doing various other results we get

the final result;-
[sin[{(n+1)x}/2]*sin(nx/2)]/sin(x/2)


this is what the right hand side of the given equation

Answer 3

yes, your question is not complete. i don't really get the clear picture there.

Answer 4

Please complete your question: The right
side of the equation is missing!!