Value of an expression involving summation of a series of complex number

Question

Let $z=cos(\theta) + i sin(\theta)$. Then the value of $\sum_{m=1}^{15}Im(z^{2m-1})$ at $\theta=2^°$ is $$(a) \,\frac{1}{sin2^°} \\(b) \,\frac{1}{3sin2^°} \\(c) \,\frac{1}{2sin2^°} \\(d) \,\frac{1}{4sin2^°} $$

My attempt: I solved the problem and got the answer as follows: Value of the expression = $$ sin(\theta) + sin(3\theta)+ ... + sin(27\theta)+ sin(29\theta)$$ $$= 2sin(15\theta)[cos(14\theta) + cos(12\theta) + ...+cos(2\theta)] + sin(15\theta)$$ $$ = 2cos(7\theta)[cos(7\theta)+cos(5\theta)+cos(3\theta) + cos(\theta)]$$ $$= 8cos(7\theta)cos(4\theta)cos(2\theta)cos(\theta) $$ $$=\frac{1}{2sin(\theta)}(sin(15\theta) + sin(\theta))$$ $$=\frac{1}{4sin(\theta)} + 0.5$$ $$=\frac{1}{4sin(2^°)} + 0.5$$

Please confirm if my method is correct. How do I get to the current answer?

See also : https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro — lab bhattacharjee, Jan 13 '18 at 11:22

score 1 · Answer 1 · answered Jan 13 '18 at 10:35

Its better to use the summation formula of Geometric Series $$\sum_{r=1}^{15}z^{2m-1}=z\cdot\dfrac{1-(z^2)^{16}}{1-z^2}=\dfrac{z^{16+1}}z\cdot\dfrac{z^{16}-z^{-16}}{z-z^{-1}}$$

$$=(\cos16\theta+i\sin16\theta)\cdot\dfrac{2i\sin16\theta}{2i\sin\theta}$$

whose imaginary part is $$\dfrac{\sin^216\theta}{\sin\theta}$$

Value of an expression involving summation of a series of complex number

1 Answers1