- #1

iDimension

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## Homework Statement

If

$$C = 1+cos\theta+...+cos(n-1)\theta,$$

$$S = sin\theta+...+sin(n-1)\theta,$$prove that

$$C=\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}} cos\frac{(n-1)\theta}{2} \enspace and \enspace S = \frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}}sin\frac{(n-1)\theta}{2}$$

## Homework Equations

## The Attempt at a Solution

$$C+iS = 1+(cos\theta+isin\theta)+...+(cos(n-1)\theta+isin(n-1)\theta)$$

$$=1+e^{i\theta}+...+e^{i(n-1)\theta}$$

$$=1+z+...+z^{n-1}, \enspace where \enspace z=e^{i\theta}$$

$$=\frac{1-z^n}{1-z}, \enspace if \enspace z\neq1$$

$$=\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in\theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{i\theta}{2}(e\frac{-i\theta}{2}-e\frac{i\theta}{2})}$$

$$=e^{i(n-1)\frac{\theta}{2}}\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}}$$

$$=(cos(n-1)\frac{\theta}{2}+isin(n-1)\frac{\theta}{2})\frac{sin\frac{n\theta}{2}}{sin\frac{\theta}{2}}$$

Is this correct? I'm not sure about the final solution. The final result is that the real parts equal the imaginary parts?