Question

Prove that $$ \cos \theta+\cos (\theta+\alpha)+\cdots+\cos (\theta+n \alpha)=\frac{\sin \frac{1}{2}(n+1) \alpha}{\sin \frac{1}{2} \alpha} \cos \left(\theta+\frac{1}{2} n \alpha\right) $$

Short answer

The sum is \[ \frac{\text{sin}((n+1)\alpha/2)}{\text{sin}(\alpha/2)} \text{cos} \left( \theta + \frac{n \alpha}{2} \right) \].

Step-by-step solution

- Express the sum in complex exponential form

Recall that \(\text{cos} \theta = \text{Re}(e^{i\theta})\). Using this, we can rewrite the given sum: \[ \text{cos} \theta + \text{cos} (\theta + \alpha) + \text{cos} (\theta + 2\alpha) + ... + \text{cos} (\theta + n\alpha) \]. This is equivalent to the real part of \[ e^{i \theta} + e^{i (\theta+\alpha)} + e^{i (\theta+2\alpha)} + ... + e^{i (\theta+n\alpha)} \].

- Recognize the geometric series

The expression \[ e^{i\theta} + e^{i(\theta+\alpha)} + e^{i(\theta+2\alpha)} + \ldots + e^{i(\theta+n\alpha)} \] forms a geometric series with the first term \[ e^{i\theta} \] and common ratio \[ e^{i\alpha} \].

- Sum the geometric series

For a geometric series \(\text{S} = a + ar + ar^2 + \ldots + ar^n \), the sum is given by \[ \text{S} = a \frac{1-r^{n+1}}{1-r} \]. Here, \[ a = e^{i\theta} \text{and} \ r = e^{i\alpha} \]. So, our series sum is \[ S = e^{i\theta} \frac{1 - (e^{i\alpha})^{n+1}}{1 - e^{i\alpha}} = e^{i\theta} \frac{1 - e^{i(n+1)\alpha}}{1 - e^{i\alpha}} \].

- Simplify the exponential terms

Use Euler's formulas: \[ 1 - e^{i(n+1)\alpha} = 1 - ( \text{cos}((n+1)\alpha) + i \text{sin}((n+1)\alpha) ) \]. Notice that \[ 1 - e^{i(n+1)\alpha} = -2i \text{sin} \frac{(n+1)\alpha}{2} e^{i\frac{(n+1)\alpha}{2}} \] and \[ 1 - e^{i\alpha} = -2i \text{sin} \frac{\alpha}{2} e^{i\frac{\alpha}{2}} \].

- Substitute back and simplify

Substituting back, our sum becomes \[ e^{i\theta} \frac{ -2i \text{sin}((n+1)\alpha/2) e^{i(n+1)\alpha/2} }{ -2i \text{sin}(\alpha/2) e^{i\alpha/2} } = e^{i\theta} e^{i\frac{n\alpha}{2}} \frac{\text{sin}((n+1)\alpha/2)}{\text{sin}(\alpha/2)} \]. Simplifying, we get \[ e^{i(\theta+n\alpha/2)} \frac{\text{sin}((n+1)\alpha/2)}{\text{sin}(\alpha/2)} \].

- Take the real part

The real part of \[ e^{i(\theta+n\alpha/2)} \frac{\text{sin}((n+1)\alpha/2)}{\text{sin}(\alpha/2)} \] is: \[ \text{Re} \left( e^{i(\theta + \frac{n\alpha}{2})} \right) = \cos \left( \theta + \frac{n \alpha}{2} \right) \]. Finally, combining our results, we get \[ \frac{\text{sin}((n+1)\alpha/2)}{\text{sin}(\alpha/2)} \text{cos} \left( \theta + \frac{n \alpha}{2} \right) \].

Key concepts

geometric series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a constant called the common ratio. For instance, in the series: \[ a, ar, ar^2, ar^3, \text{...} \], \(a\) is the first term and \(r\) is the common ratio. The sum of the first \(n+1\) terms of a geometric series is given by the formula: \[ S = a \frac{1 - r^{n+1}}{1 - r} \]. Understanding geometric series helps us sum complex exponentials effectively. Here, our series has a first term \( e^{i\theta} \) and a common ratio \( e^{i\theta} \).

Recognizing these components lets us apply the geometric series sum formula.

complex exponential form

The complex exponential form leverages Euler's formula. Euler's formula states that for any real number \( \theta \), the complex exponential \( e^{i\theta} \) can be expressed as: \[ e^{i\theta} = \text{cos}(\theta) + i \text{sin}(\theta) \]. Using this form simplifies the manipulation and summation of trigonometric expressions.

For our problem, we originally have: \[ \text{cos}(\theta) + \text{cos}(\theta + \text{...}) \]. Using the real part of the complex exponential form, we can rewrite it as: \[ e^{i\theta} + e^{i(\theta + \text{...} ) \]. This change transforms the trigonometric sum into a more manageable arithmetic series.

Euler's formulas

Euler’s formulas form the bridge between trigonometric functions and complex exponentials. The formulas are:

• \( e^{i\theta} = \text{cos}(\theta) + i \text{sin}(\theta) \)
• \( e^{-i\theta} = \text{cos}(\theta) - i \text{sin}(\theta) \)

These help in converting exponential terms back into sines and cosines.

For our geometric series: \[ e^{i(\theta + n\text{...})} \], we used Euler's formulas to break down: \[ 1 - e^{i(n+1)\text{...}} \] into trigonometric components. Such conversions are essential for simplifying our final expression.

real part of complex number

When dealing with complex numbers, the real part is crucial for getting back to trigonometric functions. A complex number \( z \) can be written as \[ z = \text{Re}(z) + i\text{Im}(z) \], where \( \text{Re}(z) \) denotes the real part and \( \text{Im}(z) \) denotes the imaginary part. For the expression \( e^{i(\theta+n\text{...}/2)} \),
the real part is \( \text{Re}(e^{i(\theta + n\text{...}/2)}) = \text{cos}(\theta + n\text{...}/2) \).

Thus, by focusing on the real part of our final sum expression of complex exponentials, we can extract the result involving \( \text{cos} \).