Help in manipulating Integrals

Question

I try to express : $\displaystyle 1+2\sum _{ k=1 }^n \cos(2k\theta ) $

as : $\dfrac { \sin\left( \theta +2\theta n \right) }{ \sin\left( \theta \right) } $

I tried to use the exponential function :

$$I=\sum _{ k=1 }^n \cos(2k\theta ) ;\quad J=\sum _{ k=1 }^n \sin(2k\theta) $$

$I+i\cdot J=\displaystyle\sum _{ k=1 }^n e^{ 2ki\theta }$ and then get only the real part, but without success.

I also tried multiplying by $\dfrac {\sin(p)}{\sin(p)}$ (with $p$ an unknown variable I would determine later), but still, no success.

If someone could help, or identify the problem (I know it's a pretty famous one)

Thank you

Answer 1

There are many equivalent ways to proceed. Below are a couple of equivalent methods.

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First method:

Let $$S = \sum_{k=1}^n \cos(2k \theta)$$ We then have $$S \sin(\theta) = \sum_{k=1}^n \sin(\theta) \cos(2k \theta)$$ Now recall that $$\sin(A) \cos(B) = \dfrac{\sin(A+B) - \sin(B-A)}2$$ Hence, we get that $$S \sin(\theta) = \sum_{k=1}^n \dfrac{\left(\sin((2k+1) \theta) - \sin((2k-1) \theta) \right)}2$$ Now you can see a telescopic cancellation \begin{align} 2S \sin(\theta) & = \sin((2n+1) \theta) - \sin((2n-1) \theta)\\ & {}+ \sin((2n-1) \theta) - \sin((2n-3) \theta)\\ & {}+ \sin((2n-3) \theta) - \sin((2n-5) \theta) + \cdots\\ & {}+ \sin(3 \theta) - \sin(\theta) \end{align} Hence, we get that $$2S \sin(\theta) = \sin((2n+1) \theta) - \sin(\theta) = 2 \sin(n \theta) \cos((n+1) \theta)$$ Hence, we get that $$S = \dfrac{\sin(n \theta) \cos((n+1) \theta)}{\sin(\theta)}$$

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Second method:

Another way is to look at the real part of $e^{2i k \theta}$ and sum the geometric progression.

$$S = \text{Re} \left( \sum_{k=1}^n e^{2ik \theta} \right)$$ $$\sum_{k=1}^n e^{2ik \theta} = e^{2i \theta} \left(\dfrac{e^{2in \theta} - 1}{e^{2i \theta} - 1}\right) = e^{2i \theta} \left(\dfrac{1-e^{2in \theta}}{1-e^{2i \theta}}\right)$$ $$1-e^{2i \theta} = 1 - \cos(2 \theta) - i \sin(2 \theta) = 2\sin^2(\theta) - 2i \sin(\theta) \cos(\theta) = -2i \sin(\theta) e^{i \theta}$$ $$1-e^{2i n\theta} = -2i \sin(n\theta) e^{i n\theta}$$ $$\sum_{k=1}^n e^{2ik \theta} = e^{2i \theta} \times \dfrac{-2i \sin(n\theta) e^{i n\theta}}{-2i \sin(\theta) e^{i \theta}} = \dfrac{\sin(n \theta)}{\sin(\theta)} e^{i(n+1) \theta}$$ Hence, $$\text{Re} \left( \sum_{k=1}^n e^{2ik \theta} \right) = \dfrac{\sin(n \theta) \cos((n+1) \theta)}{\sin(\theta)}$$

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Third method:

You can also use induction directly to prove $$\sum_{k=1}^n \cos(2k \theta) = \dfrac{\sin(n \theta) \cos((n+1) \theta)}{\sin(\theta)}$$ For $n=1$, we have $$\cos(2 \theta) = \dfrac{\sin(\theta)\cos(2 \theta)}{\sin(\theta)}$$ Assume it is true for $n=m$ i.e. $$\sum_{k=1}^m \cos(2k \theta) = \dfrac{\sin(m \theta) \cos((m+1) \theta)}{\sin(\theta)}$$ Now at the induction step, we have that \begin{align} \sum_{k=1}^{m+1} \cos(2k \theta) & = \sum_{k=1}^{m} \cos(2k \theta) + \cos(2(m+1) \theta)\\ & = \underbrace{\dfrac{\sin(m \theta) \cos((m+1) \theta)}{\sin(\theta)}}_{\text{From induction hypothesis}} + \cos(2(m+1) \theta)\\ & = \dfrac{\sin(m \theta) \cos((m+1) \theta) + \cos(2(m+1) \theta) \sin(\theta)}{\sin(\theta)} \end{align} $$\sin(\theta) \cos(2(m+1) \theta) = \dfrac{\sin((2m+3)\theta) - \sin((2m+1)\theta)}2$$ $$\sin(m \theta) \cos((m+1) \theta) = \dfrac{\sin((2m+1)\theta) - \sin(\theta)}2$$ Hence,$$\sin(\theta) \cos(2(m+1) \theta) + \sin(m\theta) \cos((m+1) \theta) = \dfrac{\sin((2m+3)\theta) - \sin(\theta)}2 = \sin((m+1)\theta) \cos((m+2) \theta)$$ $$\sum_{k=1}^{m+1} \cos(2k \theta) = \dfrac{\sin((m+1)\theta) \cos((m+2) \theta)}{\sin(\theta)}$$

Answer 2

The formula for the sum of a geometric series says $$ \begin{align} \sum_{k=1}^ne^{i2\theta k} &=e^{i2\theta}\frac{e^{i2\theta n}-1}{e^{i2\theta}-1}\\ &=e^{i(n+1)\theta}\frac{e^{i\theta n}-e^{-i\theta n}}{e^{i\theta}-e^{-i\theta}}\\ &=e^{i(n+1)\theta}\frac{\sin(n\theta)}{\sin(\theta)}\tag{1} \end{align} $$ Taking the real part of $(1)$ yields $$ \begin{align} \sum_{k=1}^n\cos(2\theta k) &=\frac{\cos((n+1)\theta)\sin(n\theta)}{\sin(\theta)}\\ &=\frac{\sin((2n+1)\theta)-\sin(\theta)}{2\sin(\theta)}\tag{2} \end{align} $$ Doubling $(2)$ and adding $1$ gives $$ 1+2\sum_{k=1}^n\cos(2\theta k)=\frac{\sin((2n+1)\theta)}{\sin(\theta)}\tag{3} $$

Answer 3

$$\begin{align*}\sum_{k=1}^n\cos(2\theta k)&=\operatorname{Re}\sum_{k=1}^n(e^{2i\theta})^k=\operatorname{Re}\left(\frac{e^{2i\theta(n+1)}-1}{e^{2i\theta}-1}-1\right)=\operatorname{Re}\frac{e^{2i\theta(n+1)}-e^{2i\theta}}{e^{i\theta}(e^{i\theta}-e^{-i\theta})}\\ &=\operatorname{Re}\frac{e^{2in\theta}-e^{i\theta}}{2i\sin\theta}=\operatorname{Re}\frac{\cos2n\theta+i\sin2n\theta-\cos\theta-i\sin\theta}{2i\sin\theta}\\ &=\operatorname{Re}\left(\frac{\sin2n\theta-\sin\theta}{2\sin\theta}+i\frac{\cos\theta-\cos2n\theta}{2\sin\theta}\right)=\frac{\sin2n\theta-\sin\theta}{2\sin\theta} \end{align*}$$ Hence $$1+2\sum _{ k=1 }^{ n }{\cos{(2k\theta )} }=1+2\frac{\sin2n\theta-\sin\theta}{2\sin\theta}=\frac{\sin2n\theta}{\sin\theta}$$