Help in manipulating Integrals 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
 A: There are many equivalent ways to proceed. Below are a couple of equivalent methods.

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
$$S = \dfrac{\sin(n \theta) \cos((n+1) \theta)}{\sin(\theta)}$$

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

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,$$\sin(\theta) \cos(2(m+1) \theta) + \sin(m\theta) \cos((m+1) \theta) = \dfrac{\sin((2m+3)\theta) - \sin(\theta)}2 = \sin((m+1)\theta) \cos((m+2) \theta)$$
$$\sum_{k=1}^{m+1} \cos(2k \theta) = \dfrac{\sin((m+1)\theta) \cos((m+2) \theta)}{\sin(\theta)}$$
A: 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
$$
1+2\sum_{k=1}^n\cos(2\theta k)=\frac{\sin((2n+1)\theta)}{\sin(\theta)}\tag{3}
$$
A: $$\begin{align*}\sum_{k=1}^n\cos(2\theta k)&=\operatorname{Re}\sum_{k=1}^n(e^{2i\theta})^k=\operatorname{Re}\left(\frac{e^{2i\theta(n+1)}-1}{e^{2i\theta}-1}-1\right)=\operatorname{Re}\frac{e^{2i\theta(n+1)}-e^{2i\theta}}{e^{i\theta}(e^{i\theta}-e^{-i\theta})}\\
&=\operatorname{Re}\frac{e^{2in\theta}-e^{i\theta}}{2i\sin\theta}=\operatorname{Re}\frac{\cos2n\theta+i\sin2n\theta-\cos\theta-i\sin\theta}{2i\sin\theta}\\
&=\operatorname{Re}\left(\frac{\sin2n\theta-\sin\theta}{2\sin\theta}+i\frac{\cos\theta-\cos2n\theta}{2\sin\theta}\right)=\frac{\sin2n\theta-\sin\theta}{2\sin\theta}
\end{align*}$$
Hence
$$1+2\sum _{ k=1 }^{ n }{\cos{(2k\theta )} }=1+2\frac{\sin2n\theta-\sin\theta}{2\sin\theta}=\frac{\sin2n\theta}{\sin\theta}$$
