Prove that $\frac12 + \cos (a) + \cos (2a) + ... + \cos (na) = \frac{\sin ((n+1/2)a)}{ 2sin(a/2)}$

Question

I need to prove that :

$$ \frac{1}{2} + \cos (\theta ) + \cos(2\theta ) + ... + \cos(n \theta) = \frac{\sin((n+\frac{1}{2})\theta))}{2\sin(\frac{\theta}{2}))} \;\;\;\;\;\; (1)$$

What I did :

• I have evaluated this series using Euler's formulas :

$\displaystyle\sum_{k=1}^{n} e^{ik\theta} =\sum_{k=0}^{n}(e^{ik\theta}) - 1 =e^{i\theta} \frac{1 - e^{i(n+1))\theta } }{1 - e^{i\theta}} - 1 =e^{i\theta} \cdot e^{i\frac{n\theta}{2}} \frac{\sin(\frac{n+1}{2}\theta)}{\sin(\frac{\theta}{2}))} -1 =e^{i(\theta+\frac{n\theta}{2})} \frac{\sin(\frac{n+1}{2}\theta)}{\sin(\frac{\theta}{2}))} - 1$

• Now, knowing that $\cos(x)=\Re(e^{ix})$ we get :

$\displaystyle\sum_{k=1}^{n} \cos(kx) = \Re(\sum_{k=1}^{n} e^{ik\theta} ) = \cos(\theta + \frac{n\theta}{2})\frac{\sin(\frac{n+1}{2}\theta)}{\sin(\frac{\theta}{2}))} - 1 $

• Re-plugging into (1) we have the following equation :

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

Am I doing something wrong ? Because I have the feeling that the argument in cos is not right. If the argument are the same in cos and sin I can use some trigonometric identities and finish the proof.

Thank you in advance.

Answer 1

Probably the fastest way:

$$2S=\sum_{k=\color{red}{-n}}^n e^{ika}=\frac{e^{i(n+1)a}-e^{-ina}}{e^{ia}-1}=\frac{e^{i(n+1/2)a}-e^{-i(n+1/2)a}}{e^{ia/2}-e^{-ia/2}}=\frac{\sin(n+\frac12)a}{\sin\frac12a}.$$

Answer 2

By mathematical induction on $n$

For $n=0$ the equation reduces to $1/2=1/2$. Now suppose it holds for $n=k$ and add $\cos((k+1)a)$:

$(1/2)+\cos(a)+\cos(2a)+...+\cos((k+1)a)=\dfrac{\sin((k+(1/2)a)}{2\sin(a/2)}+\cos((k+1)a)$

Then

$\dfrac{\sin((k+(1/2)a)}{2\sin(a/2)}+\cos((k+1)a)=\dfrac{\sin((k+(1/2))a)+2\sin(a/2)cos((k+1)a)}{2\sin(a/2)}$

And with $\cos(u)\sin(v)=(1/2)(\sin(u+v)-\sin(u-v))$:

$2\sin(a/2)\cos((k+1)a)=\sin(((k+1)+(1/2))a)-\sin((k+(1/2))a)$

Substituting this into the last term of the numerator then proves the claim for $n=k+1$:

$(1/2)+\cos(a)+\cos(2a)+...+\cos((k+1)a)=\dfrac{\sin(((k+1)+(1/2))a)}{2\sin(a/2)}$.