What trigonometric identity was used?

Question

The exercise asked to proof this identity: $$\sum_{n=0}^Ncos(nx) = \frac{1}{2} + \frac{1}{2}\frac{sin((N+\frac{1}{2})x)}{sin(\frac{x}{2})}$$

$$ $$ The solution is using Euler's identity and goes as follows:

$$\sum_{n=0}^Ne^{inx} = \frac{1-e^{i(N+1)x}}{1-e^{ix}}$$ now, taking only the real part:

$$\sum_{n=0}^Ncos(nx) = Re\frac{(1-e^{i(N+1)x})(1-e^{-ix})}{|1-e^{ix}|^2} = Re\frac{1-e^{-x}-e^{i(N+1)x}+e^{iNx}}{(1-cosx)^2+sin^2x} = Re\frac{1-cosx-cos((N+1)x)+cos(Nx)}{2-2cosx} =$$ $$ =^\star \frac{1}{2}+\frac{sin(\frac{2N+1}{2}x)sin\frac{x}{2}}{2-2cosx} =^\star \frac{1}{2}+\frac{1}{2}\frac{sin((N+\frac{1}{2})x)}{sin\frac{x}{2}}$$

Now, I understand everything except the last two equalities (I marked them with $\star$ so it will be clearer). What were the trigonometric identities used in them? I was thinking double angle but I couldn't figure it out exactly. (The $\frac{1}{2}+$ part is understood obviously, since it's just splitting the nominator).

Thanks!

Answer 1

1. $\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$

https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities

2. $1-\cos 2A=2\sin^2A$

https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle_formulae

Answer 2

The first equality comes from one of the so-called factorisation formulae: $$\cos p-\cos q=-2\sin\frac{p+q}2\sin\frac{p-q}2.$$ The second comes from a linearisation formula: $$\sin^2x=\frac{1-\cos 2x}2.$$ (There is also $\;\cos^2x=\dfrac{1+\cos 2x}2$.)

Answer 3

What about induction? After showing the base case and assuming truth for $\;k<N\;$ , show for $\;k=N\;$ :

$$\sum_{n=0}^N\cos nx=\sum_{n=0}^{N-1}\cos nx+\cos Nx\stackrel{\text{Ind. Hypothesis}}=\frac12+\frac12\frac{\sin\left[\left((N-1)+\frac12\right)x\right]}{\sin\frac x2}+\cos Nx=$$$${}$$

$$=\frac12+\frac12\frac{\sin\left[\left(N-\frac12\right)x\right]+2\cos Nx\sin\frac x2}{\sin\frac x2}$$

But

$$\color{red}{\sin\left[\left(N-\frac12\right)x\right]}+2\cos Nx\sin\frac x2=\color{red}{\sin Nx\cos\frac x2-\sin\frac x2\cos Nx}+2\cos Nx\sin\frac x2=$$$${}$$

$$=\sin Nx\cos\frac x2+\cos Nx\sin\frac x2=\sin\left(Nx+\frac x2\right)=\sin\left[\left(N+\frac12\right)x\right]$$

and we're done.