Why does $ 2-2\cos(ka) = 4 \sin^2(\frac{ka}{2})$

Question

$$2-2\cos(ka) = 4 \sin^2\left(\frac{ka}{2}\right)$$

This came up in a Fourier Transform question in my notes and I was wondering why it was true.

Thanks!

Answer 1

Hint: $$\sin^2 (\frac{\alpha}{2}) = \frac{1- \cos (\alpha)}{2}$$

Now multiply by 4

Answer 2

This has nothing with Fourier; it is a basic trigonometry identity:

Let $x= ka/2$, then

$$ 2-2\cos 2x = 2(\cos^2x+\sin^2x )-2(\cos^2x-\sin^2x) = 4\sin^2x $$

Answer 3

Hint: Use the fundamental trigonometric formula

$$\cos(2\alpha)=\cos^2 \alpha - \sin^2\alpha = 1-\sin^2\alpha -\sin^2\alpha=1-2\sin^2\alpha$$

Answer 4

This comes from one of the linearisation formulæ in trigonometry: $$\cos^2\theta=\frac{1+\cos\frac\theta 2}{2},\qquad \sin^2\theta=\frac{1-\cos\frac\theta 2}{2}.$$ (remember the duplication formulæ: $\;\cos 2x=\cos^2 x-\sin^2x=2\cos^2x-1=1-2\sin^2x$).

Answer 5

From the basics we have $$ \cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b) $$ Now using $a=b=\frac{k\alpha}{2}$ we have $$ \cos(k\alpha) = \cos\left(\frac{k\alpha}{2}\right)^2 - \sin\left(\frac{k\alpha}{2}\right)^2 = 1 - 2 \sin\left(\frac{k\alpha}{2}\right)^2 $$ since $\cos(a)^2+\sin(a)^2=1$. Rearranging and multiplying by two results on the identity you have.