limit of $1- \cos(ka)$

Question

I am reading a text which says

for $ka \ll 1$:

$1 - \cos(ka) \approx \frac{1}{2}(ka)^{2}$ but I fail to understand why this is so.

Could someone shed light on this?

Answer 1

Think of its Taylor Series

$$\cos(ka)=1-\frac{(ka)^2}{2!} + \;...$$

Can you continue?

Answer 2

If you know that $$ \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1, $$ then you can expect that $\frac{\sin x}{x}\approx 1$ for small $x$. Thus $$ 1 - \cos x = 2\left( {\sin \left( {\frac{x}{2}} \right)} \right)^2 \approx 2\left( {\frac{x}{2}} \right)^2 = \frac{{x^2 }}{2} $$ for small $x$. Now take $x=ka$ with $ka \ll 1$.