Are functions: $1$; $\cos(x)$; $\cos^2(\frac{x}{2})$ linearly independent? I used the definition.
$1$; $\cos(x)$; $\cos^2(\frac{x}{2})$
$c_1\cdot1+c_2\cdot\cos(x)+c_3\cdot\cos^2(\frac{x}{2}) = 0$
I tried converting $\cos^2(\frac{x}{2})$ into something better: $\frac{1+\cos(x)}{2}$
$c_1+c_2\cdot\cos(x)+c_3\cdot\frac{1+\cos(x)}{2} = 0$
$c_1+\cos(x)\cdot\left(c_2+c_3\cdot\frac{1+\cos(x)}{2\cos(x)}\right) = 0+0\cdot\cos(x)$
$c_1=0$
$c_2+c_3\cdot\frac{1+\cos(x)}{2\cos(x)}=0 \to c_2=-c_3\cdot\frac{1+\cos(x)}{2\cos(x)} \to \frac{c_2}{c_3}=\frac{-1-\cos(x)}{2\cos(x)}$
$c_2=-1-\cos(x)$ and $c_3=2\cos(x)$
There exist $c_i$-s outside of $c_i=0$ which can be solutions. So these functions are linearly dependent.
Is this correct?
Thanks.

Once you know that $$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2},$$ you already know that $$(1/2, 1/2, -1) \cdot (1, \cos x, \cos^2 \tfrac{x}{2}) = 0.$$
