Krdytkyu

Having three different vectors:

$u_1 = cos(2x)$ and $u_2 = sin^2(x)$ and $u_3 = cos(x)$

How can I prove that they are linearly independent?

Thank you!

Let $a,b,c$ such that $acos(2x)+bsin^2(x)+ccos(x)=0$ for all $xinmathbb{R}$.

(1) If $x=0$ then $a+c=0to a=-c$.

(2) If $x=pi$ then $a-c=0to a=cto_{(1)} a=c=0$.

(3) If $x=pi/2$ then $b=0$.

Ie, are LI

If $a cos (2x)+bsin^{2}(x)+ccos, x=0$ then $a cos (2x)+frac b 2 (1-cos (2x))+ccos, x=0$. Put $x=pi /2$ to get $b-a=0$. Put $x=0$ to get $a+c=0$ and put $x=pi /4$ to get $frac b 2 +cfrac 1 {sqrt 2}=0$. From these can you drive $a=b=c=0$?

Suppose
$$
alpha_1u_1+alpha_2u_2+alpha_3u_3=0
$$
(the constant zero function). Evaluate at suitable values of $x$, for instance $x=0$, $x=pi/4$ and…

You can do this via the definition. So suppose there are scalars $a,b,c$ such that $$acos(2x)+bsin^2(x)+ccos(x)=0qquad forall x$$ then you want to show that $a=b=c=0$. The key here is that the above equation should hold for all $x$, so try evaluating at certain $x$ values to have parts vanish and get some equations that show the desired result.

$c_1cos 2x+c_2sin^2x+c_3cos x=0\implies c_1cos 2x+frac{c_2}2(1-cos 2x)+c_3cos x=0\implies c'cos 2x+c_3cos x+c''=0$

where $c'=c_1-c_2/2,c''=c_2/2$.

If $c'$ or $c_3$ is non-zero, the $LHS$ will be periodic with period at-most $2pi$ while the same cannot be said for the $RHS$, which is the zero function. So $c'=c_3=0implies c''=c_1=c_2=c_3=0$.