A puzzle about replacing $v_1$, $v_2$, $v_3$ while retaining the linear independence of the resulting set.

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I am reading the book, Applied Linear Algebra and Matrix Analysis. When I was doing the exercise of Section3.5 Exercise 5, I was puzzled at some of it. Here is the problem description:

Exercise 5. Let $V = C [0,1]$ and $w_1 = \sin^2 x$, $w_2 = \cos x$, $v_1 = \sin x$, $v_2 = \cos^2x$, $v_3 = 1$. The set $v_1,\ v_2,\ v_3$ is linearly independent in $V$. Determine which $v_j$ ’s could be replaced by w1, and which $v_j$’s could be replaced by both $w1$ and $w2$, while retaining the linear independence of the resulting set.

It is not hard to know that $w_1$ can replace $v_3$.
But the reference answer is given as followed:

$w_1$ could replace $v_2$ or $v_3$, $w_2$ could replace any of $v_1$, $v_2$, $v_3$ and $w_1$, $w_2$ could replace any two of $v_1$, $v_2$, $v_3$.

What Puzzles me is why $w_2$ could replace any of $v_1$, $v_2$, $v_3$ and $w_1$, $w_2$.
I can't figure it out by myself.
So if not mind, could anyone help?
Thanks in advance.

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The set $\{v_1,v_2,v_3\}$ is linearly independent: evaluate a linear combination at $0$, $\pi/2$ and $\pi/4$.

Also $w_1=\sin^2x=1-\cos^2x\in\langle v_1,v_2,v_3\rangle=U$. Thus $w_1$ can replace either $v_2$ or $v_3$ and the resulting set spans the same subspace: $$ U=\langle v_1,v_2,v_3\rangle=\langle v_1,v_2,w_1\rangle= \langle v_1,w_1,v_3\rangle $$ On the other hand, $w_2\notin U$, because the set $\{v_1,v_2,w_1,w_2\}$ is linearly independent as you can check with suitable substitutions.

Hence removing either $v_1$ or $v_2$ yields a linearly independent set.

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Try out all the combinations. Mainly you want to avoid $w_1=\sin^2 x, v_2=\cos^2 x, v_3=1$, which is linearly dependent. Looks like any other combination is linearly independent, (unless there is some new trigonometric identity I'm not aware of).