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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!

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closed as off-topic by Namaste, Holo, user98602, mrtaurho, José Carlos Santos Dec 23 '18 at 0:00

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I edited your post to get the $\LaTeX$ to work better. Cheers! $\endgroup$ – Robert Lewis Dec 22 '18 at 23:21
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    $\begingroup$ @RobertLewis Thank you! :) $\endgroup$ – Miguel Ferreira Dec 22 '18 at 23:23
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Let $a,b,c$ such that $a\cos(2x)+b\sin^2(x)+c\cos(x)=0$ for all $x\in\mathbb{R}$.

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

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

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

Ie, are LI

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

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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…

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  • $\begingroup$ But, since that for $x = pi$ the sum equals 0, and with a = b = c = 1, how can they be LI? Thank you! $\endgroup$ – Miguel Ferreira Dec 22 '18 at 23:38
  • $\begingroup$ @MiguelFerreira The relation must hold for every $x$. $\endgroup$ – egreg Dec 22 '18 at 23:39
  • $\begingroup$ @egred Oh, that's right! Thank you very much! :) $\endgroup$ – Miguel Ferreira Dec 22 '18 at 23:50
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You can do this via the definition. So suppose there are scalars $a,b,c$ such that $$a\cos(2x)+b\sin^2(x)+c\cos(x)=0\qquad \forall x$$ 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.

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  • $\begingroup$ Thank you for the help! @Dave $\endgroup$ – Miguel Ferreira Dec 22 '18 at 23:50
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$c_1\cos 2x+c_2\sin^2x+c_3\cos x=0\\\implies c_1\cos 2x+\frac{c_2}2(1-\cos 2x)+c_3\cos x=0\\\implies c'\cos 2x+c_3\cos x+c''=0$

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 $2\pi$ while the same cannot be said for the $RHS$, which is the zero function. So $c'=c_3=0\implies c''=c_1=c_2=c_3=0$.

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