# Linear independence of $\cos(2x)$, $\sin^2(x)$ and $\cos(x)$. [closed]

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!

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

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• I edited your post to get the $\LaTeX$ to work better. Cheers! – Robert Lewis Dec 22 '18 at 23:21
• @RobertLewis Thank you! :) – Miguel Ferreira Dec 22 '18 at 23:23

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

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 $$a=b=c=0$$?

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…

• But, since that for $x = pi$ the sum equals 0, and with a = b = c = 1, how can they be LI? Thank you! – Miguel Ferreira Dec 22 '18 at 23:38
• @MiguelFerreira The relation must hold for every $x$. – egreg Dec 22 '18 at 23:39
• @egred Oh, that's right! Thank you very much! :) – Miguel Ferreira Dec 22 '18 at 23:50

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.

• Thank you for the help! @Dave – Miguel Ferreira Dec 22 '18 at 23:50

$$c_1\cos 2x+c_2\sin^2x+c_3\cos x=0\\\implies c_1\cos 2x+\frac{c_2}2(1-\cos 2x)+c_3\cos x=0\\\implies c'\cos 2x+c_3\cos x+c''=0$$

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 $$c'=c_3=0\implies c''=c_1=c_2=c_3=0$$.