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I've been having a problem solving this task. I originally assumed that since $\cos(2x) = 2\cos^2(x) - 1$, and since $V_4 = 2(V_3) - V_1$, one of the vectors is a linear combination of the other two and therefore, it is not linearly independent; but according to Wolfram Alpha, we, in fact, do have a linear independence. Can anyone explain why my original assumption is incorrect and how to approach solving this question?

Thank you very much in advance.

Edit $\#1$: You can see the link showing WA doing the calculation: See Wolfram Alpha

I'd also like to thank whoever edited this post for me. I'm still new here, I will try not to make those mistakes again.

Edit $\#2$: We've proven the linear dependence and figured out the issue with WolframAlpha's calculation. Major thanks to Git Gud and user133281.

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    Don't trust computers more than you should. Can you provide the WA link for the event you say occurred? – 2017-01-29
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    My apologies. I'll also edit the link in the post. http://www.wolframalpha.com/input/?i=linear+independence+%7B1,sin(x),cos(2x),(cos(x))%5E2%7D – 2017-01-29
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    See [this](http://www.wolframalpha.com/input/?i=linear+independence+%7B1,2%7D). WA doesn't interpret the question as you wish. What it is actually saying is that the set $\{(1, \sin(x), (\cos(x))^2, \cos(2x))\}$ is linearly independent (for all $x$), (and of course it is, it is a set with a single non-null vector). – 2017-01-29

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$$\cos(2x) = \cos^2 x - \sin^2 x = \cos^2 x - (1-\cos^2 x) = 2 \cos^2 x - 1,$$ so $$ 1 - 2 \cos^2 x + \cos(2x) = 0 $$ is a linear relation.