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I have this homework question that I have no idea how to do:

Show that $\{t, \sin(t), \cos(2t), \sin(t)\cos(t) \}$ is a linearly independent set of functions defined on $\mathbb{R}$. Start by assuming that $$c_1 t + c_2 \sin(t) + c_3 \cos(2t) + c_4 \sin(t)\cos(t) = 0$$ for all $t$. Choose specific values of $t$ ($t=0, 1, 2\dots$) until you get a system with enough equations to determine that all the $c_i$'s must be $0$.

My only guess is to set up a matrix based on this polynomial somehow, then row reduce it to find the pivot columns. Subbing in different $t$'s would allow for some value to be attached to each $c$. Would that be the matrix? Shouldn't it be formed by the $c$ values?

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    I would choose specific values *from the unit circle* for $t$...2011-11-01
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    What does choosing values for t do for you though? Is the matrix formed by the t terms?2011-11-01
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    possible duplicate of [Showing that $\{ 1, \cos t, \cos^2 t, \dots, \cos^6 t \}$ is a linearly independent set](http://math.stackexchange.com/questions/70920/showing-that-1-cos-t-cos2-t-dots-cos6-t-is-a-linearly-independ)2011-11-01
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    @Gortaur, I wouldn't say this question is a duplicate of that one, but I would certainly recommend that Randy have a look at the older question, as much of what is there is directly relevant to Randy's needs.2011-11-01
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    Evaluate at four points. If you are not unlucky, you will get $4$ linearly independent vectors in $\mathbb{R}^4$. Maybe try (i) $t=0$; (ii) $t=2\pi$; (iii) $t=\pi/2$; (iv) $t=\pi/4$. Look at the question referenced by @Gortaur. There are ways other than evaluating.2011-11-01
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    Since this equation should hold for **all** values of $t$, we can pick "convenient" values! For instance, with $t = 0$, we get (since sin(o) = 0) that $c_3 = 0$ right away. Then you can take the derivative (as has been mentioned in Mr. Cook's answer) and set t = 0 again. Or try t = $\pi / 2$, etc2011-11-01
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    that link helped. thanks guys!2011-11-01
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    @TheChaz2.0: The unit circle contains uncountably many points. I don't see how that helps narrow it down. (Of course there are specific points on the unit circle that are helpful to consider, like the right, top, left, bottom, maybe northeast.)2014-08-22

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