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The answer key says the following set of functions is linearly dependent: $\{5, \cos^2x, \sin^2x\}$.

Without calculating the Wronskian, I would've guessed it was independent because there's apparently no way you can form a linear combination out of any of these functions to get others: you can't multiply $5$ to get $\cos x$; you can't multiply $\cos x$ to get $\sin x$, etc. What's wrong with my reasoning?

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    $\begingroup$ It is not $\cos x$ and $\sin x$, but $(\cos x)^2$ and $(\sin x)^2$. $\endgroup$ – Marc van Leeuwen Mar 15 '15 at 17:28
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You can use $\sin^2x+\cos^2x=1$ to form a linear combination of those three functions that results in $0$.

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you can't multiply 5 to get $\cos x$, you can't multiply $cos$ to get $\sin x$ etc. What's wrong with my reasoning?

It looks like you are just checking whether any of these functions is a scalar multiple of a single one of the others.

But linear combinations are, in general, combinations — you need to consider sums of scalar multiples of the functions. And (as other answers show) there is a way to get 5 as a sum of scalar multiples of $\cos^2 x$ and $\sin^2 x$: $$ 5 = 5 \cos^2 x + 5 \sin^2 x $$ or, rearranged into the symmetric form of a linear dependency: $$ (-1) \cdot 5 + 5 \cdot \cos^2 x + 5 \cdot \sin^2 x = 0.$$

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Hint: $$ 5 \cos^2 x+ 5\sin^2x=5 $$

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You have $-\frac{1}{5}*5+\cos^2(x)+\sin^2(x)=-1+1=0$.

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