Is set {x, sinx, sin2x,}in C[0,1] linearly dependent or independent?

I understand the problem is to show if

c1v1+c2v2+c3v3=0

in this case

c1x + c2sinx + c3sin2x=0

has more than the trivial solution in the space of all continouse functions in [0,1]. In otherwords can I find at least one non-zero c coefficient to satisfy this equation?

What is the best way to approach this? Can I plug in numbers for x and see if any of the c's are non-zero. The non-trivial solution should work for all x's so if I show it doesn't work for specific x's in the interval then the solution must be trivial? Or can I define another continous function in the interval = function 1 (dot product) function 2 then say these are orthogonal hence = 0 and find c's that way?

Thanks

My intution tells me they are independent but math is all about showing proof. Its just when I look at the graph of these on 0,1 interval I can't see how any of the functions can be a linear combination of the other two.

Originally Posted by rick81

Is set {x, sinx, sin2x,}in C[0,1] linearly dependent or independent?

If $\displaystyle \{x,\sin x,\sin 2x\}$ is linearly dependent on $\displaystyle [0,1]$ then it means $\displaystyle c_1x + c_2\sin x + c_3\sin 2x = 0$ has a non-trivial solution (not all three coefficients are zero) for all $\displaystyle x\in [0,1]$. Differenciating we find $\displaystyle c_1 + c_2\cos x + 2c_3 \cos 2x = 0$ and differenciating again we find $\displaystyle 0 - c_2\sin x - 4c_3\sin 2x = 0$. Thus, we have shown that:
$\displaystyle \left\{ \begin{array}{c}c_1x + c_2\sin x + c_3\sin 2x = 0\\c_1 + c_2\cos x + 2c_3 \cos 2x = 0\\ 0c_1 - c_2\sin x - 4c_3\sin 2x = 0\end{array} \right.$
Has a non-trivial solution, and so, it must be that,
$\displaystyle \left| \begin{array}{ccc} x & \sin x & \sin 2x \\ 1 & \cos x & 2\cos 2x \\ 0 & -\sin x & - 4\sin 4x \end{array} \right| = 0 \text{ for all }x\in [0,1]$.

Try to argue that this is impossible by expanding out the determinant.

I've been reviewing trig because some of the identities maybe helpful and timesaving ways to show linear dependence for trig functions such as the functions in my original question.

Which brings me to my next question.

Can I use the identity sin2x=2sinxcosx to show that x , sinx, sin2x are linearly dependent?

Sin2x=C1(x)+C2(sinx)
I can choose C1=0 & C2=2cosx?

Sin2x=0x + 2cosxsinx
Sin2x=2cosxsinx=2sinxcosx

Is my assumption that associative properties hold incorrect? Been a while since I took a trig class.