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For $V = F(\mathbb R)$ be the space of real valued functions on the real line. $S$ is: $\ S = \operatorname{Span}\lbrace\sin\theta, \cos\theta, \sin\theta\cos\theta\rbrace. $

I have to find the $\dim S $

I think that $\sin{\theta}$, $\cos{\theta}$, and $\sin\theta\cos\theta$ are linearly independent and if so then the dimension of $S$ is 3. However I am unsure how to prove this. I do know that I have to prove that $a = b = c = 0$ in the following equation: $a \sin\theta + b\cos\theta + c \sin\theta\cos\theta = 0 $ but I don't know how to proceed from here.

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    Thanks @ArturoMagidin. I didn't know how to do the other type of equations.2012-02-14

1 Answers 1

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Hint. Remember that $a\sin\theta + b\cos\theta + c(\sin\theta)(\cos\theta)=0,$ is an equality of functions. That is, the function on the left is supposed to be identically zero. That is, you get $0$ for each and every value of $\theta$. So then you get zero when $\theta=0$; what does that tell you? And you also get $0$ when $\theta=\frac{\pi}{2}$; what does that tell you? And you also get $0$ when $\theta=\frac{\pi}{4}$. What does that tell you?

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    Thank you for the help. That helped me solve it. :)2012-02-14