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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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Please don't abuse the display equations, and avoid using them in the title. – Arturo Magidin Feb 14 '12 at 6:14
Thanks @ArturoMagidin. I didn't know how to do the other type of equations. – Kyra Feb 14 '12 at 6:40
up vote 7 down vote accepted

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. :) – Kyra Feb 14 '12 at 6:49

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