Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
add comment

1 Answer 1

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?

share|improve this answer
    
Thank you for the help. That helped me solve it. :) –  Kyra Feb 14 '12 at 6:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.