So I have the terms 1 , cos(x), and sin(x). Am I right in saying they are linear dependent as sin squared x add cos squared x = 1?
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Linear dependence would mean that we could find constants $A,B\in \mathbb R$ (not both $0$) such that $Acos(x)+Bsin(x)=0;\forall x$ – lulu Jun 23 '16 at 13:21
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That relation $\sin^2 x + \cos^2 x = 1$ is not a linear dependence relation because it involves squaring the terms on the left hand side. – hardmath Jun 25 '17 at 20:31
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No, because you can not multiply vectors. Yes, there are definitions of how to multiply certain vectors, such as polynomial multiplication, matrix multiplication, or inner product, but when it comes to linear independence, you are not allow to square or multiply vectors. You are only allowed to multiply them by scalars and add them together. Ask yourself, is there any real numbers $a,b,c$ such that: $$a\cdot 1+b\cdot \cos x+c\cdot \sin x=0$$ Can you make that equation $0$ for all $x \in \Bbb{R}$ for some $a,b,c \in \Bbb{R}$? That is the question here. No squaring allowed, only multiplying by scalars and adding.
Noble Mushtak
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If you know that in the space of continuous functions over $[0,2\pi]$ the position $$ \langle f,g\rangle=\int_0^{2\pi}f(x)g(x)\,dx $$ defines an inner product, then you can show that the three given functions form an orthogonal set.
By the way, this inner product is the main tool for starting Fourier analysis.
Now \begin{align} &\int_0^{2\pi}1\cdot\sin x\,dx=\Bigl[-\cos x\Bigr]_0^{2\pi}=0 \\ &\int_0^{2\pi}1\cdot\cos x\,dx=\Bigl[\sin x\Bigr]_0^{2\pi}=0 \\ &\int_0^{2\pi}\sin x\cos x\,dx= \frac{1}{2}\int_0^{2\pi}\sin2x\,dx= \frac{1}{2}\Bigl[-\frac{1}{2}\cos 2x\Bigr]_0^{2\pi}=0 \end{align}
An orthogonal set consisting of nonzero vectors is linearly independent. So you see your conjecture is false.
egreg
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