Finding dimensions of functions Hey guys I need help showing if a function is a vector space or not. I believe we show is addition and multipication holds. but I don't know. Also how do I find out dimensions of such functions.
The question
Show that the function $a_1 + a_2\sin x + a_3\cos x$ form a vector space. If it is what are the dimensions for it like how did you figure it out. Because I think it would be 3 because we are looking at $\{1, \sin x, \cos x\}$. But I don't know if it is right. 
Also what about $b_1 + b_2\sin^2 x + b_3\cos^2 x$
Any help or hints would be greatly appreciated.
Thank you
 A: We are looking at the set $S$ of functions of the shape $a+b\sin x+c\cos x$, where the numbers $a, b, c$ range over the reals.
We need to show that the sum of any two functions in $S$ is also in $S$, and that if $f$ is in $S$, so is $k$, for any real number $k$.  Both these things are straightforward to verify.  
We also in principle need to verify some more basic things, for example that the $0$-function is in $S$, and that if $f$ is in $S$ so is $-f$. Also, we need (in principle) to verify the other vector space properties.
As to the dimension of this vector space, you are right, it is $3$, and the set $\{1,\sin x,\cos x\}$ is a basis. Certainly all the elements of our space are, by definition, linear combinations of $1,\sin x,\cos x$.
We need to show also that our three functions are linearly independent. So we need to check that if $a+b\sin x+c\cos x=0$ (identically) then all of $a,b,c$ are $0$.
If $a+b\sin x+c\cos x$ is identically $0$, then its  derivative $b\cos x-c\sin x$ is identically $0$. Put $x=0$. We conclude $b=0$. Put $x=\pi/2$. We conclude $c=0$. And finally since $a+b\sin x+c\cos x=0$, we get $a=0$. This completes the proof that $1,\sin x,\cos x$ are a linearly independent set.

We turn now to the set of all functions of the form $a+b\sin^2 x+c\cos^2 x$. These form a vector space, basically for the same reason as in our previous calculation. But $1,\sin^2 x,\cos^2 x$ are not linearly independent, since $1-\sin^2 x-\cos^2 x=0$.
You can verify without much trouble that $\{1,\sin^2 x\}$ is a basis of our vector space. Indeed any two of $1,\sin^2 x,\cos^2 x$ form a basis for the space.
