Linear independence of sin (x), cos (x) and 1, proof

1. Apr 23, 2012

Luka

What would be the best way to show that functions $f(x)=1$, $g(x)=sin(x)$ and $h(x)=cos(x)$ are linearly independent elements of the vector space $\mathbb{R}^{\mathbb{R}}$?

I know that the linear independence means that an expression like $\alpha \mathbb{x}_1 + \beta \mathbb{x}_2 + \gamma \mathbb{x}_3 = \mathbb{0}$ is true only for $\alpha = \beta = \gamma = 0$ where $x_1,...,x_3$ are vectors and $\alpha$, $\beta$ and $\gamma$ are scalars of the vector space.

I think that the proof might look like this:

$\alpha sin(x)+ \beta cos(x)+ \gamma 1=0$

If $x=0$ then $sin(x)=0$. Therefore, $\beta=0$ and $\gamma=0$, but $\alpha$ might be different than zero, and the above-mentioned expression still equal to zero.

2. Apr 23, 2012

micromass

Staff Emeritus
Your attempt is a good one. So assume that there are $\alpha,\beta,\gamma$ such that

$$\alpha f + \beta g+\gamma h=0$$

That means that for ALL x must hold that

$$\alpha+\beta\sin(x)+\gamma \cos(x)=0$$

This holds for all x, so try to pick some good values for x.

You already tried x=0, this gives us that necessarily

$$\alpha+\gamma=0$$

(and not $\alpha=0,\gamma=0$ as you claimed).

Now try some other values for x. For example pi or pi/2 ??

PS excuse me for using other $\alpha,\beta,\gamma$ as in your post.

3. Apr 23, 2012

Luka

For $x=\pi$, we get $\gamma - \beta = 0$ which means that $\alpha$ can be of any value, and the expression still equal to zero. Then those elements ($f(x)$, $g(x)$ and $h(x)$) would not be linearly independent according to the definition of linear independence. I think that we need all three scalars to be zero to prove the linear independence: $\alpha =0$, $\beta =0$ and $\gamma = 0$. In other words, $sin(x)\neq 0$ and $cos(x)\neq 0$.

For $x=\frac{\pi}{3}$, we get $\frac{\sqrt{3}}{2}\alpha +\frac{1}{2}\beta + \gamma = 0$, which means that $\alpha$, $\beta$ and $\gamma$ must be equal to zero for the expression to be true.

4. Apr 23, 2012

micromass

Staff Emeritus
Why should this imply that $\alpha,\beta,\gamma$ are all zero?? It doesn't.

5. Apr 23, 2012

Luka

It does if we want to prove the linear independence (because of the definition itself). I'm worried about the fact that not all $x$ satisfy the conditions $sin(x)\neq 0$, $cos(x)\neq 0$ that allow us to prove it.

6. Apr 23, 2012

HallsofIvy

Staff Emeritus
Because you want them all equal to 0, you simply declare that
$$\frac{\sqrt{3}}{2}\alpha+ \frac{1}{2}\beta+ \gamma= 0$$?
Looks like you are assuming what you want to prove.

What about $\alpha= 0$, $\beta= 2$, $\gamma= -1$?

To prove that 1, sin(x), and cos(x) are independent, you want to prove that the only way you can have $\alpha (1)+ \beta(sin(x))+ \gamma(cos(x))= 0$ for all x is to have $\alpha= \beta= \gamma= 0$. But that is what we want to prove- we cannot assume it.

Since that is true for all x, it is, in particular, true for x= 0, we must have
$\alpha+ \gamma= 0$
And, for $x= \pi/2$, we must have
$\alpha+ \beta= 0$

Finally, for $x= \pi$, we must have
$\alpha- \gamma= 0$

Solve those three equations for $\alpha$, $\beta$, and $\gamma$.