# 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 $\gamma$.