Why cannot $A\sin\alpha x +B\cos \alpha x$ be zero?

Question

I was going through solving wave equations using fourier and I came across a note saying $A\sin\alpha x +B\cos \alpha x \neq 0$

I believe this applies to $\alpha ,A,B\neq 0$

I was solving $$ \frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}$$

which was reduced to $$u(x,t)=(A\sin\alpha x +B\cos \alpha x).(C\sin\alpha t +D\cos \alpha t)$$

The boundary condition given as $u(x,0)=0 \implies 0=D(A\sin\alpha x +B\cos \alpha x)$

And here it said $D=0 \because A\sin\alpha x +B\cos \alpha x \neq 0$

How is this possible?

The next boundary considition states $$u(x,\pi)=0 \implies u(x,t)= \sum_{n=1}^\infty C\sin(nx)(A\sin nx +B\cos nx)$$

How did the $\Sigma$ come here?

All the boundary conditions are as follows:-

$$u(x,0)=0 \\ u(x,\pi )=0 \\ u(0,t)=sin t \\ u_x(0,t)=t^2$$

Answer 1

That is because $\alpha$,$A$ and $B$ are constants, and when since $x$ can be arbitrary, the expression $Acos(\alpha x)+B sin(\alpha x)$ can not be zero for all $x$.

Answer 2

Here's a quick proof I made up:

Say $A \cos \alpha x + B \sin \alpha x = 0$ with $A, B, \alpha \neq 0$. Then $\cos \alpha x = -(B/A) \sin \alpha x$. But $\cos \alpha x$ attains its maximum at $0$ while $-(B/A) \sin \alpha x$ attains its maximum at $\pi/(2\alpha)$, and not at $0$. This is a contradiction, so $A \cos \alpha x + B \sin \alpha x \neq 0$.

Answer 3

Let $f(x)= A\sin(\alpha x)+B\cos(\alpha x)$. We will assume that $f=0$ and arrive at a contradiction.

If $f=0$ for all $x$, then so is its derivative $f'$. Therefore, we have both

$$A\sin(\alpha x)+B\cos(\alpha x)=0 \tag 1$$

and

$$\alpha (A\cos (\alpha x)-B\sin(\alpha x))=0 \tag 2$$

for all $x$.

If $\alpha \ne 0$, then $(1)$ and $(2)$ imply $A^2+B^2=0$. Since both $A$ and $B$ are not zero, then we have a contradiction.

Therefore, $f\ne 0$ and we are done!

Answer 4

The functions $ f(x) = \cos \alpha x $ and $ g(x) = \sin \alpha x$ are linearly independent if and only if their Wronskian is nonzero everywhere. But the Wronskian is $\alpha$ in this particular case.