Is it true that $a\cos \alpha \theta = b \cos \beta \theta \implies a=b$ and $\alpha = \beta$

Question

Is it true that $\forall \theta:a\cos \alpha \theta = b \cos \beta \theta \implies a=b$ and $\alpha = \beta$?

If so, how do I prove it? I know it isn't true for the sine case since we could have $a=-b$ and $\alpha = - \beta$

Answer 1

Fill in $\theta=0$ to get $a=b$.

The $\alpha$ and $\beta$ part isn't quite true, since $\alpha=-\beta$ will give the same functions.

Answer 2

Plug in $\theta=0$, which gives $a=b$. If $a=b=0$ all values of $\alpha$ and $\beta$ are good.

If $a=b\ne0$, then we have $\cos(\alpha\theta)=\cos(\beta\theta)$ and, taking the derivative with respect to $\theta$, we get $$ -\alpha\sin(\alpha\theta)=-\beta\sin(\beta\theta). $$ Another derivative gives $$ -\alpha^2\cos(\alpha\theta)=-\beta^2\cos(\beta\theta). $$ Again, for $\theta=0$, we obtain $\alpha^2=\beta^2$, so $\alpha=\beta$ or $\alpha=-\beta$.

Both cases are good, because $\cos(-x)=\cos x$.

So from your hypothesis you can conclude that one of these conditions holds:

1. $a=b=0$;
2. $a=b\ne0$ and $\alpha=\beta$;
3. $a=b\ne0$ and $\alpha=-\beta$.

Answer 3

As @uncountable notes, $a = b$ in all circumstances, but $\alpha = -\beta$ satisfies $a\cos(\alpha \theta) = b\cos(\beta \theta) \,\forall \theta$, as does $a = b = 0$ and arbitrary $\alpha, \beta$. To show that this is the only way one can have $\alpha \neq \beta$, note that equality for all $\theta$ means the derivatives and second derivatives (etc.) are equal. In particular,

$$ \alpha^2 \cos(\alpha \theta) = \beta^2 \cos(\beta \theta) $$

and plugging in $\theta = 0$ shows $\alpha^2 = \beta^2$.

One can establish similar results for sine: if $\alpha = \beta = 0$, then arbitrary $a,b$ satisfy $a\sin(\alpha\theta) = b\sin(\beta\theta)$, the same for $a = b = 0$ and arbitrary $\alpha,\beta$, and likewise if $\alpha = -\beta$ and $a = -b$.

If $a = \beta = 0$, then $0 = b\cos(0)$, and we have covered this case. Likewise for $b = \alpha = 0$. Finally, it is clear that we can't have only one of $a,b,\alpha,\beta = 0$ (any value of $\theta$ other than $0$ will demonstrate this).

Taking a derivative, you have

$$ a\alpha\cos(\alpha\theta) = b\beta\cos(\beta\theta) $$

and taking $\theta = 0$ shows $a\alpha = b\beta$. We have already dealt with the case where some of them are zero, so assume they aren't.

Taking two more derivatives, and $\theta = 0$ shows

$$(a\alpha)\alpha^2 = a\alpha^3 = b\beta^3 = (b\beta)\beta^2$$

so $\alpha^2 = \beta^2$, i.e. $\alpha = \pm \beta$. This, along with $a\alpha = b\beta$ shows $a = \pm b$.