Is the function $A\cos(\lambda x) + B\sin(\lambda x)$ periodical?

Question

I am trying to check the function for periodicity...

$y(x) = A\cos(\lambda x) + B\sin(\lambda x) $

I think that it's possible to rewrite the function as $y(x) = Csin (\lambda x + t)$, where $C = \sqrt{A^2 + B^2}$

But I can't prove that new function is periodical or is not periodical.

How can I do it?

Answer 1

Hint:

The period of the sine and cosine functions are well known to be $2\pi$ for both. Hence $\dfrac{2\pi}\lambda$ is a period of the linear combination, for the argument $\lambda x$.

Remains to show that it is the smallest.

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Setting $t:=\lambda x$, let $T=\lambda X$ be the period.

$$A\cos(t+T)+B\sin(t+T)=A\cos(t)+B\sin(t)$$

implies, using the sum-to-product formula,

$$-2A\sin\left(t+\frac T2\right)\sin\left(\frac T2\right)+2A\cos\left(t+\frac T2\right)\sin\left(\frac T2\right)=0.$$

This expression is identically zero for the smallest nonzero value $T=2\pi$.

Answer 2

$$f(x)=A\cos(\lambda x) + B\sin(\lambda x)$$ Suppose we have $$f(x+p)=f(x)$$ for all $x$, so we have also for $x=0$:

$$A\cos(\lambda p) + B\sin(\lambda p)= A\cos(0) + B\sin(0) =A$$ So if we take $\lambda p = 2\pi $ we get $p={2\pi\over \lambda}$ which is period since it is not difficult to see that $f(x+p)-f(x)=...=0$.