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?

  • $\begingroup$ Study $y(x+T)$. $\endgroup$
    – user65203
    Commented Nov 8, 2017 at 11:51
  • $\begingroup$ Hint: $\cos x$ and $\sin x$ have the period $2\pi$. $\endgroup$ Commented Nov 8, 2017 at 11:54
  • $\begingroup$ Yes, but you're running down the wrong road. Instead you should try to rewrite $C\sin(\lambda x+t)$ using trigonometric identities and make it match $A\cos(\lambda x)+B\sin(\lambda x)$. It just being periodic is not enough to show that the functions are the same (but if you show them being the same it's obvious that it would be periodic). $\endgroup$
    – skyking
    Commented Nov 8, 2017 at 11:56
  • $\begingroup$ Thank you, but I have difficulties anyway... I have sum of two periodical functions $A \cos (\lambda x)$ and $B \sin(\lambda x)$ . How can I use it for my solve? $\endgroup$
    – Nikolai
    Commented Nov 8, 2017 at 11:59
  • $\begingroup$ @skyking: your comment is misleading/ambiguous. You can show periodicity in both representations. $\endgroup$
    – user65203
    Commented Nov 8, 2017 at 12:04

2 Answers 2



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.

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


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 $T=2\pi$.


$$f(x)=A\cos(\lambda x) + B\sin(\lambda x)$$ 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 $f(x+p)-f(x)=...=0$.

  • $\begingroup$ Thank you very much! Is it possible to prove that $\frac{2\pi}{\lambda}$ is least period? How did you guess to get $p = \frac{2\pi}{\lambda} $ as period? $\endgroup$
    – Nikolai
    Commented Nov 8, 2017 at 12:27
  • $\begingroup$ 1. Yes it is, since $2\pi$ is least period for $\sin x$ and $\cos x$. 2. We know that $2\pi$ is period for these two function so I put $\lambda p = 2\pi$. $\endgroup$
    – nonuser
    Commented Nov 8, 2017 at 12:31
  • $\begingroup$ @JohnWatson: this argument is insufficient. Think of the functions $\cos 10x-\cos x$ and $\cos 10x+\cos x$. Both have period $2\pi$ but their sum has period $\pi/5$. $\endgroup$
    – user65203
    Commented Nov 8, 2017 at 14:44

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