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
    Nov 8, 2017 at 11:51
  • $\begingroup$ Hint: $\cos x$ and $\sin x$ have the period $2\pi$. $\endgroup$ 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
    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
    Nov 8, 2017 at 11:59
  • $\begingroup$ @skyking: your comment is misleading/ambiguous. You can show periodicity in both representations. $\endgroup$
    – user65203
    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
    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
    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
    Nov 8, 2017 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.