5
$\begingroup$

I want to find the period of the function $f(x) = \sin 2x + \cos 3x$. I tried to rewrite it using the double angle formula and addition formula for cosine. However, I did not obtain an easy function. Another idea I had was to calculate the zeros and find the difference between the zeros. But that is only applicable if the function oscillates around $y = 0$, right? My third approach was to calculate the extrema using calculus and from that derive the period. Does anybody have another approach? Thanks

$\endgroup$
1
  • $\begingroup$ Hint: You know the period of $\sin 2x$ and of $\cos 3x$. You should compare them already and see if something can pop up. It should give you an idea of at least one period of the sum. $\endgroup$
    – Martigan
    May 21, 2015 at 13:30

3 Answers 3

12
$\begingroup$

Hint

The period of $\sin(2x)$ is $\pi$, and the period of $\cos(3x)$ is $2\pi/3$.

Can you find a point where both will be at the start of a new period?

$\endgroup$
2
  • $\begingroup$ So the answer is $lcm(\pi,2\pi/3)=2\pi$? $\endgroup$
    – user30523
    May 21, 2015 at 13:45
  • 1
    $\begingroup$ Yes, that will be it $\endgroup$ May 21, 2015 at 17:43
4
$\begingroup$

In general, if $T$ is the period of a function $f(x)$ then the period of the function $f(ax)$ is $\frac{T}{a}$

In general, if two periodic functions $f_1(x)$ & $f_2(x)$ have periods $T_1$ & $T_2$ then the period of the function $g(x)=f_1(x)\pm f_2(x)$ is L.C.M. (Least common multiple) of $T_1$ & $T_2$.

As per your question, the periods of $\sin2x$ & $\cos3x$ are calculated as $\frac{2\pi}{2}=\pi$ & $\frac{2\pi}{3}$ respectively.

Hence the period of the function $f(x)=\sin2x+\cos3x$ is L.C.M. of $\pi$ & $\frac{2\pi}{3}$ which is given by generalized formula of L.C.M. of fractions $$=\frac{\text{L.C.M. (least common multiple) of numerators}}{\text{H.C.F. (highest common factor) of denominators}}=\frac{2\pi}{1}=2\pi$$ Hence, the required period is $2\pi$

$\endgroup$
3
$\begingroup$

$\sin(2x)=\sin(2(x+k\pi))$ and $\cos(3x)=\cos(3(x+l2\pi/3))$, then when $k=\dfrac{2l}3$, $3k=2l=6m$, the function repeats itself, and the period is at most $2\pi$.

Anyway, remains to prove that there is no shorter period.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .