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My first attempt is under this, i can work out the primitive period of both of the $\cos(4x)$ and $\sin(2x)$ but how do I calculate the primitive period of $\cos(4x)+\sin(2x)$?

My attempt:

Let $u=4x$ then $x=\frac{u}{4}$ and $\cos(u)$ is $2\pi$ periodic thus $T=\frac{2\pi}{4}$ hence $\cos(4x)$ is periodic with primitive period $T=\frac{\pi}{4}$.

Now Let $H=2x$ and thus $x=\frac{H}{2}$ and $\sin(H)$ is also $2\pi$ periodic Thus $\sin(2x)$ is periodic with primitive period $T=\frac{2\pi}{2}=\pi$ but i dont know how to combine these results to calculate the primitive period of the sum of both $\cos(4x)$ and $\sin(2x)$

3 Answers 3

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Your function is $$f(x)=\cos4x+\sin2x=1-2\sin^22x+\sin2x$$

Since it is a function of $\sin2x$ which has period $\pi$, the period of $f$ is also $\pi$

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    +1 Clever; I'd still add something to show that it isn't $\pi/2$. – 2017-01-29
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    @BrianTung. Are you sure that would be necessary? If we accept that $\sin2x$ has a period of $\pi$ then so does $f$. – 2017-01-29
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    This proves clearly that $\pi$ is a period for $f$, but not that it is the lowest positive one ... @David Quinn – 2017-01-29
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If the period of $f(x)$ is $2T$, and the period of $g(x)$ is $T$, then their sum, $h(x) = f(x)+g(x)$, is periodic with period no shorter than $2T$, as you have essentially shown: For any $x$,

$$ h(x+2T) = f(x+2T)+g(x+2T) = f(x)+g(x) = h(x) $$

It then remains to show that $h(x)$ is not periodic with period equal to $T$. Find two values, separated by $T$, for which $h(x)$ is not equal: that is, find $x_0$ such that $h(x_0) \not= h(x_0+T)$.

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    Why couldn't the period be $3T/2$, for instance? – 2017-01-29
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    @YoTengoUnLCD: You're quite right; in order to be rigorous, the OP would have to exclude all periods less than $2T$. In the case of $3T/2$, that could be excluded by showing that any function that is periodic with period $3T/2$ *and* $2T$ would also have to be periodic with period $T/2$, and therefore also $T$. – 2017-01-29
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Your attempt seems fine. Choose $T = \pi$, because that is a multiple of $T_1 = \pi /4$ and $T_2 = \pi$. Then, \begin{align} f(x + T) & = \cos (4(x+T)) + \sin (2(x+T)) \\ & = \cos (4x + 4T) + \sin (2x + 2T) \\ & = \cos(4x+4\pi) + \sin (2x + 2 \pi) \\ & = \cos (4x) + \sin (2x) \\ & = f(x). \end{align}

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    i got $\frac{\pi}{2}$ is this correct? – 2017-01-29