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The period of $\sin(x)$ is $2\pi$. Thus the period of $\sin(\pi x)$ will become $T_1=2$, similarly the period of $\sin(2\pi x)$ is $T_2=1$ and for $\sin(5\pi x)$, the period is $T_3=\frac{2}{5}$.

To find the period of $\sin(\pi x)-\sin(2\pi x)+\sin(5\pi x)$, we can have $$\frac{T_1}{T_{2}}=\frac{2}{1}\Rightarrow \,\, T^*=T_1=2T_2=2$$ Now, We can also have $$\frac{T^*}{T_{3}}=\frac{2}{\frac{2}{5}}\Rightarrow \,\, T=2T^*=10T_3=4$$

So the period is 2. Is this correct?

zhk
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2 Answers2

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You have $$ (T_1,T_2,T_3)=\left(\tfrac{10}5,\tfrac55,\tfrac25\right) $$ where $\operatorname{LCM}(10,5,2)=10$ (least common multiple). Thus the minimal common integer multiple where those fractions coincide must be $$ \frac{10}5=2\cdot \frac 55=5\cdot\frac25=2 $$ So your suggestion is correct.

String
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  • The material from where I came up with the trigonometric functions says that the period is 10. – zhk Nov 14 '16 at 12:51
  • @MMM: That must be an error. – String Nov 14 '16 at 12:54
  • Is this has to do with the interval $(0,5)$ of study? Please check page E-2 - 22 of this document. http://www.math.psu.edu/tseng/class/Math251/Notes-PDE%20pt2.pdf – zhk Nov 14 '16 at 12:57
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I think you are right, since $2$ divided by $\frac{1}{3}$ is an integer, as well as the $1$. You can have a check by plotting

The image of the function

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AsianDuckKing
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