Determining the period of sin⁡2xcos⁡3x I would like to compute the period of this function which is a fraction of two trigonometric functions. sin⁡2xcos⁡3x Is there a theorem for this? what trick to use to easily find the period? I started by reducing the fraction but Im

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Determining the period of sin⁡2xcos⁡3x  I would like to compute the period of this function which is a fraction of two trigonometric functions.  sin⁡2xcos⁡3x  Is there a theorem for this? what trick to use to easily find the period? I started by reducing the fraction but Im

David Clayton · Accepted Answer

Suppose the period is p, and suppose the domain of the function is suitably defined, then for all values of x in the domain, we must have  sin⁡2xcos⁡3x=sin⁡(2(x+p))cos⁡(3(x+p))  ⇒sin⁡2xcos⁡(3x+3p)=cos⁡3xsin⁡(2x+2p)   If we set x=0, we have   sin⁡(2p)=0⇒2p−kπ, k∈Z  On the other hand, noting that sin⁡2x=cos⁡3x when x=π10, amongst other possible values, if we set x=π10, we get  cos⁡(3π10+3p)=sin⁡(π5+2p)=cos⁡(3π10−2p)   ⇒3π10+3p=±(3π10−2p)+n⋅2π, n∈Z  From this we get p=n⋅2π or p=3π5+n⋅2π  In order to satisfy this and the previous result for p we have to choose k and n such that both equations p=kπ2 and p=n⋅2π are satisfied and p has minimum value, so we choose k=4 and n=1.  Therefore the period is 2π

MoxboasteBots5h · Accepted Answer

If F(x) has a period T Then F(ax+b) has a period |(T/a)| If the expression is separated by + or - signs take the LCM of periods One of important concepts for determining period of function involving multiple periodic terms is that individual function should repeat simultaneously. This gives rises to LCM rule. It is defined for terms which appear as sum or difference in the function. However, LCM concept can be extended to division or multiplication of periodic terms as well.   Exceptions to LCM rule are important. Even function and function comprising of cofunctions are two notable exceptions to LCM rule. Besides, LCM of irrational periods of different kinds is not possible. This fact is used to determine periodic nature of function involving irrational periods. If LCM can not be determined, then given function is not periodic in the first place.

nick1337 · Accepted Answer

sin⁡2xcos⁡3x=2sin⁡xcos⁡x4cos3⁡x−3cos⁡x=2sin⁡x4cos2⁡x−3 sin⁡x and cos⁡x has a period of 2π. Therefore 2sin⁡(x)4cos2⁡(x)−3 has a period of 2π The fundamental period must be 2πn where n∈N. We have 2sin⁡x4cos2⁡x−3=2sin⁡(x+2πn)4cos2⁡(x+2πn)−3 Substituting x=(n−2)πn we get 2sin⁡((n−2)πn)4cos2⁡((n−2)πn)−3=2sin⁡(π)4cos2⁡(π)−3=0 ⇒sin⁡((n−2)πn)=0 This is true only if n=1 or 2. If n=2 then 2sin⁡x4cos2⁡x−3=2sin⁡(x+π)4cos2⁡(x+π)−3=−2sin⁡x4cos2⁡x−3 Therefore n≠2. So the fundamental period is 2π

Determining the period of \frac{\sin 2x}{\cos 3x}

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