How do you find the period of $y = 4 sin (\frac{1}{2}) (x + \frac{π}{4})$ $y = 4sin (1/2)(x + pi/4)$ ?

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How do you find the period of $y = 4 sin (\frac{1}{2}) (x + \frac{π}{4})$ $y = 4sin (1/2)(x + pi/4)$ ?

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Answer 1 · 2016-11-02T00:07:15

By getting your equation in the form $A sin (B x + C)$ $Asin(Bx+C)$ and solving $\frac{2 π}{| B |}$ $(2pi)/|B|$

Explanation:

First of all, you want to get your equation in the form of $(B x + C)$ $(Bx+C)$ . it's not necessary in this case, but just for the sake of learning and avoiding mistakes, I'm including it anyway. You'll want to multiply $(\frac{1}{2})$ $(1/2)$ and $(x + \frac{π}{4})$ $(x+pi/4)$ to get the equation in the form $A sin (B x + C)$ $Asin(Bx+C)$ as $4 sin (\frac{1}{2} x + \frac{π}{8})$ $4sin(1/2 x+ pi/8)$ . Now, we are certain that B equals $\frac{1}{2}$ $1/2$ . Next, we'll plug in to to $\frac{2 π}{| B |}$ $(2pi)/|B|$ $\to$ $->$ $\frac{2 π}{\frac{1}{2}}$ $(2pi)/(1/2)$ = $4 π$ $4pi$
The period equals $4 π$ $4pi$

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How do you find the period of $y = 4 sin (\frac{1}{2}) (x + \frac{π}{4})$ $y = 4sin (1/2)(x + pi/4)$ ?

1 Answer

Explanation:

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How do you find the period of $y = 4 sin (\frac{1}{2}) (x + \frac{π}{4})$ $y = 4sin (1/2)(x + pi/4)$ ?