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I hope you can help me out because this problem has been bothering me all day long.

Graph of sine function

The tasks:

A) The graph belongs to the function $f(x)=a\sin(bx+c)$. Determine the parameters a, b and c according to the graph.

B) Determine it's roots $x_1, x_2$ and $x_3$.

Obviously $a=3$ and $b=2$, but what about the phase shift $c$? I looked up the solutions and they state that $c=\frac{5\pi}{3}$ (positive value, so they shifted it to the left). I have no clue how they arrived at that though.

Thank you.

scoopfaze
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QWERTYZ
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  • Hint: Do you know what value $x_1$ has? If you know $x_1$, the first positve root of the function, how can you determine $c$ when you already determined $b$? – Ingix Jul 05 '19 at 22:04
  • Ok, i figured out that x1=pi/6, x2= 2pi/3 and x3=7pi/6. Still thinking about the phase shift though. – QWERTYZ Jul 05 '19 at 22:18

2 Answers2

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We know $x_1 = \frac{\pi}6$. We also can see that $y=\sin (2x)$ (the non-shifted version) has the "same kind" of y-intercept at $x = \pi$. So our function must be shifted $\pi-\frac{\pi}6 = \frac{5\pi}6$ units to the left.

The general equation for a sine function is $$y = a\sin(b(x+c))+d$$

We know $$y = 3\sin(2(x+c))$$ where $c$ is the phase shift, $\frac{5\pi}6$ in this case.

So $$y = 3\sin(2x+\frac{5\pi}3)$$

RayDansh
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The amplitude is obviously $3$. One half a period is $11 \pi/12 - 5 \pi/12 = \pi/2$. It is then simplest to write down the $\cos$ function and then convert to $\sin$.

$$y = 3 \cos[2 (x - 5\pi/12)] = 3 \sin[2 (x - 5 \pi/12) - \pi/2] = 3 \sin [2 x + 5 \pi/3]$$

enter image description here

David G. Stork
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