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I need some help with how to work out the period of this sine graph equation.

$$h(x)= 50-40\sin\left(\frac{4\pi x}{3}\right)$$

I have a question that asks me to graph it, and I have worked out the amplitude to be $50$.

I understand that if it is $3x$ on its own, the period will be $2\pi/3$, and that if it is $x/3$ for example, the period will be $6\pi$.

What I am unsure of is how to work it out when multiple steps are involved as well as when $\pi$ is used.

Could someone please explain in simple steps how to solve this. Thankyou :)

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  • $\begingroup$ i have worked out the amplitude to be 50 That doesn't sound right, but you should mention what definition of amplitude you are working with, anyway. I need some help with how to work out the period Find the minimum $\,T\,$ such that $\,h(x)=h(x+T)\,$ holds for $\,\forall x\,$. $\endgroup$ – dxiv May 30 '17 at 4:06
  • $\begingroup$ The number $50$ represents the vertical shift, not the period. $\endgroup$ – N. F. Taussig May 30 '17 at 9:22
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig May 30 '17 at 9:27
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The trigonometric function $$f(x) = A\sin(Bx - C) + D$$ has amplitude $|A|$, period $2\pi/|B|$, phase shift $C/B$, and vertical shift $D$.

We can express the function $$h(x) = 50 - 40\sin\left(\frac{4\pi x}{3}\right)$$ in the form $$h(x) = -40\sin\left(\frac{4\pi}{3}x - 0\right) + 50$$ so $A = -40$, $B = \dfrac{4\pi}{3}$, $C = 0$, and $D = 50$. Using the stated formulas yields an amplitude of $$|A| = |-40| = 40$$ a period of $$\frac{2\pi}{|B|} = \frac{2\pi}{\frac{4\pi}{3}} = 2\pi \cdot \frac{3}{4\pi} = \frac{3}{2}$$ a phase shift of $$\frac{C}{B} = \frac{0}{\frac{4\pi}{3}} = 0$$ and a vertical shift of $$D = 50$$

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After one period $T$ the argument of the sine function has gone from 0 to $2\pi$

Solve $\frac{4\pi T}{3}=2\pi$

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As you mentioned, we know that the period when we have $\sin(x)$ is $2\pi$. When multiplying by a constant inside the $\sin$, you could see this as "accelerating" the function. And so, if we multiply by a constant $c>1$, our function $\sin(cx)$ will go faster, and thus the period will be smaller (and similarly, if $0<c<1$ we are slowing down the function, and the period of $\sin(cx)$ will be larger).

In this case, we are acceleration the function by a constant $c=\frac{4\pi}{3}$. The period will then be $\frac{2\pi}{c}=\frac{3}{2}$.

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  • $\begingroup$ How to you go from c=4π/ 3 to 2π /c ? $\endgroup$ – A.Mahony May 30 '17 at 4:17

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