Sine model of height as a function of distance

Question

I actually assumed the sine model to be: $$y=A\sin(B(x-C))+D$$

We have: $$A=\frac{64-0}{2}=32$$

Now from the data we see that: $$23=32\sin(40B-BC)+D$$ and $$23=32\sin(160B-BC)+D$$ $\implies$ $$160B-BC=2\pi+40B-BC$$ Which gives $B=\frac{\pi}{60}$. But the book answer is $B=\frac{\pi}{100}$. Am i missing something?

Answer 1

If $\sin (\alpha) = \sin (\beta)$, we can't conclude that $\alpha = 2\pi + \beta$.

The general solution is $\alpha = (2n+1)\pi - \beta $ or $\alpha = 2n \pi + \beta$

Now, the minimum is attained at $0$ and the maximum is attained at $100$. We conclude that $C$ (horizontal shifting) is $50$ and $D=A=32$.

Now, let $x=100$, $64=32\sin(B(x-50))+32$,

$$\sin(50B)=1$$

We let $50B=\frac{\pi}{2}$, hence $B=\frac{\pi}{100}$.

Remark: Let's evaluate $40B-BC=\frac{\pi}{100}\left( 40-50\right)=-\frac{\pi}{10}$. Also $160B-BC=\frac{\pi}{100}\left( 160-50\right)=\frac{11\pi}{10}$, when we add them together, we get an odd multiple of $\pi$.