Sine curve that passes through (1,1) (2,2) and (3,3) Question: Find a sine function in the form $a\sin(bx) + c$ that passes through points (1,1) (2,2) and (3,3)
Working so far:


*

*We have three points for three unknown variables in the function, so we can use simultaneous equations to solve for them.

*Simultaneous equations:

*

*$a\sin(b) + c = 1$

*$a\sin(2b) + c = 2$

*$a\sin(3b) + c = 3$



*Equation 1, solving for c:

*

*$c = 1 - a\sin(b)$



*Equation 2: solving for a:

*

*$a\sin(2b) + 1 - a\sin(b) = 2$

*$a\sin(2b) - a\sin(b) = 1$

*$a(\sin(2b) - \sin(b)) = 1$

*$a = \frac{1}{\sin(2b) - \sin(b)}$



*Equation 3: solving for b:

*

*$(\frac{1}{\sin(2b) - \sin(b)})(\sin(3b)) + 1 - (\frac{1}{\sin(2b) - \sin(b)})(\sin(b)) = 3$

*$\frac{\sin(3b)}{\sin(2b) - \sin(b)} - \frac{\sin(b)}{\sin(2b) - \sin(b)} = 2$

*$\frac{\sin(3b) - \sin(b)}{\sin(2b) - \sin(b)} = 2$

*$\sin(3b) - \sin(b) = 2\sin(2b) - 2\sin(b)$

*$\sin(3b) + \sin(b) = 2\sin(2b)$

*$3\sin(b) - 4\sin^{3}(b) + \sin(b) = 4\sin(b)\cos(b)$

*$4\sin(b) - 4\sin^{3}(b) = ±4\sin(b)\sqrt{1 - \sin^{2}(b)}$

*Substitute $u = \sin(b)$

*$4u - 4u^3 = ±4u\sqrt{1-u^2}$

*$u - u^3 = ±u\sqrt{1-u^2}$

*$1 - u^2 = ±\sqrt{1-u^2}$

*$1 - 2u^2 + u^4 = 1 - u^2$

*$u^4 - u^2 = 0$, $u = 0$

*$u^2 - 1 = 0$

*$u^2 = 1$

*$u = 1$, $u = -1$

*For $u = -1$:

*

*$\sin(b) = -1$

*$b = \arcsin(-1) = (\frac{4n - 1}{2})\pi$ and $(\frac{4n + 3}{2})\pi$, n is an integer.



*For $u = 0$:

*

*$b = \arcsin(0) = n\pi$, n is an integer.



*For $u = 1$:

*

*$b = \arcsin(1) = (\frac{4n + 1}{2})\pi$, n is an integer.





*$b = n\pi$ and $b = n\pi - \frac{\pi}{2}$

*Therefore $a = \frac{1}{\sin(2(n\pi)) - \sin((n\pi))}$

*$a = \frac{1}{0 - 0} = \frac{1}{0}$, let's see if the other solution works.

*$a = \frac{1}{\sin(2(n\pi - \frac{\pi}{2})) - \sin(n\pi - \frac{\pi}{2})}$

*$a = \frac{1}{1} = 1$ when n is an even integer and $a = \frac{1}{-1} = -1$ when n is an odd integer.

*$c = 1 - (1)(\sin(n\pi)) = 1$, $c = 1 - (-1)(\sin(n\pi)) = 1$, $c = 1 - (1)(\sin(n\pi - \frac{\pi}{2})) = 2$ when n is an even integer and $c = 1 - (1)(\sin(n\pi - \frac{\pi}{2})) = 0$ when n is an odd integer, $c = 1 - (-1)(\sin(n\pi - \frac{\pi}{2})) = 0$ when n is an even integer and finally $c = 1 - (-1)(\sin(n\pi - \frac{\pi}{2})) = 2$ when n is an odd integer

*With these results, let's try n = 0:

*

*$a = 1$, $b = 0$ and $b = -\frac{\pi}{2}$, $c = 1$, $c = 2$, $c = 0$

*$y = \sin(-\frac{\pi}{2}x) + 1$ touches none of the three points.




A way to get the curve to approximately touch the 3 points would be to equate $a$ to a very large number, and equate $b$ to the reciprocal of $a$. And have $c$ equal to 0, in this case the graph mimics the straight-line function $f(x) = x$ in the desired range. But this doesn't account for the possibility that the graph could pass through several 'waves' between $x = 0$ and $x = 3$ to pass through all three points in those waves.
What could the values of a, b and c be for the sine curve to pass through all three points?

The solutions are
$$ f(x) = -\sin\left(\left(\frac{\pi}{2} + 2\pi k\right)x\right) + 2$$
and
$$ f(x) = \sin\left(\left(\frac{3\pi}{2} + 2\pi k\right)x\right) + 2$$

We can add the first and third equations to get
$$a\ \sin(b) + a\ \sin(3b)+2c=4 \\ a\ \sin(2b-b)+a\ \sin(2b+B)+2c=4 \\ 2a\ \sin(2b)\cos(b) + 2c = 4 \\a\ \sin(2b)\cos(b) = 2 - c$$
But the second equation tells us that
$$a\ \sin(2b) = 2 - c \\ a\ \sin(2b) = a\ \sin(2b)\cos(b)\\ a\ \sin(2b)(1-\cos(b)) = 0$$
It is simple to verify that $a \ne 0$ and $\cos(b) \ne 1$ by plugging the values back into the original equations.
Thus $\sin(2b) = 0$ which implies that $b = 0, \frac{\pi}{2}, \pi,$ or $\frac{3\pi}{2}\ (+\ 2\pi k)$.
For the same reasons as before, $b = 0$ and $\pi\ (+\ 2\pi k)$ do not work.
Luckily, $b = \frac{\pi}{2}$ and $\frac{3\pi}{2}\ (+\ 2\pi k)$ do work and plugging the values back into the original equation yields the other variables.
