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?
 A: A little geometric intuition is helpful here. Your desired points are odd-symmetric around $(2,2)$, so you can imagine shifting your function up 2 steps and then scaling it horizontally to have period $4$ (instead of $2\pi$).

It would be straightforward then to shift the origin to $(2,2)$ and scale $x$ by $\pi/2$, but the resulting form of the expression is $c + \sin(ax+b)$, which is not quite what you asked for. There are tricks for transforming it in the way you need, but once you've drawn the graph another much easier idea is clear anyway; shift up $2$ units and scale $x$ by negative $\pi/2$ to exploit the symmetry of $sin$.
$$ f(x) = 2-\sin(\pi x/2) $$
A: 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.
