# Thread: trig function that passes through (x,y)

1. ## [Solved] trig function that passes through (x,y)

hey guys! I need some help with this:

Find the function of the form y = sin x + c that passes through the given
points:

1. (225°, 3)
2. (30°, -2)

i'm lost. all i know is 225° is 5pi/4 and 30° is pi/6 in radians

2. 1. Function that passes through ( $\displaystyle \frac{5\pi}{4}$,3)

Answer: $\displaystyle y=3\sin(x-\frac{3\pi}{4})$

Explaination:

If you look at the normal sin graph:

You'll notice it reaches it peak at ($\displaystyle \frac{\pi}{2}$,1)

The amplitude is basically the height = 1/2(max-min) so here its 1/2(1-(-1)) = 1.

You want it to reach to your y value 3, so if we look at a function, e.g. y = 3sin(x) the amplitude is always the absolute value of the # in front of the sin. That means it max is going to be 3, while its min is going to be -3.

y=3sin(x):

My x-scale is set to $\displaystyle \frac{2\pi}{4}$.$\displaystyle \frac{5\pi}{4}$is half-way between$\displaystyle \frac{\pi}{2}$ (where it crosses the x-axis) and $\displaystyle \frac{3\pi}{2}.$We need to SHIFT this line to the right $\displaystyle \frac{3\pi}{4}$ to get it to where we need.

So you answer will be $\displaystyle y=3\sin(x-\frac{3\pi}{4})$

2. You will need to put a (-) in front of sin to flip the graph. y=-2sin(x+$\displaystyle \frac{2\pi}{6}$)

Here is the graph, divided into 6ths.

3. thanks, but i don't think that's right. the equation needs to be y = sin x + c, so the c would be a vertical shift of the graph y = sin x

if you do y = sin(x) on your graphing calc, and you set the X-Scale to 5pi/4, you'll see a sin(x) graph, and the 5pi/4 point. The graph needs to pass through the point (5pi/4, 3). the mark on the x-axis will show you where 5pi/4 is, but the y = sin(x) graph needs to have a vertical shift (upwards) to pass through that point.

EDIT: okay, i solved the first one. the equation must pass through (5pi/4, 3) but the regular y=sin(x) graph at x = 5pi/4 is (5pi/4, -1/sqrt[2]). So if you add a vertical shift of 3 units upwards + 1/sqrt[2], the graph will pass through the point, (5pi/4, 3).

4. You could also use y= 3sin(x+13pi/4) as it gives you the same result.

Graph broken down into 4ths..

5. Same with the 2nd one..instead of doing the -sin you could do 2sin(x+8pi/6) and it will pass through the right point.

6. no no no lol you don't understand. the equation must be in the form of y = sin x + c. that means the coefficint of sin is 1, it cannot be anything else.

but i actually figured out the second one, logically.

- for the graph y = sin(x): when x = pi/6, y = 0.5
- this means there is a displacement of 0.5
- the equation needs to pass through (pi/6, -2)
- therefore, you need to do a vertical shift of (2 + displacement) * (-1 since its a downward shift)

so sin(x) - 2.5 will pass through the point (pi/6, -2)

it works because if you do sin(pi/6) you get 0.5

I'm going to thank you anyway because you took the trouble of actually graphing it on a ticalc emulator and posted it up. thanks for your help!!!

7. crazyness..haha at least you got it!