The full question: A point on a wheel has an equation y = 10 sin (x - 45°) + 20 that models the height as the wheel rotates. Answer the following questions.

1) What is the height if the wheel has rotated 135°?

2) What are the possible values of rotation (i.e., the value of x) if the height is 15m?

3) If the hub of the wheel were moved down 5m, what would the values of rotation (the values of x) be if the height were 15m?

I've answered 1 just fine on my own, and this is what I have for 2 so far:

y=10sin(x-45°)+20
15-20=10sin(x-45°)+20
-5/10=10sin(x-45°)/10
-0.5=sin(x-45°)

I'm having trouble figuring out how to remove the 45° from the equation. I looked at the examples in the lesson, but it doesn't include an example involving the phase shift, and I'm just confused. I did a bit of Googling out of frustration and found the answers are supposed to be 210° and 330°, but I don't know how to get to those answers. If anyone could point me in the right direction, I'd be super grateful.

y=10sin(x-45°)+20 y is the height, so

15-20=10sin(x-45)
sin(x-45)= -.5
sinTheta= -.5
theta= 330deg, or 210
but x-45=210, or 330 so
x= 255 or 375 (15 deg)

To find the values of rotation (x) when the height is 15m, you need to solve the equation -0.5 = sin(x - 45°).

Here is how you can proceed:
1) Rewrite the equation as sin(x - 45°) = -0.5.
2) Recall that the sine function has a range between -1 and 1. The angle whose sine is -0.5 can be found in the second and fourth quadrants of the unit circle.
3) In the second quadrant, sin(x) is positive, so you have sin(x) = 0.5. Find the reference angle whose sine is 0.5; it is 30°.
4) To find the angle (x) in the second quadrant, subtract the reference angle from 180°: x - 45° = 180° - 30° = 150°.
5) In the fourth quadrant, sin(x) is negative, so you have sin(x) = -0.5. Again, find the reference angle whose sine is 0.5; it is 30°.
6) To find the angle (x) in the fourth quadrant, add the reference angle to 360°: x - 45° = 360° + 30° = 390°.

Therefore, the possible values of rotation (x) when the height is 15m are 150° and 390°.

For question 3, if the hub of the wheel is moved down 5m, the equation for the height becomes y = 10sin(x - 45°) + 15, since the constant term "20" becomes "15". To find the values of rotation (x) when the height is 15m, you need to solve the equation -0.5 = sin(x - 45°) + 15 in the same steps as explained above.

Hope this helps! Let me know if you have any further questions.

To find the values of rotation (x) when the height is 15m (question 2), you correctly started by setting up the equation:

-0.5 = sin(x - 45°)

To remove the phase shift of 45°, you need to isolate the sine function. Here's how you can do it:

1) Start by getting rid of the negative sign. Multiply both sides of the equation by -1:

0.5 = -sin(x - 45°)

2) Since sine is an odd function, you can apply the identity sin(-θ) = -sin(θ) to simplify the equation further:

0.5 = sin(45° - x)

3) Now, to find the values of x, you need to take the inverse sine (sin⁻¹) of both sides of the equation. Be aware that the inverse sine function only gives values in the range of -90° to 90° (or -π/2 to π/2 in radians), so you might get some values outside of this range.

sin⁻¹(0.5) = sin⁻¹(sin(45° - x))

4) Evaluate the inverse sine of 0.5 using a calculator, which will give you the principal angle:

30° = 45° - x

5) Solve for x by isolating it in the equation:

45° - 30° = x

15° = x

So, one possible value of rotation (x) when the height is 15m is 15°.

However, there's another possible value because the sine function is periodic with a period of 360° (or 2π radians). To find the second possible value, you can add the period to the previous value:

15° + 360° = 375°

Therefore, the possible values of rotation (x) for a height of 15m are 15° and 375°.

Please note that the values 210° and 330° you found online may be mentioned in a different reference frame or equation, so it's essential to follow the steps outlined above based on the given equation.