Question

For `0 <= x <= 360^@`, how would you find the coordinates of the points of intersection with the coordinate axes of `y = sin(x - 45^@)`?

Answer 1

Find the intersection points of the function `f(x)=sin(x-45°)` with the lines y=0 (x-axis) and x=0 (y-axis)

Explanation:

1. Intersection with y-axis
Let `x=0` in `f(x)`
`f(x)=sin(0-45°)`
`f(x)=sin(-45°)`

On the function `sinx`, where does `x=-45°` or `315°`?

Using the unit circle, `sin45°=sqrt(2)/2` so since we are going in the opposite direction -45° instead of +45° and `sinx` is the vertical height, the sign must be the opposite, thus `sin-45°=-sqrt(2)/2`
Therefore intersection with y-axis is at point `(0,-sqrt(2)/2)`

2. Intersection with x-axis
Let `y=0` in `f(x)`
`0=sin(x-45°)`

On the function `sinx`, when does `sinx=0`?

Using the unit circle, `sinx=0` when `x` is 0°, 180°, or 360°

Therefore `x-45°=0, x-45°=180, x-45°=360`
`x=45°, x=225°, x=405°`
However, 405° is not inside the domain so x cannot be greater than 360° so this number is excluded. All other numbers lie in between 0 and 360 so they're fine.

Therefore intersection with x-axis is at points `(45°, 0)` and `(225°,0)`

Hope this helped!