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

Mar 3, 2017

#### Answer:

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 $f \left(x\right)$
f(x)=sin(0-45°)
f(x)=sin(-45°)

On the function $\sin x$, 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 $\sin x$ is the vertical height, the sign must be the opposite, thus sin-45°=-sqrt(2)/2
Therefore intersection with y-axis is at point $\left(0 , - \frac{\sqrt{2}}{2}\right)$

2. Intersection with x-axis
Let $y = 0$ in $f \left(x\right)$
0=sin(x-45°)

On the function $\sin x$, when does $\sin x = 0$?

Using the unit circle, $\sin x = 0$ 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!