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

1 Answer
zgurney
Mar 3, 2017

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!

Related questions
See all questions in Translating Sine and Cosine Functions
Impact of this question
3312 views around the world
You can reuse this answer
Creative Commons License