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!

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