How do you find the x intercepts of #y=sinpix+cospix#?
1 Answer
Feb 20, 2018
Explanation:
Note that:
#y = sin pi x + cos pi x#
#color(white)(y) = sqrt(2)(sqrt(2)/2 sin pi x + sqrt(2)/2 cos pi x)#
#color(white)(y) = sqrt(2)(sin (pi/4) sin pi x + cos (pi/4) cos pi x)#
#color(white)(y) = sqrt(2) sin (pi/4 + pi x)#
Also note that:
#sin theta = 0" "# if and only if#theta = k pi# for some integer#k#
So we require:
#pi/4 + pi x = k pi#
Dividing both sides by
#1/4 + x = k#
Then subtracting
#x = k - 1/4" "# for any integer#k#