How do you sketch one cycle of #y=-cos(1/2x)#?

1 Answer
Jul 31, 2018

Answer:

See details and graph

Explanation:

#y = - cos ( 1/2 x ) in [ -1, 1 ]#

Amplitude: #abs ( - 1 ) = 1#

Period: #(2pi)/(1/2)= 4pi#

Zeros ( x-intercepts ): #-1/2x = ( 2k +1 ) pi/2,#

#k = 0, +-1, +-2, +-3,...#, giving

#x = ( 2 k - 1 ) (2 pi )# = an odd multiple of # 2 pi #.

Graph for one cycle, #x in [ -2 pi, 2 pi ]#:
graph{ (y + cos ( x/2 ) )( x - 2pi+0.0001y)(x+2pi+0.0001y)(y^2-1)((x-pi)^2+y^2-0.01)((x+pi)^2+y^2-0.01)=0[-6.3 6.3 -3.15 3.15]}