Question

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

Answer 1

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]}