How do you find the interval of decrease and increase of a sin function?
Given this function
f(x) = sin2x-90degrees
What is the interval of increase and decrease and how do you find the zeros?
Your help is greatly appreciated! Thanks
The function f(x) is incresing when f'(x) is >0.
f'(x) = (sin2x -90)' = 2cos2x
2co2x > 0 implies cos2x > 0
Cos2x > 0 when 2npi-pi/2 < 2x < 2npi-pi.
Cos2x > 0 when -pi/4 < x <pi/4. That is when x is in (npi/2-pi/4 , npi/2+pi/4), n = 0,1,2,..
Therefore sin2x - 90 is increasing in (npi-pi/4 ,npi+ pi/4).
When x is in the interval (180n-45 deg ,180n+45 deg ) for n =0,1,2,....
Sin2x -90 is decreasing when (sin2x-90)' < 0. Or
2cos2x < 0 or cos2x < 0.
cos2x < 0 when 2npi+ pi/2 < 2x < 2npi+3pi/2.
npi+Pi/4 < x < npi+3pi/4 is same as when x is in (180n+45deg to 135 deg.), n =0,1,2,3....
Therefore sin2x-90 is decreasing in (180n+45 deg , 180n+135deg), for n = 0,1,2,3....