2015年7月22日 · Period of y = csc (x - pi/2) Ans: 2pi y = 1/sin (x - pi/2) The period of sin x is 2pi, then the period of y is also 2pi.

2015年8月8日 · sec(pi/4) = sqrt(2) Note, however, that sqrt(2) is not in radians. It is a ratio (not in degrees or ...

2018年2月3日 · Here is a graph of #y= sec(pix/4)#, using Desmos: Desmos has the ability to export a png file type and Socratic has the ability to import a png file type. You can, also, you the Socratic.org graphing feature in a Socratic.org scratchpad but desmos has many more features.

2016年11月30日 · They are the same as the ones for y=tan x Note that: cot(x-pi/2) = frac cos (x-pi/2) sin (x-pi/2) = - frac cos( pi/2-x) sin(pi/2-x) = - frac sinx cosx = -tan x The range of tan x is (-oo,+oo) so it is not affected by the change in sign. Same for the domain of tan x that is symmetrical with respect to x=0 Also the asymptotes do not change, only the approach to the asymptotes is reversed. graph ...

2018年2月20日 · sec x/tan x (1/cosx )/(sinx/cosx) 1/cancelcosx xx cancelcosx/sinx 1/sinx = cscx

2016年10月17日 · 4/pi The average value of a function y=f(x) over an interval [a,b] is given by 1/(b-a)int_a^b f(x) dx So for y=sec^2 x we have: Avg = 1/(pi/4-0) int_0^(pi/4) sec^2x dx = 1/(pi/4) int_0^(pi/4) sec^2x dx = 4/pi [tanx]_0^(pi/4) = 4/pi {tan (pi/4) - tan0 } = 4/pi {tan (pi/4) - …

2017年11月20日 · Given: #(sec(x) + csc(x))/(1+ tan(x)) = # Substitute #1/cos(x)# for #sec(x)#, #1/sin(x)# for #csc(x)#, and #sin(x)/cos(x)# for #tan(x)#:

2015年6月19日 · Amplitude= oo Period= (pi^2+pi)/3 The amplitude is infinity. Because the tan function is increasing over its whole domain of definition. graph{tanx [-10, 10, -5, 5]} The period of any tan is the value of x when the "inside" of the tancolor(red)() function equals pi.

2016年9月23日 · #(tan2x)/(2tanx)-sec^2x/(1-tan^2x)# #=(2tanx)/((2tanx)(1-tan^2x))-sec^2x/(1-tan^2x)# #=1/(1-tan^2x)-sec^2x/(1-tan^2x)#

2016年7月14日 · How do you use inverse trigonometric functions to find the solutions of the equation that are in