How do you graph #y=2sect# by first sketching the related sine and cosine graphs?

1 Answer
Jul 10, 2018

Remember that #sec t = 1/cost#
Then use your knowledge of reciprocal functions to sketch!

Explanation:

To sketch #y_1 = 2sect#, the only related graph you need think about is the cosine graph.

The identity relating secant and cosine is as follows:
#sec t = 1/cost AA {t : cost != 0}#
Hence:
# 2 sec t = 1/(1/2cost)#

Which is just the reciprocal of a regular cosine graph dilated by a factor of #1/2# from the #t#-axis (i.e. an amplitude of #1/2#).

Things to remember for sketching reciprocal functions:

  • #f(t) = 1/f(t)# whenever #f(t) = 1#
    This will never occur with the secant graph! because #y_2 = 1/2cost# never equals one.

  • Whenever #f'(t) = 0#, #d/dt(1/f(t))# will also be zero.
    This means that the maximum and minimum values of both graphs will occur at the same #t#-values

  • #ran y_2 = [-1/2,1/2]#, so the range of the reciprocal will be given by #ran y_1 = (-oo,-2] uu [2, oo)#.

  • Maximum and minimum points will occur at#t in {-2pi, -pi, 0, pi, 2pi ...}#
    (Check this with the derivative:
    #d/dt(2sect) = 2 sect tant = (2sint)/cos^2t#
    #d/dt(1/2cost) = -sint/2#
    Of course these can only be zero when #sint = 0#.

  • Whenever #y_1# is decreasing, #y_2# is increasing, and vice-versa.

I'm not sure what else to say... I hope this helps! Use a graphing calculator to do some of these things if you need.