Question

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

Answer 1

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.