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

Jul 10, 2018

Remember that $\sec t = \frac{1}{\cos} t$
Then use your knowledge of reciprocal functions to sketch!

#### Explanation:

To sketch ${y}_{1} = 2 \sec t$, the only related graph you need think about is the cosine graph.

The identity relating secant and cosine is as follows:
$\sec t = \frac{1}{\cos} t \forall \left\{t : \cos t \ne 0\right\}$
Hence:
$2 \sec t = \frac{1}{\frac{1}{2} \cos t}$

Which is just the reciprocal of a regular cosine graph dilated by a factor of $\frac{1}{2}$ from the $t$-axis (i.e. an amplitude of $\frac{1}{2}$).

Things to remember for sketching reciprocal functions:

• $f \left(t\right) = \frac{1}{f} \left(t\right)$ whenever $f \left(t\right) = 1$
This will never occur with the secant graph! because ${y}_{2} = \frac{1}{2} \cos t$ never equals one.

• Whenever $f ' \left(t\right) = 0$, $\frac{d}{\mathrm{dt}} \left(\frac{1}{f} \left(t\right)\right)$ will also be zero.
This means that the maximum and minimum values of both graphs will occur at the same $t$-values

• $r a n {y}_{2} = \left[- \frac{1}{2} , \frac{1}{2}\right]$, so the range of the reciprocal will be given by $r a n {y}_{1} = \left(- \infty , - 2\right] \cup \left[2 , \infty\right)$.

• Maximum and minimum points will occur at$t \in \left\{- 2 \pi , - \pi , 0 , \pi , 2 \pi \ldots\right\}$
(Check this with the derivative:
$\frac{d}{\mathrm{dt}} \left(2 \sec t\right) = 2 \sec t \tan t = \frac{2 \sin t}{\cos} ^ 2 t$
$\frac{d}{\mathrm{dt}} \left(\frac{1}{2} \cos t\right) = - \sin \frac{t}{2}$
Of course these can only be zero when $\sin t = 0$.

• 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.