Question

The diagram shows how cos θ, sin θ, and tan θ relate to the unit circle. Copy the diagram and show how sec θ, csc θ, and cot θ relate to the unit circle.
a. First, find in the diagram a segment whose length is sec θ. Explain why its length is sec θ.
b. Next, find cot θ. To do this you must add to the diagram. Use the representation of tangent as a clue for what to show for cotangent. Justify your claim for cot θ.
c. find csc θ in your diagram.
BIG idea Function
You can derive some functions from a basic parent function by a particular transformation. Functions related through these transformations

Answer 1

Copy the diagram and show how sec θ, csc θ, and cot θ relate to the unit circle. 

The representation of the diagram is shown if Figure 1. There's a relationship between sec θ, csc θ, and cot θ related the unit circle. Lines green, blue and pink show the relationship. 

a.1 First, find in the diagram a segment whose length is sec θ. 

The segment whose length is sec θ is shown in Figure 2, this length is the segment $\overline{OF}$, that is, the line in green.

a.2 Explain why its length is sec θ.

We know these relationships:

(1) $sin \theta=\frac{\overline{BD}}{\overline{OB}}=\frac{\overline{BD}}{r}=\frac{\overline{BD}}{1}=\overline{BD}$

(2) $cos \theta=\frac{\overline{OD}}{\overline{OB}}=\frac{\overline{OD}}{r}=\frac{\overline{OD}}{1}=\overline{OD}$

(3) $tan \theta=\frac{\overline{FD}}{\overline{OC}}=\frac{\overline{FC}}{r}=\frac{\overline{FC}}{1}=\overline{FC}$

Triangles ΔOFC and ΔOBD are similar, so it is true that:

$\frac{\overline{FC}}{\overline{OF}}= \frac{\overline{BD}}{\overline{OB}}$

∴ $\overline{OF}= \frac{\overline{FC}}{\overline{BD}}= \frac{tan \theta}{sin \theta}= \frac{1}{cos \theta} \rightarrow \boxed{sec \theta= \frac{1}{cos \theta}}$

b.1 Next, find cot θ

The segment whose length is cot θ is shown in Figure 3, this length is the segment $\overline{AR}$, that is, the line in pink.

b.2 Use the representation of tangent as a clue for what to show for cotangent. 

It's true that:

$\frac{\overline{OS}}{\overline{OC}}= \frac{\overline{SR}}{\overline{FC}}$

But:

$\overline{SR}=\overline{OA}$
$\overline{OS}=\overline{AR}$

Then:

$\overline{AR}= \frac{1}{\overline{FC}}= \frac{1}{tan\theta} \rightarrow \boxed{cot \theta= \frac{1}{tan \theta}}$

b.3  Justify your claim for cot θ.

As shown in Figure 3, θ is an internal angle and ∠A = 90°, therefore ΔOAR is a right angle, so it is true that:

$cot \theta= \frac{\overline{AR}}{\overline{OA}}=\frac{\overline{AR}}{r}=\frac{\overline{AR}}{1} \rightarrow \boxed{cot \theta=\overline{AR}}$

c. find csc θ in your diagram.

The segment whose length is csc θ is shown in Figure 4, this length is the segment $\overline{OR}$, that is, the line in green.