Chapter 4.5, Problem 64E

Solution

a.

To explain: Why Cosine means sine of the complement.

The sine of the complement is the cosine.

Given information:

The unit circle $x^2+y^2=1$ .

Calculation:

If the sum of the angles is ${90}^{\circ }$ , then the two angle are said to be complementary angles.

Taking the cosine of an angle $\theta$ and sine of its complementary angle $\left({90}^{\circ }-\theta \right)$ to get,

  $\cos \theta =\sin \left({90}^{\circ }-\theta \right)$

Therefore, from the above equation it is shown that the sine of the complement is the cosine.

b.

To find: The coordinates of $A$ as a function of $t$ .

The coordinates of $A$ is $\left(\sin t,\cos t\right)$ .

Given information:

The unit circle $x^2+y^2=1$ .

Calculation:

  

Let’s take the point $A$ as $(x,y)$ ,

Consider the right angle triangle ADO,

  $\begin{array}{c}\sin \theta =\frac{\text{Opposite}}{\text{Hypotenuse}}\\ \sin t=\frac{x}{1}\\ x=\sin t\end{array}$

Using the cosine function,

  $\begin{array}{c}\cos \theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}\\ =\frac{y}{1}\\ y=\cos t\end{array}$

Therefore, the coordinates of $A$ is $\left(\sin t,\cos t\right)$ .

c.

To find: The length BC as a function of $t$ .

The length with respect to the function of t is $BC=\tan t$ .

Given information:

The unit circle $x^2+y^2=1$ .

Calculation:

  

Consider the triangle ODA and OCD,

  $\frac{AD}{DO}=\frac{BC}{CO}$

From part (b) it is known that,

  $\begin{array}{c}AD=\sin t\\ DO=\cos t\\ CO=1\end{array}$

Substituting the values,

  $\begin{array}{c}\frac{\sin t}{\cos t}=\frac{BC}{1}\\ BC=\tan t\end{array}$

Therefore, the length with respect to the function of t is $BC=\tan t$ .

d.

To find: The length OB as a function of $t$ .

The length with respect to the function of t is $OB=\sec t$ .

Given information:

The unit circle $x^2+y^2=1$ .

Calculation:

  

Consider the triangle ODA and OCD,

  $\frac{OA}{OB}=\frac{OD}{OC}$

From part (b) it is known that,

  $\begin{array}{c}OA=1\\ OD=\cos t\\ OC=1\end{array}$

Substituting the values,

  $\begin{array}{c}\frac{1}{OB}=\frac{\cos t}{1}\\ OB=\sec t\end{array}$

Therefore, the length with respect to the function of t is $OB=\sec t$ .

e.

To explain: Where the names tangent, cotangent, secant and cosecant came from.

Given information:

The unit circle $x^2+y^2=1$ .

Explanation:

A tangent is named after a Latin term that means "to touch." The tangent line touches the unit circle, hence the name.

Since a cotangent is the tangent of a complementary angle, its name is derived from this fact,

  $\tan \left(90-x\right)=\cot x$

The word "secant" is derived from the Latin word "secare," which meaning "to cut," because a secant must intersect at least two separate locations.