Chapter 4.5, Problem 41E

Solution

a.

To prove: Why the coordinates of $p_1$ on the circle and the terminal side of angle $t-\pi$ are $\left(-a,-b\right)$ .

Given information:

The coordinates of point,

  $\begin{array}{c}{p}_1=\left(-a,-b\right)\\ p_2=\left(a,b\right)\end{array}$

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

Proof:

The line $\overline{p_1{p}_2}$ passes through the center of the circle.

The two points are in the quadrant with the diagonal line at point $\overline{P_1}$ is $\left(-a,-b\right)$ .

The straight line has the angle $\pi$ .

Therefore, the terminal side of the point $\overline{P_1}$ will have the angle $t-\pi$ .

b.

To prove: Why the value of $t=\frac{b}{a}$ .

Given information:

The coordinates of point,

  $\begin{array}{c}{p}_1=\left(-a,-b\right)\\ p_2=\left(a,b\right)\end{array}$

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

Proof:

Graph:

  

Considering the right angle triangle,

By using tangent function, it is known that

  $\begin{array}{c}\tan \theta =\frac{\text{Opposite}}{\text{Adjacent}}\\ \tan \left(t-\pi \right)=\frac{-b}{-a}\\ \tan \left(-\left(\pi -t\right)\right)=\frac{b}{a}\\ -\tan \left(\pi -t\right)=\frac{b}{a}\\ \tan t=\frac{b}{a}\end{array}$

Therefore, it is shown that $\tan t=\frac{b}{a}$ .

c.

To find: The value of $\tan \left(t-\pi \right)$ and show that $\tan t=\tan \left(t-\pi \right)$ .

The value of $\tan \left(t-\pi \right)=\frac{b}{a}$ .

Given information:

The coordinates of point,

  $\begin{array}{c}{p}_1=\left(-a,-b\right)\\ p_2=\left(a,b\right)\end{array}$

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

Calculation:

Graph:

  

Considering the right angle triangle,

By using tangent function, it is known that

  $\begin{array}{c}\tan \theta =\frac{\text{Opposite}}{\text{Adjacent}}\\ \tan \left(t-\pi \right)=\frac{-b}{-a}\\ \tan \left(t-\pi \right)=\frac{b}{a}\end{array}$

Now to prove,

  $\begin{array}{c}\tan \left(t-\pi \right)=\tan \left(-\left(\pi -t\right)\right)\\ =-\tan \left(\pi -t\right)\\ =-\left(-\tan t\right)\\ =\tan t\end{array}$

Therefore, $\tan \left(t-\pi \right)=\tan t$ has been proved.

Conclusion:

The value of $\tan \left(t-\pi \right)=\frac{b}{a}$ .

d.

To prove: Why the period of the tangent function is $\pi$ .

Given information:

The coordinates of point,

  $\begin{array}{c}{p}_1=\left(-a,-b\right)\\ p_2=\left(a,b\right)\end{array}$

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

Proof:

For all integers t in the domain,

  $\left(t\pm \pi \right)=\tan t$

The points on opposing sides of the unit circle determine the same tangent ratio.

Triangles with varying tangent ratios are produced by other points on the unit circle.

Therefore, the tangent function repeats for every $\pi$ unit.

e.

To prove: Why the period of the cotangent function is $\pi$ .

Given information:

The coordinates of point,

  $\begin{array}{c}{p}_1=\left(-a,-b\right)\\ p_2=\left(a,b\right)\end{array}$

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

Proof:

For all integers t in the domain,

  $\left(t\pm \pi \right)=\tan t$

The tan and cot function repeat for every $\pi$ units.

Every $\pi$ unit is a repetition of the tangent function. Therefore, the reciprocal of the tangent function is the cotangent.