Chapter 4.2, Problem 55E

a.

To determine

Identify the symmetry of the points $\left(x_1,y_1\right)and\left(x_2,y_2\right)$ .

Answer to Problem 55E

  $\boxed{y-axis}$

Explanation of Solution

Given information:

Let $\left(x_1,y_1\right)and\left(x_2,y_2\right)$ be points on the unit circle corresponding to $t=t_{1}andt=\pi -t_1$ , respectively.

Identify the symmetry of the points $\left(x_1,y_1\right)and\left(x_2,y_2\right)$ .

Calculation:

Consider the given point to find the symmetry of the points, $\left(x_1,y_1\right)and\left(x_2,y_2\right)$ , we know that each real number $t$ corresponds to a point $\left(x,y\right)$ on the unit circle.

If the point $t=t_1$ lies in first quadrant then point $t=\pi -t_1$ lies in second quadrant and $t=t_1$ lies in third quadrant $t=\pi -t_1$ lies in forth quadrant.

So that point $t=t_1$ and $t=\pi -t_1$ are symmetric to $y-axis$ .

Hence points $\left(x_1,y_1\right)and\left(x_2,y_2\right)$ corresponds to $t=t_1$ and $t=\pi -t_1$ also has symmetry to $\boxed{y-axis}$ .

b.

To determine

Make a conjecture about any relationship between $\sin t_1$ and $\sin \left(\pi -t_1\right)$ .

Answer to Problem 55E

  $\boxed{\sin t_1=\sin \left(\pi -t_1\right)}$

Explanation of Solution

Given information:

Let $\left(x_1,y_1\right)and\left(x_2,y_2\right)$ be points on the unit circle corresponding to $t=t_{1}andt=\pi -t_1$ , respectively.

Make a conjecture about any relationship between $\sin t_1$ and $\sin \left(\pi -t_1\right)$ .

Calculation:

the points, $\left(x_1,y_1\right)and\left(x_2,y_2\right)$ , $t$ corresponds to a point $t=t_1$ and $t=\pi -t_1$ are symmetric to $y-axis$ .

The value of $y$ for $t=t_1$ and $t=\pi -t_1$ is same. we know that each real number $t$ corresponds to a point $\left(x,y\right)$ on the unit circle and the coordinate $x$ and $y$ are two functions of real variable $t$ .

These coordinates to be used to identify the six trigonometric functions of $t$ .

  $\sin t=y$ , $\cos t=x$ and $\tan t=\frac{y}{x},x\ne 0$

  $\csc t=\frac{1}{y},y\ne 0$ , $sect=\frac{1}{x},x\ne 0$ , $\cot t=\frac{x}{y},y\ne 0$

So, $\sin t=y$ and the value of $y$ for $t=0$ , $t=t_1$ and $t=\pi -t_1$ is same, thus $\sin t_1=\sin \left(\pi -t_1\right)=y$

Hence, $\boxed{\sin t_1=\sin \left(\pi -t_1\right)}$ .

c.

To determine

Make a conjecture about any relationship between $\cos t_1$ and $\cos \left(\pi -t_1\right)$ .

Answer to Problem 55E

  $\boxed{\cos \left(\pi -t_1\right)=-\cos t}$

Explanation of Solution

Given information:

Let $\left(x_1,y_1\right)and\left(x_2,y_2\right)$ be points on the unit circle corresponding to $t=t_{1}andt=\pi -t_1$ , respectively.

Make a conjecture about any relationship between $\cos t_1$ and $\cos \left(\pi -t_1\right)$ .

Calculation:

the points, $\left(x_1,y_1\right)and\left(x_2,y_2\right)$ , $t$ corresponds to a point $t=t_1$ and $t=\pi -t_1$ are symmetric to $y-axis$ .

The value of $x$ for $t=t_1$ and $t=\pi -t_1$ is differ by negative sign. We know that each real number $t$ corresponds to a point $\left(x,y\right)$ on the unit circle and the coordinate $x$ and $y$ are two functions of real variable $t$ .

These coordinates to be used to identify the six trigonometric functions of $t$ .

  $\sin t=y$ , $\cos t=x$ and $\tan t=\frac{y}{x},x\ne 0$

  $\csc t=\frac{1}{y},y\ne 0$ , $sect=\frac{1}{x},x\ne 0$ , $\cot t=\frac{x}{y},y\ne 0$

So, $\cos t=x$ and the value of $y$ for $t=0$ , $t=t_1$ and $t=\pi -t_1$ is different by negative sign. Thus $\boxed{\cos \left(\pi -t_1\right)=-\cos t}$

Hence, $\boxed{\cos \left(\pi -t_1\right)=-\cos t}$ .