Chapter 5.2, Problem 32E

To determine

Find $\sin t,\cos t$ and $\tan t$ .

Answer to Problem 32E

  $\begin{array}{l}\sin t=\frac{y}{r}=-\frac{2\sqrt{2}}{3}\\ \cos t=\frac{x}{r}=-\frac{1}{3}\\ \tan t=\frac{y}{x}=2\sqrt{2}\end{array}$

Explanation of Solution

Given information:

  $\left(-\frac{1}{3},-\frac{2\sqrt{2}}{3}\right)$

Formula used:

Pythagoras theorem:

  $\boxed{H^2=P^2+B^2}$

  $\boxed{\sin t=\frac{P}{H},\cos t=\frac{B}{H},\tan t=\frac{P}{B}}$

Where, P=perpendicular, B=base, H=hypothenuse

The points $\left(-\frac{1}{3},-\frac{2\sqrt{2}}{3}\right)$ forms a circle which passes through the origin

  

  $r^2=\sqrt{x^2+y^2}$

Here we can see that x is negative and y is also negative, so it is forming circle in third quadrant therefore tan will be positive; Sin andcos will be negative.

Now in coordinates $\left(-\frac{1}{3},-\frac{2\sqrt{2}}{3}\right)$ , $x=-\frac{1}{3}$ , and $y=-\frac{2\sqrt{2}}{3}$

Calculating r:

  $\begin{array}{l}{r}^2=\sqrt{x^2+y^2}\\ r^2=\sqrt{{\left(-\frac{1}{3}\right)}^2+{\left(\frac{-2\sqrt{2}}{3}\right)}^2}\\ r^2=\sqrt{\frac{1}{9}+\frac{8}{9}}\\ r^2=1\\ r=1\end{array}$

Now,

  $\begin{array}{l}\sin t=\frac{y}{r}=-\frac{2\sqrt{2}}{3}\\ \cos t=\frac{x}{r}=-\frac{1}{3}\\ \tan t=\frac{y}{x}=2\sqrt{2}\end{array}$