Chapter 12.3, Problem 6WE

To determine

Give the values of six trigonometric functions for $\theta =90°$

Answer to Problem 6WE

  $\sin \theta =1$

  $\cos \theta =0$

  $\tan \theta =undefined$

  $\csc \theta =1$

  $\sec \theta =undefined$

  $\cot \theta =0$

Explanation of Solution

Formula Used:

  $\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}=\frac{y}{r}$

  $\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}=\frac{x}{r}$

  $\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }=\frac{y}{x}$

  $\csc \theta =\frac{1}{\sin \theta }=\frac{r}{y}$

  $\sec \theta =\frac{1}{\cos \theta }=\frac{r}{x}$

  $\cot \theta =\frac{1}{\tan \theta }=\frac{x}{y}$

Calculation:

Draw the angle $\theta =90°$ , quadrantal angle on a unit circle . That is , circle with radius 1.

So, the terminal side passes through the point $P(0,1)$

So,

x = length of the side adjacent to the angle $\theta$ = 0

y = length of the side opposite to the angle $\theta$ = 1

r = length of the hypotenuse =the distance between (0,0) and (0,1)= $\sqrt{x^2+y^2}=\sqrt{1+0}=1$

  $\sin \theta =\frac{y}{r}=\frac{1}{1}=1$

  $\cos \theta =\frac{x}{r}=\frac{0}{1}=0$

  $\tan \theta =\frac{y}{x}=\frac{1}{0}=undefined$

  $\csc \theta =\frac{r}{y}=\frac{1}{1}=1$

  $\sec \theta =\frac{r}{x}=\frac{1}{0}=undefined$

  $\cot \theta =\frac{x}{y}=\frac{0}{1}=0$