Chapter 12.3, Problem 15OE

To determine

Give the values of six trigonometric functions of $\theta$ . If any value is undefined , say so.

Answer to Problem 15OE

  $\sin \theta =\frac{\sqrt{2}}{2}$

  $\cos \theta =\frac{\sqrt{2}}{2}$

  $\tan \theta =1$

  $\csc \theta =\sqrt{2}$

  $\sec \theta =\sqrt{2}$

  $\cot \theta =1$

Explanation of Solution

Given:

  

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:

Given angle $\theta$ is in Quadrant I , and the point on the terminal side of the standard angle is $P(1,1)$

So,

x = length of the side adjacent to the angle θ= 1

y = length of the side opposite to the angle θ= 1

r = length of the hypotenuse = $\sqrt{x^2+y^2}=\sqrt{1+1}=\sqrt{2}$

  $\sin \theta =\frac{y}{r}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

  $\cos \theta =\frac{x}{r}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

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

  $\csc \theta =\frac{r}{y}=\frac{\sqrt{2}}{1}=\sqrt{2}$

  $\sec \theta =\frac{r}{x}=\frac{\sqrt{2}}{1}=\sqrt{2}$

  $\cot \theta =\frac{x}{y}=1$