Question

A point on the terminal side of an angle $\theta$in standard position is given. Find the exact value of each of the six trigonometric functions of$\theta$:$(-1,1)$

Short answer

The exact values of the six trigonometric functions are,

$$\begin{gathered} \sin \theta =\frac{1}{\sqrt{2}} \\ \cos \theta =\frac{-1}{\sqrt{2}} \\ \tan \theta =-1 \\ \csc \theta =\sqrt{2} \\ \sec \theta =-\sqrt{2} \\ \cot \theta =-1 \end{gathered}$$

Step-by-step solution

Step 1. Given information  

We are given a point $(-1,1)$.

We need to find the exact value of each of the six trigonometric functions.

Step 2. Concept   

Let $(x,y)$be a point on the terminal side of the angle $\theta$in the standard position, which is also on the circle $x^2+y^2=r^2$. Then,

$$\begin{gathered} \sin \theta =\frac{y}{r} \\ \cos \theta =\frac{x}{r} \\ \tan \theta =\frac{y}{x} \\ \csc \theta =\frac{r}{y} \\ \sec \theta =\frac{r}{x} \\ \cot \theta =\frac{x}{y} \end{gathered}$$

Step 3. Finding the value of r

The given point is $(-1,1)$. Here, $x=-1,y=1$.

To find $r$, we have

$$\begin{gathered} r=\sqrt{x^2+y^2} \\ r=\sqrt{{(-1)}^2+1^2} \\ r=\sqrt{2} \end{gathered}$$

Step 4. Finding the values of the six trigonometric functions

We are given a point, $(-1,1)$where $x=-1,y=1$. Also, we know that $r=\sqrt{2}$.

Therefore, the values of the six trigonometric functions are

role="math" localid="1646619679554" $$\begin{gathered} \sin \theta =\frac{y}{r}=\frac{1}{\sqrt{2}} \\ \cos \theta =\frac{x}{r}=\frac{-1}{\sqrt{2}} \\ \tan \theta =\frac{y}{x}=\frac{1}{-1}=-1 \\ \csc \theta =\frac{r}{y}=\frac{\sqrt{2}}{1}=\sqrt{2} \\ \sec \theta =\frac{r}{x}=\frac{\sqrt{2}}{-1}=-\sqrt{2} \\ \cot \theta =\frac{x}{y}=\frac{-1}{1}=-1 \end{gathered}$$

Step 5. Final answer

The values of the six trigonometric functions are,

$$\begin{gathered} \sin \theta =\frac{1}{\sqrt{2}} \\ \cos \theta =\frac{-1}{\sqrt{2}} \\ \tan \theta =-1 \\ \csc \theta =\sqrt{2} \\ \sec \theta =-\sqrt{2} \\ \cot \theta =-1 \end{gathered}$$