SOLUTION: Line Quadrant y = −x II sin θ = cos θ = tan θ = csc θ = sec θ = cot θ =

Question 1165687: Line Quadrant
y = −x
II
sin θ =

cos θ =

tan θ =

csc θ =

sec θ =

cot θ =

Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
The line $y = -x$ passes through Quadrant II. To find the trigonometric values for an angle $\theta$ whose terminal side lies on this line in Quadrant II, we can choose any point on that segment of the line.
### 1. Identify a Point $(x, y)$
The line is $y = -x$. For the line to be in **Quadrant II**, the $x$-coordinate must be **negative** and the $y$-coordinate must be **positive**.
* Let $x = -1$.
* Then $y = -(-1) = 1$.
* The point on the terminal side of the angle is $\mathbf{(-1, 1)}$.
### 2. Calculate the Radius ($r$)
The distance from the origin $(0, 0)$ to the point $(x, y)$ is $r$, calculated using the distance formula:
$$r = \sqrt{x^2 + y^2}$$
$$r = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$$
### 3. Determine the Trigonometric Ratios
Using the point $(-1, 1)$ and $r=\sqrt{2}$:
$$
\sin \theta = \frac{y}{r} \quad \cos \theta = \frac{x}{r} \quad \tan \theta = \frac{y}{x}
$$
| Trigonometric Function | Ratio | Value | Rationalized Value |
| :---: | :---: | :---: | :---: |
| **$\sin \theta$** | $y/r$ | $1/\sqrt{2}$ | $\mathbf{\sqrt{2}/2}$ |
| **$\cos \theta$** | $x/r$ | $-1/\sqrt{2}$ | $\mathbf{-\sqrt{2}/2}$ |
| **$\tan \theta$** | $y/x$ | $1/(-1)$ | $\mathbf{-1}$ |
| **$\csc \theta$** | $r/y$ | $\sqrt{2}/1$ | $\mathbf{\sqrt{2}}$ |
| **$\sec \theta$** | $r/x$ | $\sqrt{2}/(-1)$ | $\mathbf{-\sqrt{2}}$ |
| **$\cot \theta$** | $x/y$ | $-1/1$ | $\mathbf{-1}$ |
***
*Note: The angle $\theta$ represented here is $135^\circ$ or $3\pi/4$ radians.*
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