We are tasked with sketching the least positive angle $\theta$ in standard position where the terminal side of the angle lies on the line defined by the equation:

$-\sqrt{2}x + y = 0 \quad \text{with} \quad x \leq 0$

Step 1: Understanding the Line Equation

We start by rearranging the equation to express $y$ in terms of $x$:

$y = \sqrt{2}x$

This line has a slope of $\sqrt{2}$, which means it rises steeply as $x$ decreases. Since the condition $x \leq 0$ restricts us to the second and third quadrants, the terminal side of the angle will lie in the second quadrant (where $x \leq 0$ and $y \geq 0$).

Step 2: Slope of the Line

To find the angle $\theta$, we calculate the angle of the line with respect to the positive $x$-axis. The slope of the line is $m = \sqrt{2}$, and the tangent of the angle between the terminal side and the positive $x$-axis is given by the slope:

$\tan(\theta) = \sqrt{2}$

Thus, $\theta = \tan^{-1}(\sqrt{2})$. Using a calculator:

$\theta = 65.54^\circ$

This angle corresponds to a reference angle of $65.54^\circ$. Since the line is in the second quadrant, the actual angle from the positive $x$-axis is:

$\theta = 180^\circ - 65.54^\circ = 114.46^\circ$

Thus, the least positive angle $\theta$ is approximately $114.46^\circ$.

Step 3: Finding the Six Trigonometric Functions of $\theta$

To find the values of the six trigonometric functions, we need a point on the terminal side of the angle. From the equation $y = \sqrt{2}x$, we can pick a simple point. Let’s choose $x = -1$, which gives:

$y = \sqrt{2}(-1) = -\sqrt{2}$

Thus, the point is $(-1, -\sqrt{2})$. Now, we can find the radius (hypotenuse) $r$:

$r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (-\sqrt{2})^2} = \sqrt{1 + 2} = \sqrt{3}$

Now, we can calculate the six trigonometric functions:

1. Sine: $\sin(\theta) = \frac{y}{r} = \frac{-\sqrt{2}}{\sqrt{3}} = -\frac{\sqrt{6}}{3}$

2. Cosine: $\cos(\theta) = \frac{x}{r} = \frac{-1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

3. Tangent: $\tan(\theta) = \frac{y}{x} = \frac{-\sqrt{2}}{-1} = \sqrt{2}$

4. Cosecant (reciprocal of sine): $\csc(\theta) = \frac{r}{y} = \frac{\sqrt{3}}{-\sqrt{2}} = -\frac{\sqrt{6}}{2}$

5. Secant (reciprocal of cosine): $\sec(\theta) = \frac{r}{x} = \frac{\sqrt{3}}{-1} = -\sqrt{3}$

6. Cotangent (reciprocal of tangent): $\cot(\theta) = \frac{x}{y} = \frac{-1}{-\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Conclusion

The six trigonometric functions of the least positive angle $\theta = 114.46^\circ$ are:

• $\sin(\theta) = -\frac{\sqrt{6}}{3}$
• $\cos(\theta) = -\frac{\sqrt{3}}{3}$
• $\tan(\theta) = \sqrt{2}$
• $\csc(\theta) = -\frac{\sqrt{6}}{2}$
• $\sec(\theta) = -\sqrt{3}$
• $\cot(\theta) = \frac{\sqrt{2}}{2}$

Would you like further details or have any questions? Here are 5 related questions for deeper exploration:

1. How do you derive trigonometric values from a unit circle?
2. What are the implications of negative trigonometric values in various quadrants?
3. How can you verify trigonometric identities using the functions calculated?
4. How does the slope of a line relate to the angle formed with the x-axis?
5. Can you generalize these trigonometric calculations to any line equation?

Tip: Memorizing the unit circle angles helps quickly identify trigonometric values for many common angles.