Problem 34 - A Graphical Approach to Precalculus with Limits

Question

Sketch an angle \(\theta\) in standard position such that \(\theta\) has the least possible positive measure, and the given point is on the terminal side of \(\theta .\) Find the values of the six trigonometric functions for each angle. Rationalize denominators when applicable. Do not use a calculator. $$(-\sqrt{2},-\sqrt{2})$$

Step-by-Step Solution

Answer

Angle is 225° or \(\frac{5\pi}{4}\) radians. Trig functions: \(\sin(\theta) = -\frac{\sqrt{2}}{2}\), \(\cos(\theta) = -\frac{\sqrt{2}}{2}\), \(\tan(\theta) = 1\), \(\csc(\theta) = -\sqrt{2}\), \(\sec(\theta) = -\sqrt{2}\), \(\cot(\theta) = 1\).

Step 1: Determine the Quadrant

The point \((-\sqrt{2},-\sqrt{2})\) is in the third quadrant because both the x-coordinate and the y-coordinate are negative.

Step 2: Find the Reference Angle

The reference angle \( \theta_{ref} \) can be found using the symmetry in the coordinate plane. Since \((-\sqrt{2},-\sqrt{2})\) indicates a 45-degree angle from the horizontal axis (due to equal negative lengths in x and y), the reference angle is \(45^\circ\) or \(\frac{\pi}{4}\) radians.

Step 3: Determine the Actual Angle

The angle \( \theta \) in standard position is found by adding the reference angle to \(180^\circ\) or \(\pi\) radians (because it's in the third quadrant). Hence, the angle is \(180^\circ + 45^\circ = 225^\circ\) or \(\pi + \frac{\pi}{4} = \frac{5\pi}{4}\) radians.

Step 4: Calculate the Radius or Hypotenuse

Use the Pythagorean theorem to find the radius \( r \). Here, \( r = \sqrt{(-\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \).

Step 5: Evaluate the Sine Function

The value of \( \sin(\theta) \) is \( \frac{y}{r} = \frac{-\sqrt{2}}{2} \).

Step 6: Evaluate the Cosine Function

The value of \( \cos(\theta) \) is \( \frac{x}{r} = \frac{-\sqrt{2}}{2} \).

Step 7: Evaluate the Tangent Function

The value of \( \tan(\theta) \) is \( \frac{y}{x} = \frac{-\sqrt{2}}{-\sqrt{2}} = 1 \).

Step 8: Evaluate the Cosecant Function

The value of \( \csc(\theta) \) is the reciprocal of sine: \( \csc(\theta) = \frac{r}{y} = \frac{2}{-\sqrt{2}} = -\sqrt{2} \).

Step 9: Evaluate the Secant Function

The value of \( \sec(\theta) \) is the reciprocal of cosine: \( \sec(\theta) = \frac{r}{x} = \frac{2}{-\sqrt{2}} = -\sqrt{2} \).

Step 10: Evaluate the Cotangent Function

The value of \( \cot(\theta) \) is the reciprocal of tangent: \( \cot(\theta) = \frac{x}{y} = \frac{-\sqrt{2}}{-\sqrt{2}} = 1 \).

Key Concepts

Standard position

In trigonometry, an angle is in its standard position if its vertex is placed at the origin of the coordinate plane and its initial side lies along the positive x-axis. This setup allows for easier identification of angle measures within the four quadrants of the plane.
When working with angles in standard position, it helps to understand how they wind around the origin:

• Positive angles are measured counterclockwise from the initial side.
• Negative angles are measured clockwise.

By standardizing the angle’s position, we create a common groundwork that supports the derivation of trigonometric functions and their relationships across different quadrants.

Reference angle

The reference angle is a smaller version of the given angle, lying between the angle in standard position and the horizontal axis.
To find the reference angle, consider which quadrant the terminal side of the initial angle falls in:

• If in the first quadrant, the reference angle is the angle itself.
• In the second quadrant, subtract the angle from 180° or π radians.
• In the third quadrant, subtract 180° or π radians from the angle.
• In the fourth quadrant, subtract the angle from 360° or 2π radians.

In our problem, because we have the point \((-\sqrt{2},-\sqrt{2})\) which aligns with the third quadrant, the reference angle here is 45° (or \(\frac{\pi}{4}\) radians). The reference angle always helps in identifying trigonometric function values since their standard ratios are defined using these angles.

Third quadrant

The third quadrant of the coordinate plane includes angles between 180° to 270° (or π to \(\frac{3\pi}{2}\) radians). In this quadrant, the signs of trigonometric functions are particularly important:

• Both sine and cosine are negative because both the x and y coordinates involve negative values.
• Their quotient, the tangent function, remains positive because a negative divided by a negative is positive.

The angle we've discussed is in this third quadrant, and its terminal point \((-\sqrt{2}, -\sqrt{2})\) reaffirms this. The measures put us comfortably at 225° or \(\frac{5\pi}{4}\) radians. Understanding the quadrant helps in determining the correct sign for each trigonometric function related to the angle.

Pythagorean theorem

The Pythagorean theorem is a principle that connects the sides of a right triangle: \[c^2 = a^2 + b^2\]where \(c\) is the hypotenuse and \(a\) and \(b\) are the other two sides. In trigonometry, it assists in calculating the hypotenuse (or radius) when dealing with points in the coordinate plane, particularly when the points relate to trigonometric functions.
In the provided problem, for point \((-\sqrt{2}, -\sqrt{2})\), the radius \(r\) is found using:

• \((-\sqrt{2})^2 + (-\sqrt{2})^2 = 2 + 2 = 4\)
• Thus, \(r = \sqrt{4} = 2\)

This radius then plays a crucial role in evaluating the trigonometric functions like sine, cosine, and their reciprocals by connecting each function's value with a right triangle's sides.