Problem 35 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the values of the six trigonometric functions of \(\theta\). Constraint \(\theta\) lies in Quadrant II. \(\theta\) lies in Quadrant III. \(\sin \theta < 0\) \(\cot \theta < 0\) \(0 \leq \theta \leq \pi\) \(\frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\) \(\frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\) \(\pi \leq \theta \leq 2 \pi\) Function Value $$\sec \theta=-2$$

Step-by-Step Solution

Answer

The six trigonometric functions of \( \theta \) are \( \cos \theta = -1/2 \), \( \sin \theta = \sqrt{3}/2 \) and \( \sin \theta = -\sqrt{3}/2 \), \( \tan \theta = \sqrt{3} \) and \( \tan \theta = -\sqrt{3} \), \( \cot \theta = -\sqrt{3} \) and \( \cot \theta = \sqrt{3} \), \( \csc \theta = 2/\sqrt{3} \) and \( \csc \theta = - 2/\sqrt{3} \), \( \sec \theta = -2 \).

Step 1: Find the value of the cosine function

Given that \( \sec \theta \) = -2. we know that \( \sec \theta \) is the reciprocal of the cosine function. Calculating the reciprocal of -2 gives the cosine function as \( \cos \theta \) = -1/2.

Step 2: Use the Pythagorean identity

We can use the Pythagorean identity, \( \sin^2 \theta + \cos^2 \theta = 1 \), to find the value of \( \sin \theta \). By substituting the calculated cosine value into the identity, we get \( \sin^2 \theta + (-1/2)^2 = 1 \). Solving it yields \( \sin \theta = \pm \sqrt{3}/2 \).

Step 3: Determine the sign of sine function

Since \( \theta \) is given to be in the second and third quarters, where the sine function is positive and negative respectively, the sine function takes the values \( \sin \theta = \sqrt{3}/2 \) and \( \sin \theta=-\sqrt{3}/2 \) respectively.

Step 4: Calculate the value of the remaining trigonometric functions

The remaining trigonometric functions can be calculated using the known sine and cosine values. The tangent function is the sine function divided by the cosine function, the cotangent function is the reciprocal of the tangent function, and the cosecant function is the reciprocal of the sine function. So, we get \( \tan \theta = \sqrt{3} \) and \( \tan \theta = -\sqrt{3} \), \( \cot \theta = -\sqrt{3} \) and \( \cot \theta = \sqrt{3} \), \( \csc \theta = 2/\sqrt{3} \) and \( \csc \theta = - 2/\sqrt{3} \).

Key Concepts

Pythagorean Identity

Understanding the Pythagorean identity is essential for mastering trigonometry. It represents a fundamental relationship between the sine and cosine functions of any angle \theta. The identity is expressed as
\[\begin{equation} \[ \sin^2 \theta + \cos^2 \theta = 1 \] \[ \] \[ \] \end{equation}\]This equation simplifies the process of finding one trigonometric function when another is known. For instance, knowing the value of \[ \cos \theta \] allows us to find \[ \sin \theta \] by rearranging the identity into
\[\begin{equation} \[ \sin^2 \theta = 1 - \cos^2 \theta \] \[ \] \[ \] \end{equation}\]and then taking the square root. Remembering that the square root gives both positive and negative solutions is key, as the signs of these functions change depending on which quadrant the angle \[ \theta \] lies in. An angle in Quadrant II will have a positive sine value and a negative cosine value, while for an angle in Quadrant III, the sine value will be negative, and the cosine value will be positive.

Quadrant II and III Trigonometry

In trigonometry, the sign of a trigonometric function's value is determined by the angle's position in the coordinate plane. The standard coordinate plane is divided into four quadrants:

• Quadrant I: All trigonometric functions are positive.
• Quadrant II: Sine and cosecant are positive; cosine, secant, tangent, and cotangent are negative.
• Quadrant III: Tangent and cotangent are positive; sine, cosecant, cosine, and secant are negative.
• Quadrant IV: Cosine and secant are positive; sine, cosecant, tangent, and cotangent are negative.

In Quadrant II, angles range from \[ \frac{\pi}{2} \] to \[ \pi \], and because the x-coordinate (related to cosine) is negative while the y-coordinate (related to sine) is positive, the sine function will output a positive value, and the cosine and its reciprocal the secant function will output negative values. Conversely, in Quadrant III, where angles range from \[ \pi \] to \[ \frac{3\pi}{2} \], the situation is reversed—sine outputs a negative value, while cosine remains negative. This understanding is crucial when solving for unknown trigonometric functions using given constraints, like \[ \sec \theta = -2 \] in the provided exercise.

Trigonometric Function Values

When working with trigonometric function values, knowing the exact values can be quite powerful. Trigonometric functions are based on the properties of right triangles and the unit circle. For example, \[ \cos \theta \] represents the ratio of the adjacent side to the hypotenuse in a right triangle, or the x-coordinate on the unit circle, while \[ \sin \theta \] represents the opposite side to hypotenuse ratio or the y-coordinate on the unit circle. To find these values from secant, one must remember that secant is the reciprocal of cosine, so if \[ \sec \theta = -2 \], we have \[ \cos \theta = -\frac{1}{2} \].
Similarly, the function values for tangent \[ (\tan \theta) \], cotangent \[ (\cot \theta) \], and cosecant \[ (\csc \theta) \] can be determined as follows:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
\[ \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \]
\[ \csc \theta = \frac{1}{\sin \theta} \]
These relationships allow for the calculation of the remaining trigonometric functions once the values for sine and cosine are known. Overall, trigonometric function values are interconnected, and understanding how to compute one from another, especially in the context of different quadrants, is crucial for solving trigonometry problems.