Problem 6 - Precalculus: Functions and Graphs

Question

Find the exact values of \(\sin (\theta / 2), \cos (\theta / 2),\) and \(\tan (\theta / 2)\) for the given conditions. $$\csc \theta=-\frac{5}{3} ; \quad-90^{\circ}<\theta<0^{\circ}$$

Step-by-Step Solution

Answer

\( \sin(\theta/2) = -\frac{\sqrt{10}}{10}, \cos(\theta/2) = \frac{3\sqrt{10}}{10}, \tan(\theta/2) = -\frac{1}{3} \).

Step 1: Determine the Value of \( \sin \theta \)

Recall that \( \csc \theta = \frac{1}{\sin \theta} \). Given that \( \csc \theta = -\frac{5}{3} \), we can find \( \sin \theta \) by finding the reciprocal:\[ \sin \theta = -\frac{3}{5} \]

Step 2: Find \( \cos \theta \) Using the Pythagorean Identity

Utilize the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Substitute \( \sin \theta = -\frac{3}{5} \):\[ (\sin \theta)^2 = \left( -\frac{3}{5} \right)^2 = \frac{9}{25} \]\[ \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \]Since \(-90^{\circ}<\theta<0^{\circ}\), \(\cos\theta \) is positive, hence:\[ \cos \theta = \frac{4}{5} \]

Step 3: Use Half-Angle Formulas for \( \sin(\theta/2) \)

The half-angle formula for sine is given by:\[ \sin \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - \cos \theta}{2}} \]Substitute \( \cos \theta = \frac{4}{5} \):\[ \sin \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - \frac{4}{5}}{2}} = \pm \sqrt{\frac{1}{10}} = \pm \frac{1}{\sqrt{10}} \]Since \(-90^{\circ}<\theta<0^{\circ}\), \(\theta/2\) is in the range \(-45^{\circ}<\theta/2<0^{\circ}\), implying \( \sin(\theta/2) < 0 \):\[ \sin \left( \frac{\theta}{2} \right) = -\frac{1}{\sqrt{10}} = -\frac{\sqrt{10}}{10} \]

Step 4: Use Half-Angle Formulas for \( \cos(\theta/2) \)

The half-angle formula for cosine is given by:\[ \cos \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} \]Substitute \( \cos \theta = \frac{4}{5} \):\[ \cos \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 + \frac{4}{5}}{2}} = \pm \sqrt{\frac{9}{10}} = \pm \frac{3}{\sqrt{10}} \]In the range \(-45^{\circ}<\theta/2<0^{\circ}\), \( \cos(\theta/2) > 0 \):\[ \cos \left( \frac{\theta}{2} \right) = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10} \]

Step 5: Calculate \( \tan(\theta/2) \)

\( \tan(\theta/2) \) can be calculated using the formula:\[ \tan \left( \frac{\theta}{2} \right) = \frac{\sin(\theta/2)}{\cos(\theta/2)} \]Substitute the values found:\[ \tan \left( \frac{\theta}{2} \right) = \frac{-\frac{\sqrt{10}}{10}}{\frac{3\sqrt{10}}{10}} = -\frac{1}{3} \]

Key Concepts

Sine Function

The sine function is a fundamental concept in trigonometry. It helps us describe the relationship between the angles and sides of a right triangle. Given that we have a secant (csc) identity related to sine, we find that if \( \csc \theta = -\frac{5}{3} \), then the reciprocal gives us \( \sin \theta = -\frac{3}{5} \). This transformation is critical as it connects the trigonometric function with its reciprocal, providing insights into the angle's behavior. Remember, the sine function remains negative in the fourth quadrant, which aligns with our given interval \(-90^{\circ}<\theta<0^{\circ}\). Through an understanding of sine, we're able to explore complex problems involving half-angles using the formula \( \sin \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - \cos \theta}{2}} \).

Cosine Function

The cosine function, much like its sine counterpart, plays a pivotal role in trigonometry. The Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) is the bridge between \( \sin \theta \) and \( \cos \theta \). By knowing \( \sin \theta = -\frac{3}{5} \), you calculate \( \cos^2 \theta = \frac{16}{25} \) leading to \( \cos \theta = \frac{4}{5} \), considering the angle's quadrant. Such values allow you to work with half-angle formulas like \( \cos \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} \). This determination is crucial for accurate graphing and solving of trigonometric problems.

Tangent Function

The tangent function is expressed as the ratio between sine and cosine: \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). This relationship simplifies complex trigonometric problems. When dealing with half-angles, the formula expands to \( \tan \left( \frac{\theta}{2} \right) = \frac{\sin(\theta/2)}{\cos(\theta/2)} \). Substituting our values \( \sin \left( \frac{\theta}{2} \right) = -\frac{\sqrt{10}}{10} \) and \( \cos \left( \frac{\theta}{2} \right) = \frac{3\sqrt{10}}{10} \), gives \( \tan \left( \frac{\theta}{2} \right) = -\frac{1}{3} \). This analytical approach showcases the interlinking nature of trigonometric functions and assists in solving angled problems efficiently.

Pythagorean Identity

The Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) serves as a cornerstone of trigonometry. It provides a direct relationship between the functions of sine and cosine. By substituting the value of \( \sin \theta = -\frac{3}{5} \) into the identity, we ascertain \( \cos^2 \theta = \frac{16}{25} \). This simplicity allows us to find \( \cos \theta \) and efficiently use it in further calculations, especially in contexts that involve half-angle formulas. Recognizing this identity's power aids in understanding how different trigonometric functions intermingle to describe the properties of angles and circles comprehensively.