Chapter 6: Problem 8
Find the exact values of \(\sin (\theta / 2), \cos (\theta / 2),\) and \(\tan (\theta / 2)\) for the given conditions. $$\csc \theta=-\frac{5}{3} ;-90^{\circ}<\theta<0^{\circ}$$
Short Answer
Expert verified
\(\sin(\theta/2) = \frac{1}{\sqrt{10}}, \cos(\theta/2) = \frac{3}{\sqrt{10}}, \tan(\theta/2) = -\frac{1}{3}\).
Step by step solution
01
Interpret the Given Condition
We are given that \( \csc \theta = -\frac{5}{3} \) and \( -90^{\circ} < \theta < 0^{\circ} \). This means \( \theta \) is in the fourth quadrant where \( \sin \theta = -\frac{1}{\csc \theta} \).
02
Find \( \sin \theta \)
Since \( \csc \theta = -\frac{5}{3} \), we have \( \sin \theta = -\frac{1}{\csc \theta} = -\frac{1}{-\frac{5}{3}} = -\frac{3}{5} \).
03
Find \( \cos \theta \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we substitute \( \sin \theta = -\frac{3}{5} \): \((-\frac{3}{5})^2 + \cos^2 \theta = 1 \). \Thus, \( \frac{9}{25} + \cos^2 \theta = 1 \), so \( \cos^2 \theta = \frac{16}{25} \), giving \( \cos \theta = \pm \frac{4}{5} \). Since \( \theta \) is in the fourth quadrant, \( \cos \theta \) is positive: \( \cos \theta = \frac{4}{5} \).
04
Use Half-Angle Identities
We use half-angle identities: - \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}} \) - \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \) - \( \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} \) or \( \frac{1 - \cos \theta}{\sin \theta} \).
05
Calculate \( \sin \frac{\theta}{2} \)
Given \( \cos \theta = \frac{4}{5} \), calculate \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \frac{4}{5}}{2}} = \pm \sqrt{\frac{1}{10}} = \pm \frac{1}{\sqrt{10}} \). Fortunately, \( -90^{\circ} < \theta < 0^{\circ} \) implies \( \frac{\theta}{2} \) is in the second quadrant, so \( \sin \frac{\theta}{2} > 0 \): \( \sin \frac{\theta}{2} = \frac{1}{\sqrt{10}} \).
06
Calculate \( \cos \frac{\theta}{2} \)
Using \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} = \pm \sqrt{\frac{9}{10}} = \pm \frac{3}{\sqrt{10}} \). Since \( \frac{\theta}{2} \) is in the second quadrant where \( \cos > 0 \), \( \cos \frac{\theta}{2} = \frac{3}{\sqrt{10}} \).
07
Calculate \( \tan \frac{\theta}{2} \)
Using \( \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} = \frac{-\frac{3}{5}}{1 + \frac{4}{5}} = \frac{-\frac{3}{5}}{\frac{9}{5}} = -\frac{1}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Half-Angle Identities
Half-angle identities are powerful tools in trigonometry that help us find the values of sine, cosine, and tangent at half of a given angle. These identities are derived from the well-known double-angle identities and are crucial when solving problems involving specific trigonometric values. Here's a closer look at each identity:
- For sine, the half-angle identity is \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}} \). This allows us to find the sine of half the angle when we know the cosine of the full angle.
- For cosine, the formula is \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \). Similarly, this takes the known value of cosine for an angle to determine the cosine for half of that angle.
- For tangent, the identity can be expressed as \( \tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta} \) or \( \frac{\sin \theta}{1 + \cos \theta} \). This helps find the tangent of half an angle in terms of sine and cosine of the original angle.
Quadrants of the Unit Circle
The unit circle is a circle with a radius of one centered at the origin of a coordinate plane. It is divided into four quadrants, each having specific properties that determine the signs of trigonometric functions:
- Quadrant I: Ranges from \(0^\circ\) to \(90^\circ\). In this quadrant, both sine and cosine are positive.
- Quadrant II: From \(90^\circ\) to \(180^\circ\), sine remains positive while cosine becomes negative.
- Quadrant III: Extending from \(180^\circ\) to \(270^\circ\), both sine and cosine are negative.
- Quadrant IV: Ranging from \(270^\circ\) to \(360^\circ\), sine is negative while cosine returns to positive.
Trigonometric Functions
Trigonometric functions are fundamental in mathematics, describing the relationship between angles and sides in right-angled triangles. The three primary functions—sine, cosine, and tangent—are often derived or utilized through identities, like in this problem.
- **Sine** \(\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}\): Indicates the ratio of the side opposite an angle to the hypotenuse in a right triangle.
- **Cosine** \(\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}\): Represents the ratio of the adjacent side to the hypotenuse.
- **Tangent** \(\tan \theta = \frac{\sin \theta}{\cos \theta}\): Exposes the ratio of the opposite to the adjacent side, or effectively, sine over cosine.