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Magazine Chapter 7: Problem 8
Find the exact values of \(\sin (\theta / 2), \cos (\theta / 2)\), and \(\tan (\theta / 2)\) for the given conditions. $$ \sec \theta=-4 ; \quad 180^{\circ}<\theta<270^{\circ} $$
Short Answer
Expert verified
\(\sin (\theta/2) = \frac{\sqrt{10}}{4}\), \(\cos (\theta/2) = -\frac{\sqrt{6}}{4}\), \(\tan (\theta/2) = -\frac{\sqrt{15}}{3}\).
Step by step solution
01
Convert sec to cos
Given that \(\sec \theta = -4\), we know that \(\cos \theta = \frac{1}{\sec \theta} = -\frac{1}{4}\). This is because the secant function is the reciprocal of the cosine function.
02
Calculate sin using Pythagorean identity
Knowing that \(180^{\circ}<\theta<270^{\circ}\), \(\theta\) is in the third quadrant, where sine and cosine are negative. Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we find \(\sin^2 \theta = 1 - \left(-\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16}\). Hence, \(\sin \theta = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}\) as \(\theta\) is in the third quadrant.
03
Use half-angle identity for sin
The half-angle identity for sine is \(\sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}}\). With \(\cos \theta = -\frac{1}{4}\), we have:\[ \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 + \frac{1}{4}}{2}} = \pm \sqrt{\frac{5}{8}} = \pm \frac{\sqrt{10}}{4} \].Since \(\frac{\theta}{2}\) falls in the second quadrant, \(\sin \frac{\theta}{2}\) is positive. Thus, \(\sin \frac{\theta}{2} = \frac{\sqrt{10}}{4}\).
04
Use half-angle identity for cos
The half-angle identity for cosine is \(\cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}\). With \(\cos \theta = -\frac{1}{4}\):\[ \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 - \frac{1}{4}}{2}} = \pm \sqrt{\frac{3}{8}} = \pm \frac{\sqrt{6}}{4} \].Since \(\frac{\theta}{2}\) is in the second quadrant, \(\cos \frac{\theta}{2}\) is negative. Thus, \(\cos \frac{\theta}{2} = -\frac{\sqrt{6}}{4}\).
05
Use identity for tan
Using the identity \(\tan \frac{\theta}{2} = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\) from earlier steps, knowing \(\sin \frac{\theta}{2} = \frac{\sqrt{10}}{4}\) and \(\cos \frac{\theta}{2} = -\frac{\sqrt{6}}{4}\), we have:\[ \tan \frac{\theta}{2} = \frac{\frac{\sqrt{10}}{4}}{-\frac{\sqrt{6}}{4}} = -\frac{\sqrt{10}}{\sqrt{6}} = -\frac{\sqrt{5}}{\sqrt{3}} = -\frac{\sqrt{15}}{3} \].
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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 used in trigonometry to find the values of trigonometric functions at half of a given angle. These identities are particularly useful when you know the trigonometric values for one angle and wish to find those of half that angle. The most common half-angle formulas are:
- Sine: \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}} \)
- Cosine: \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \)
- Tangent: \( \tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta} = \frac{\sin \theta}{1 + \cos \theta} \)
Pythagorean Identity
The Pythagorean identity in trigonometry is a fundamental relation between the sine and cosine of an angle. It states that for any angle \( \theta \):
In the given problem, with \( \cos \theta = -\frac{1}{4} \), we used this identity to find \( \sin \theta \). Rewriting it, we have \( \sin^2 \theta = 1 - \cos^2 \theta \). Substituting in the given value, we find \( \sin^2 \theta = 1 - \left(-\frac{1}{4}\right)^2 = \frac{15}{16} \), leading to \( \sin \theta = -\frac{\sqrt{15}}{4} \) since \( \theta \) is in the third quadrant where sine is negative.
- \( \sin^2 \theta + \cos^2 \theta = 1 \)
In the given problem, with \( \cos \theta = -\frac{1}{4} \), we used this identity to find \( \sin \theta \). Rewriting it, we have \( \sin^2 \theta = 1 - \cos^2 \theta \). Substituting in the given value, we find \( \sin^2 \theta = 1 - \left(-\frac{1}{4}\right)^2 = \frac{15}{16} \), leading to \( \sin \theta = -\frac{\sqrt{15}}{4} \) since \( \theta \) is in the third quadrant where sine is negative.
Trigonometric Functions
Trigonometric functions are mathematical functions related to the angles and sides of triangles, especially right-angled triangles. The most commonly used are the sine \( \sin \), cosine \( \cos \), and tangent \( \tan \).
In this exercise, knowing \( \sec \theta = -4 \), we quickly found that \( \cos \theta = -\frac{1}{4} \), using the trigonometric function relationships.
- Sine: Relates the opposite side to the hypotenuse.
- Cosine: Relates the adjacent side to the hypotenuse.
- Tangent: Relates the opposite side to the adjacent side, \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
In this exercise, knowing \( \sec \theta = -4 \), we quickly found that \( \cos \theta = -\frac{1}{4} \), using the trigonometric function relationships.
Third Quadrant
The unit circle is divided into four quadrants, each corresponding to a range of angles and specific signs for sine, cosine, and tangent functions.
Thus, for \( \theta \) in the third quadrant, \( \sin \theta = -\frac{\sqrt{15}}{4} \) given \( \cos \theta = -\frac{1}{4} \).
- First Quadrant: Both sine and cosine are positive.
- Second Quadrant: Sine is positive, cosine is negative.
- Third Quadrant: Both sine and cosine are negative.
- Fourth Quadrant: Sine is negative, cosine is positive.
Thus, for \( \theta \) in the third quadrant, \( \sin \theta = -\frac{\sqrt{15}}{4} \) given \( \cos \theta = -\frac{1}{4} \).
