1. If 90° << 180° and cos=-, find sin.
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
Related questions
Question
![**Trigonometry Problem: Finding Sine of Half the Angle**
**Problem Statement:**
If \( 90^\circ < \theta < 180^\circ \) and \(\cos \theta = -\frac{7}{11}\), find \(\sin \frac{\theta}{2}\).
**Answer Choices:**
1. \(-\frac{3\sqrt{11}}{11}\)
2. \(\frac{3\sqrt{11}}{11}\)
3. \(-\frac{11\sqrt{3}}{3}\)
4. \(\frac{11\sqrt{3}}{3}\)
To solve the problem, use the half-angle formula for sine:
\[
\sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}}
\]
Given that \( 90^\circ < \theta < 180^\circ \), this means \( \frac{\theta}{2} \) will be in the range \( 45^\circ < \frac{\theta}{2} < 90^\circ \). In this range, sine is positive.
Using the half-angle formula:
\[
\sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}}
\]
Substitute \(\cos \theta = -\frac{7}{11}\) into the formula:
\[
\sin \frac{\theta}{2} = \sqrt{\frac{1 - \left(-\frac{7}{11}\right)}{2}} = \sqrt{\frac{1 + \frac{7}{11}}{2}} = \sqrt{\frac{\frac{11}{11} + \frac{7}{11}}{2}} = \sqrt{\frac{\frac{18}{11}}{2}} = \sqrt{\frac{18}{22}} = \sqrt{\frac{9}{11}} = \frac{3\sqrt{11}}{11}
\]
So, the correct answer is:
**Answer:**
\(\frac{3\sqrt{11}}{11}\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F918c6600-bcd3-4fd5-a080-8ff573dbe9f8%2F6af84c8d-6c1a-4971-b20c-bae9c3128369%2Fsk5ns7e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Trigonometry Problem: Finding Sine of Half the Angle**
**Problem Statement:**
If \( 90^\circ < \theta < 180^\circ \) and \(\cos \theta = -\frac{7}{11}\), find \(\sin \frac{\theta}{2}\).
**Answer Choices:**
1. \(-\frac{3\sqrt{11}}{11}\)
2. \(\frac{3\sqrt{11}}{11}\)
3. \(-\frac{11\sqrt{3}}{3}\)
4. \(\frac{11\sqrt{3}}{3}\)
To solve the problem, use the half-angle formula for sine:
\[
\sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}}
\]
Given that \( 90^\circ < \theta < 180^\circ \), this means \( \frac{\theta}{2} \) will be in the range \( 45^\circ < \frac{\theta}{2} < 90^\circ \). In this range, sine is positive.
Using the half-angle formula:
\[
\sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}}
\]
Substitute \(\cos \theta = -\frac{7}{11}\) into the formula:
\[
\sin \frac{\theta}{2} = \sqrt{\frac{1 - \left(-\frac{7}{11}\right)}{2}} = \sqrt{\frac{1 + \frac{7}{11}}{2}} = \sqrt{\frac{\frac{11}{11} + \frac{7}{11}}{2}} = \sqrt{\frac{\frac{18}{11}}{2}} = \sqrt{\frac{18}{22}} = \sqrt{\frac{9}{11}} = \frac{3\sqrt{11}}{11}
\]
So, the correct answer is:
**Answer:**
\(\frac{3\sqrt{11}}{11}\)
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