What is the exact value of cos 157.5° using half-angle identities?
Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter2: Right Triangle Trigonometry
Section: Chapter Questions
Problem 6GP
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![### Question 12
What is the exact value of \( \cos 157.5^\circ \) using half-angle identities?
### Explanation:
This problem involves calculating the exact value of \( \cos 157.5^\circ \) through the use of half-angle identities in trigonometry. The half-angle identity for cosine states that:
\[ \cos \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} \]
To solve for \( \cos 157.5^\circ \), recognize that \( 157.5^\circ \) is half of \( 315^\circ \):
\[ \theta = 315^\circ \]
\[ \frac{\theta}{2} = 157.5^\circ \]
Thus:
\[ \cos 157.5^\circ = \cos \left( \frac{315^\circ}{2} \right) \]
Using the half-angle identity:
\[ \cos 157.5^\circ = \sqrt{\frac{1 + \cos 315^\circ}{2}} \text{ since } 157.5^\circ \text{ is in the second quadrant where cosine is negative} \]
We know that:
\[ \cos 315^\circ = \cos (360^\circ - 45^\circ) = \cos 45^\circ = \frac{\sqrt{2}}{2} \]
Therefore:
\[ \cos 157.5^\circ = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \]
Simplify this expression:
\[ \cos 157.5^\circ = \sqrt{\frac{2 + \sqrt{2}}{4}} \]
Since \( 157.5^\circ \) lies in the second quadrant and cosine is negative there:
\[ \cos 157.5^\circ = -\sqrt{\frac{2 + \sqrt{2}}{4}} \]
\[ \cos 157.5^\circ = -\frac{\sqrt{2 + \sqrt{2}}}{2} \]
Thus, the exact value of \( \cos 157.5^\circ \) is:
\[ \boxed{-\frac{\sqrt{2 + \sqrt{2}}}{2}} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F050b7896-96e1-4a53-ae28-da4cd34956eb%2F47eeb1a0-d38a-448a-acb5-5eafc514d722%2Fqbde02g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 12
What is the exact value of \( \cos 157.5^\circ \) using half-angle identities?
### Explanation:
This problem involves calculating the exact value of \( \cos 157.5^\circ \) through the use of half-angle identities in trigonometry. The half-angle identity for cosine states that:
\[ \cos \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} \]
To solve for \( \cos 157.5^\circ \), recognize that \( 157.5^\circ \) is half of \( 315^\circ \):
\[ \theta = 315^\circ \]
\[ \frac{\theta}{2} = 157.5^\circ \]
Thus:
\[ \cos 157.5^\circ = \cos \left( \frac{315^\circ}{2} \right) \]
Using the half-angle identity:
\[ \cos 157.5^\circ = \sqrt{\frac{1 + \cos 315^\circ}{2}} \text{ since } 157.5^\circ \text{ is in the second quadrant where cosine is negative} \]
We know that:
\[ \cos 315^\circ = \cos (360^\circ - 45^\circ) = \cos 45^\circ = \frac{\sqrt{2}}{2} \]
Therefore:
\[ \cos 157.5^\circ = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \]
Simplify this expression:
\[ \cos 157.5^\circ = \sqrt{\frac{2 + \sqrt{2}}{4}} \]
Since \( 157.5^\circ \) lies in the second quadrant and cosine is negative there:
\[ \cos 157.5^\circ = -\sqrt{\frac{2 + \sqrt{2}}{4}} \]
\[ \cos 157.5^\circ = -\frac{\sqrt{2 + \sqrt{2}}}{2} \]
Thus, the exact value of \( \cos 157.5^\circ \) is:
\[ \boxed{-\frac{\sqrt{2 + \sqrt{2}}}{2}} \]
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