Chapter 6: Problem 19
Find the exact value of the expression whenever It is defined. (a) \(\sin \left[2 \arccos \left(-\frac{3}{5}\right)\right]\) (b) \(\cos \left(2 \sin ^{-1} \frac{15}{17}\right)\) (c) \(\tan \left(2 \tan ^{-1} \frac{3}{4}\right)\)
Short Answer
Expert verified
(a) \(\frac{-24}{25}\), (b) \(\frac{-161}{289}\), (c) \(\frac{24}{7}\)
Step by step solution
01
Solving sub-problem (a)
First, we need to evaluate the expression \( \sin[2 \arccos(-\frac{3}{5})] \). Start by letting \( \theta = \arccos(-\frac{3}{5}) \). This means \( \cos \theta = -\frac{3}{5} \). Use the Pythagorean identity, \( \sin^2 \theta + \cos^2 \theta = 1 \), to find \( \sin \theta \):\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(-\frac{3}{5}\right)^2 \]\[ \sin^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \]\[ \sin \theta = \pm \frac{4}{5} \]Since \( \theta = \arccos(-\frac{3}{5}) \) is in quadrant II where sine is positive, \( \sin \theta = \frac{4}{5} \).Now, use the double-angle identity for sine: \( \sin(2\theta) = 2 \sin \theta \cos \theta \):\[ \sin(2\theta) = 2 \cdot \frac{4}{5} \cdot -\frac{3}{5} \]\[ \sin(2\theta) = 2 \cdot \frac{-12}{25} = \frac{-24}{25} \]
02
Solving sub-problem (b)
Next, solve \( \cos(2 \sin^{-1}(\frac{15}{17})) \). Start by letting \( \theta = \sin^{-1}(\frac{15}{17}) \). So, \( \sin \theta = \frac{15}{17} \). Again use the Pythagorean identity to find \( \cos \theta \):\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{15}{17}\right)^2 \]\[ \cos^2 \theta = 1 - \frac{225}{289} = \frac{64}{289} \]\[ \cos \theta = \pm \frac{8}{17} \]Since \( \theta = \sin^{-1}(\frac{15}{17}) \) is in quadrant I where cosine is positive, \( \cos \theta = \frac{8}{17} \).Use the double-angle identity for cosine: \( \cos(2\theta) = 2 \cos^2 \theta - 1 \):\[ \cos(2\theta) = 2 \left(\frac{8}{17}\right)^2 - 1 \]\[ \cos(2\theta) = 2 \cdot \frac{64}{289} - 1 \]\[ \cos(2\theta) = \frac{128}{289} - 1 \]\[ \cos(2\theta) = \frac{128}{289} - \frac{289}{289} = \frac{-161}{289} \]
03
Solving sub-problem (c)
Finally, solve \( \tan(2\tan^{-1}(\frac{3}{4})) \). Let \( \theta = \tan^{-1}(\frac{3}{4}) \). Thus \( \tan \theta = \frac{3}{4} \).Use the double-angle identity for tangent: \( \tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} \):\[ \tan(2\theta) = \frac{2 \cdot \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2} \]\[ \tan(2\theta) = \frac{\frac{6}{4}}{1 - \frac{9}{16}} \]\[ \tan(2\theta) = \frac{\frac{3}{2}}{\frac{7}{16}} \]Multiply by the reciprocal:\[ \tan(2\theta) = \frac{3}{2} \times \frac{16}{7} = \frac{48}{14} = \frac{24}{7} \]
04
Conclusion
The exact values for each sub-problem expression are as follows:(a) \( \sin \left[2 \arccos \left(-\frac{3}{5}\right)\right] = \frac{-24}{25} \)(b) \( \cos \left(2 \sin^{-1} \frac{15}{17}\right) = \frac{-161}{289} \)(c) \( \tan \left(2 \tan^{-1} \frac{3}{4}\right) = \frac{24}{7} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double-Angle Identities
Double-angle identities are powerful tools used in trigonometry to simplify the process of finding the values of trigonometric functions for double angles (or twice a given angle). They transform expressions with a doubled angle into expressions with single angles, making calculations easier.
For sine, the double-angle identity is given by:
For sine, the double-angle identity is given by:
- \( \sin(2\theta) = 2 \sin \theta \cos \theta \)
- \( \cos(2\theta) = 2 \cos^2 \theta - 1 \)
- \( \tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} \)
Inverse Trigonometric Functions
Inverse trigonometric functions are the inverse operations of the basic trigonometric functions (sine, cosine, and tangent). These functions are used to find the angle that corresponds to a given trigonometric value. The main inverse trigonometric functions are:
These inverse functions only return angles within certain ranges, ensuring each function provides a unique output:
- \( \arcsin(x) \)
- \( \arccos(x) \)
- \( \arctan(x) \)
These inverse functions only return angles within certain ranges, ensuring each function provides a unique output:
- \( \arcsin(x) \) outputs angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\)
- \( \arccos(x) \) outputs angles between 0 and \(\pi\)
- \( \arctan(x) \) outputs angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\)
Pythagorean Identity
The Pythagorean identity is a fundamental property of trigonometry that connects the squares of the sine and cosine of an angle. It states:
In trigonometric problems, particularly those involving inverse functions, this identity is indispensable. For instance, if you know \( \cos \theta \), you can find \( \sin \theta \) through:
This identity not only aids in finding unknown trigonometric values but also forms the basis of several other trigonometric identities, making it a cornerstone of solving related mathematical problems.
- \( \sin^2 \theta + \cos^2 \theta = 1 \)
In trigonometric problems, particularly those involving inverse functions, this identity is indispensable. For instance, if you know \( \cos \theta \), you can find \( \sin \theta \) through:
- \( \sin \theta = \sqrt{1 - \cos^2 \theta} \)
This identity not only aids in finding unknown trigonometric values but also forms the basis of several other trigonometric identities, making it a cornerstone of solving related mathematical problems.