Chapter 8: Problem 35
35-40 Find \(\sin \frac{x}{2}, \cos \frac{x}{2},\) and \(\tan \frac{x}{2}\) from the given information. \(\sin x=\frac{3}{5}, \quad 0^{\circ} < x < 90^{\circ}\)
Short Answer
Expert verified
\( \sin \frac{x}{2} = \sqrt{\frac{1}{10}}, \cos \frac{x}{2} = \sqrt{\frac{9}{10}}, \tan \frac{x}{2} = \frac{1}{3} \).
Step by step solution
01
Determine the Value of \( \cos x \)
Given \( \sin x = \frac{3}{5} \) and using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can find \( \cos x \).\Plug \( \sin x \) into the identity: \( \left(\frac{3}{5}\right)^2 + \cos^2 x = 1 \).\Calculate: \( \frac{9}{25} + \cos^2 x = 1 \).\Then, \( \cos^2 x = 1 - \frac{9}{25} = \frac{16}{25} \).\So, \( \cos x = \frac{4}{5} \) (since \( \cos x > 0 \) in the first quadrant).
02
Use Double Angle Identity for \( \sin \frac{x}{2} \)
For \( \sin \frac{x}{2} \), use the identity \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \).\Plug \( \cos x = \frac{4}{5} \) into the formula: \( \sin \frac{x}{2} = \sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{1}{10}} \).\Since \( 0^\circ < x < 90^\circ \), \( \frac{x}{2} \) is also in the first quadrant, so \( \sin \frac{x}{2} > 0 \).\Hence, \( \sin \frac{x}{2} = \sqrt{\frac{1}{10}} \).
03
Use Double Angle Identity for \( \cos \frac{x}{2} \)
For \( \cos \frac{x}{2} \), use the identity \( \cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}} \).\Plug \( \cos x = \frac{4}{5} \) into the formula: \( \cos \frac{x}{2} = \sqrt{\frac{1 + \frac{4}{5}}{2}} = \sqrt{\frac{9}{10}} \).\Since \( \frac{x}{2} \) is in the first quadrant, \( \cos \frac{x}{2} > 0 \).\Therefore, \( \cos \frac{x}{2} = \sqrt{\frac{9}{10}} \).
04
Use Identity for \( \tan \frac{x}{2} \)
For \( \tan \frac{x}{2} \), use the identity \( \tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} \).\Substitute \( \sin x = \frac{3}{5} \) and \( \cos x = \frac{4}{5} \) into the formula: \( \tan \frac{x}{2} = \frac{\frac{3}{5}}{1 + \frac{4}{5}} = \frac{\frac{3}{5}}{\frac{9}{5}} \).\Simplify it: \( \tan \frac{x}{2} = \frac{3}{9} = \frac{1}{3} \).\Thus, \( \tan \frac{x}{2} = \frac{1}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pythagorean identity
The Pythagorean identity is a fundamental equation in trigonometry that relates the squares of the sine and cosine of an angle. It is given by the formula:
In the original exercise, \( \sin x = \frac{3}{5} \) was given, and by applying the Pythagorean identity, you can find the value of \( \cos x \). You'll first square \( \frac{3}{5} \) to get \( \frac{9}{25} \).
- \( \sin^2 x + \cos^2 x = 1 \)
In the original exercise, \( \sin x = \frac{3}{5} \) was given, and by applying the Pythagorean identity, you can find the value of \( \cos x \). You'll first square \( \frac{3}{5} \) to get \( \frac{9}{25} \).
- Then, subtract this from 1 because: \( \cos^2 x = 1 - \sin^2 x = 1 - \frac{9}{25} \)
Double angle identities
Double angle identities help compute trigonometric functions for half angles. In this context, we use them to find \( \sin \frac{x}{2} \) and \( \cos \frac{x}{2} \). These identities are crucial when dealing with expressions involving half angles:
Plug \( \cos x = \frac{4}{5} \) into the formulas:
- For sine: \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \)
- For cosine: \( \cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}} \)
Plug \( \cos x = \frac{4}{5} \) into the formulas:
- \( \sin \frac{x}{2} = \sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{1}{10}} \)
- \( \cos \frac{x}{2} = \sqrt{\frac{1 + \frac{4}{5}}{2}} = \sqrt{\frac{9}{10}} \)
First quadrant
The first quadrant in the coordinate plane is crucial for understanding the signs of trigonometric functions. In the first quadrant, where both x and y coordinates are positive, the sine, cosine, and tangent of an angle are also positive.
In trigonometry, the first quadrant corresponds to angles between \( 0^\circ \) and \( 90^\circ \). It is the only quadrant where all primary trigonometric functions assume positive values:
In trigonometry, the first quadrant corresponds to angles between \( 0^\circ \) and \( 90^\circ \). It is the only quadrant where all primary trigonometric functions assume positive values:
- \( \sin x > 0 \)
- \( \cos x > 0 \)
- \( \tan x > 0 \)