Answer :
To solve these trigonometric problems, we will rely on well-known trigonometric identities and properties of angles.
Given:
[tex]\[ \sin \theta = \frac{2}{3}, \quad 0 < \theta < \frac{\pi}{2} \][/tex]
Step-by-Step Solution:
### (a) [tex]\(\sin (2\theta)\)[/tex]
The double-angle identity for the sine function is:
[tex]\[ \sin (2\theta) = 2 \sin (\theta) \cos (\theta) \][/tex]
First, we need to find [tex]\(\cos (\theta)\)[/tex].
Since [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex]:
[tex]\[ \cos^2(\theta) = 1 - \sin^2(\theta) \][/tex]
Plugging in [tex]\(\sin(\theta) = \frac{2}{3}\)[/tex]:
[tex]\[ \cos^2(\theta) = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9} \][/tex]
Therefore,
[tex]\[ \cos (\theta) = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \][/tex]
Now we substitute [tex]\(\sin (\theta)\)[/tex] and [tex]\(\cos (\theta)\)[/tex] into the double-angle identity:
[tex]\[ \sin (2\theta) = 2 \left(\frac{2}{3}\right) \left(\frac{\sqrt{5}}{3}\right) = \frac{4\sqrt{5}}{9} \][/tex]
From the provided answer:
[tex]\[ \sin (2\theta) \approx 0.993808 \][/tex]
### (b) [tex]\(\cos (2\theta)\)[/tex]
The double-angle identity for the cosine function is:
[tex]\[ \cos (2\theta) = \cos^2 (\theta) - \sin^2 (\theta) \][/tex]
We already have [tex]\(\cos (\theta) = \frac{\sqrt{5}}{3}\)[/tex] and [tex]\(\sin (\theta) = \frac{2}{3}\)[/tex]:
[tex]\[ \cos^2 (\theta) = \left(\frac{\sqrt{5}}{3}\right)^2 = \frac{5}{9} \][/tex]
[tex]\[ \sin^2 (\theta) = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \][/tex]
Substituting these into the double-angle identity:
[tex]\[ \cos (2\theta) = \frac{5}{9} - \frac{4}{9} = \frac{1}{9} \][/tex]
From the provided answer:
[tex]\[ \cos (2\theta) \approx 0.111111 \][/tex]
### (c) [tex]\(\sin (\frac{\theta}{2})\)[/tex]
The half-angle identity for the sine function is:
[tex]\[ \sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos (\theta)}{2}} \][/tex]
We already know [tex]\(\cos (\theta) = \frac{\sqrt{5}}{3}\)[/tex]:
[tex]\[ \sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{2}} = \sqrt{\frac{\frac{3 - \sqrt{5}}{3}}{2}} = \sqrt{\frac{3 - \sqrt{5}}{6}} \][/tex]
From the provided answer:
[tex]\[ \sin \left(\frac{\theta}{2}\right) \approx 0.356822 \][/tex]
### (d) [tex]\(\cos (\frac{\theta}{2})\)[/tex]
The half-angle identity for the cosine function is:
[tex]\[ \cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos (\theta)}{2}} \][/tex]
Using [tex]\(\cos (\theta) = \frac{\sqrt{5}}{3}\)[/tex]:
[tex]\[ \cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \frac{\sqrt{5}}{3}}{2}} = \sqrt{\frac{\frac{3 + \sqrt{5}}{3}}{2}} = \sqrt{\frac{3 + \sqrt{5}}{6}} \][/tex]
From the provided answer:
[tex]\[ \cos \left(\frac{\theta}{2}\right) \approx 0.934172 \][/tex]
Therefore, the exact values are:
[tex]\[ \sin (2\theta) \approx 0.993808 \][/tex]
[tex]\[ \cos (2\theta) \approx 0.111111 \][/tex]
[tex]\[ \sin \left(\frac{\theta}{2}\right) \approx 0.356822 \][/tex]
[tex]\[ \cos \left(\frac{\theta}{2}\right) \approx 0.934172 \][/tex]
Given:
[tex]\[ \sin \theta = \frac{2}{3}, \quad 0 < \theta < \frac{\pi}{2} \][/tex]
Step-by-Step Solution:
### (a) [tex]\(\sin (2\theta)\)[/tex]
The double-angle identity for the sine function is:
[tex]\[ \sin (2\theta) = 2 \sin (\theta) \cos (\theta) \][/tex]
First, we need to find [tex]\(\cos (\theta)\)[/tex].
Since [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex]:
[tex]\[ \cos^2(\theta) = 1 - \sin^2(\theta) \][/tex]
Plugging in [tex]\(\sin(\theta) = \frac{2}{3}\)[/tex]:
[tex]\[ \cos^2(\theta) = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9} \][/tex]
Therefore,
[tex]\[ \cos (\theta) = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \][/tex]
Now we substitute [tex]\(\sin (\theta)\)[/tex] and [tex]\(\cos (\theta)\)[/tex] into the double-angle identity:
[tex]\[ \sin (2\theta) = 2 \left(\frac{2}{3}\right) \left(\frac{\sqrt{5}}{3}\right) = \frac{4\sqrt{5}}{9} \][/tex]
From the provided answer:
[tex]\[ \sin (2\theta) \approx 0.993808 \][/tex]
### (b) [tex]\(\cos (2\theta)\)[/tex]
The double-angle identity for the cosine function is:
[tex]\[ \cos (2\theta) = \cos^2 (\theta) - \sin^2 (\theta) \][/tex]
We already have [tex]\(\cos (\theta) = \frac{\sqrt{5}}{3}\)[/tex] and [tex]\(\sin (\theta) = \frac{2}{3}\)[/tex]:
[tex]\[ \cos^2 (\theta) = \left(\frac{\sqrt{5}}{3}\right)^2 = \frac{5}{9} \][/tex]
[tex]\[ \sin^2 (\theta) = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \][/tex]
Substituting these into the double-angle identity:
[tex]\[ \cos (2\theta) = \frac{5}{9} - \frac{4}{9} = \frac{1}{9} \][/tex]
From the provided answer:
[tex]\[ \cos (2\theta) \approx 0.111111 \][/tex]
### (c) [tex]\(\sin (\frac{\theta}{2})\)[/tex]
The half-angle identity for the sine function is:
[tex]\[ \sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos (\theta)}{2}} \][/tex]
We already know [tex]\(\cos (\theta) = \frac{\sqrt{5}}{3}\)[/tex]:
[tex]\[ \sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{2}} = \sqrt{\frac{\frac{3 - \sqrt{5}}{3}}{2}} = \sqrt{\frac{3 - \sqrt{5}}{6}} \][/tex]
From the provided answer:
[tex]\[ \sin \left(\frac{\theta}{2}\right) \approx 0.356822 \][/tex]
### (d) [tex]\(\cos (\frac{\theta}{2})\)[/tex]
The half-angle identity for the cosine function is:
[tex]\[ \cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos (\theta)}{2}} \][/tex]
Using [tex]\(\cos (\theta) = \frac{\sqrt{5}}{3}\)[/tex]:
[tex]\[ \cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \frac{\sqrt{5}}{3}}{2}} = \sqrt{\frac{\frac{3 + \sqrt{5}}{3}}{2}} = \sqrt{\frac{3 + \sqrt{5}}{6}} \][/tex]
From the provided answer:
[tex]\[ \cos \left(\frac{\theta}{2}\right) \approx 0.934172 \][/tex]
Therefore, the exact values are:
[tex]\[ \sin (2\theta) \approx 0.993808 \][/tex]
[tex]\[ \cos (2\theta) \approx 0.111111 \][/tex]
[tex]\[ \sin \left(\frac{\theta}{2}\right) \approx 0.356822 \][/tex]
[tex]\[ \cos \left(\frac{\theta}{2}\right) \approx 0.934172 \][/tex]
