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Sagot :
Let's solve each part of the question step-by-step.
### Part A: Determine the exact value of [tex]\(\cos 2\theta\)[/tex]
Given:
[tex]\[ \sin \theta = \frac{2 \sqrt{2}}{5} \][/tex]
and [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex].
First, use the double-angle formula for cosine:
[tex]\[ \cos 2\theta = 1 - 2\sin^2 \theta \][/tex]
First, we need to find [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \sin^2 \theta = \left( \frac{2 \sqrt{2}}{5} \right)^2 = \frac{8}{25} \][/tex]
Next, plug [tex]\(\sin^2 \theta\)[/tex] into the double-angle formula:
[tex]\[ \cos 2\theta = 1 - 2 \cdot \frac{8}{25} = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \][/tex]
Therefore:
[tex]\[ \cos 2\theta = \frac{9}{25} \][/tex]
### Part B: Determine the exact value of [tex]\(\sin \left( \frac{\theta}{2} \right)\)[/tex]
For the half-angle formula for sine:
[tex]\[ \sin \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - \cos \theta}{2}} \][/tex]
First, we need to find [tex]\(\cos \theta\)[/tex]. Using the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
[tex]\[ \left( \frac{2 \sqrt{2}}{5} \right)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{8}{25} + \cos^2 \theta = 1 \][/tex]
[tex]\[ \cos^2 \theta = 1 - \frac{8}{25} = \frac{17}{25} \][/tex]
[tex]\[ \cos \theta = \pm \sqrt{\frac{17}{25}} \][/tex]
Given [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex], [tex]\(\cos \theta\)[/tex] is negative. So:
[tex]\[ \cos \theta = -\sqrt{\frac{17}{25}} = -\frac{\sqrt{17}}{5} \][/tex]
Now, use this to find [tex]\(\sin \left( \frac{\theta}{2} \right)\)[/tex]:
[tex]\[ \sin \left( \frac{\theta}{2} \right) = \sqrt{\frac{1 - \cos \theta}{2}} = \sqrt{\frac{1 - \left( -\frac{\sqrt{17}}{5} \right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{17}}{5}}{2}} \][/tex]
[tex]\[ \sin \left( \frac{\theta}{2} \right) = \sqrt{\frac{5 + \sqrt{17}}{10}} \][/tex]
Considering [tex]\(\sin \left( \frac{\theta}{2} \right)\)[/tex] in the interval [tex]\(\frac{\pi}{2}<\theta<\pi\)[/tex], [tex]\(\frac{\theta}{2}\)[/tex] falls in [tex]\(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\)[/tex]. Thus, [tex]\(\sin \left( \frac{\theta}{2} \right)\)[/tex] is positive, so:
[tex]\[ \sin \left( \frac{\theta}{2} \right) \approx 0.2961 \][/tex]
In conclusion:
Part A:
[tex]\[ \cos 2\theta = \frac{9}{25} \][/tex]
Part B:
[tex]\[ \sin \left( \frac{\theta}{2} \right) \approx 0.2961 \][/tex]
### Part A: Determine the exact value of [tex]\(\cos 2\theta\)[/tex]
Given:
[tex]\[ \sin \theta = \frac{2 \sqrt{2}}{5} \][/tex]
and [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex].
First, use the double-angle formula for cosine:
[tex]\[ \cos 2\theta = 1 - 2\sin^2 \theta \][/tex]
First, we need to find [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \sin^2 \theta = \left( \frac{2 \sqrt{2}}{5} \right)^2 = \frac{8}{25} \][/tex]
Next, plug [tex]\(\sin^2 \theta\)[/tex] into the double-angle formula:
[tex]\[ \cos 2\theta = 1 - 2 \cdot \frac{8}{25} = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \][/tex]
Therefore:
[tex]\[ \cos 2\theta = \frac{9}{25} \][/tex]
### Part B: Determine the exact value of [tex]\(\sin \left( \frac{\theta}{2} \right)\)[/tex]
For the half-angle formula for sine:
[tex]\[ \sin \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - \cos \theta}{2}} \][/tex]
First, we need to find [tex]\(\cos \theta\)[/tex]. Using the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
[tex]\[ \left( \frac{2 \sqrt{2}}{5} \right)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{8}{25} + \cos^2 \theta = 1 \][/tex]
[tex]\[ \cos^2 \theta = 1 - \frac{8}{25} = \frac{17}{25} \][/tex]
[tex]\[ \cos \theta = \pm \sqrt{\frac{17}{25}} \][/tex]
Given [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex], [tex]\(\cos \theta\)[/tex] is negative. So:
[tex]\[ \cos \theta = -\sqrt{\frac{17}{25}} = -\frac{\sqrt{17}}{5} \][/tex]
Now, use this to find [tex]\(\sin \left( \frac{\theta}{2} \right)\)[/tex]:
[tex]\[ \sin \left( \frac{\theta}{2} \right) = \sqrt{\frac{1 - \cos \theta}{2}} = \sqrt{\frac{1 - \left( -\frac{\sqrt{17}}{5} \right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{17}}{5}}{2}} \][/tex]
[tex]\[ \sin \left( \frac{\theta}{2} \right) = \sqrt{\frac{5 + \sqrt{17}}{10}} \][/tex]
Considering [tex]\(\sin \left( \frac{\theta}{2} \right)\)[/tex] in the interval [tex]\(\frac{\pi}{2}<\theta<\pi\)[/tex], [tex]\(\frac{\theta}{2}\)[/tex] falls in [tex]\(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\)[/tex]. Thus, [tex]\(\sin \left( \frac{\theta}{2} \right)\)[/tex] is positive, so:
[tex]\[ \sin \left( \frac{\theta}{2} \right) \approx 0.2961 \][/tex]
In conclusion:
Part A:
[tex]\[ \cos 2\theta = \frac{9}{25} \][/tex]
Part B:
[tex]\[ \sin \left( \frac{\theta}{2} \right) \approx 0.2961 \][/tex]
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