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Sagot :
To find the values of [tex]\(\theta\)[/tex] in the interval [tex]\(0 \leq \theta \leq 8\pi\)[/tex] where the function [tex]\(y = \sin \theta\)[/tex] equals 1, we need to understand the behavior of the sine function.
The sine function reaches its maximum value of 1 at specific points within each period. For the sine function, these points are given by:
[tex]\[ \theta = \frac{\pi}{2} + 2k\pi \][/tex]
where [tex]\(k\)[/tex] is any integer. Essentially, this formula tells us that the function [tex]\(y = \sin \theta\)[/tex] is equal to 1 at [tex]\(\frac{\pi}{2}\)[/tex] plus any integer multiple of the period of the sine function [tex]\(2\pi\)[/tex].
We need to find all such [tex]\(\theta\)[/tex] values within the interval [tex]\(0 \leq \theta \leq 8\pi\)[/tex]. Let's calculate these step-by-step by substituting values for [tex]\(k\)[/tex]:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[ \theta = \frac{\pi}{2} + 2 \cdot 0 \cdot \pi = \frac{\pi}{2} \][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{2} + 2 \cdot 1 \cdot \pi = \frac{\pi}{2} + 2\pi = \frac{\pi}{2} + \frac{4\pi}{2} = \frac{5\pi}{2} = 2 \pi + \frac{\pi}{2} \][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[ \theta = \frac{\pi}{2} + 2 \cdot 2 \cdot \pi = \frac{\pi}{2} + 4\pi = \frac{\pi}{2} + \frac{8\pi}{2} = \frac{9\pi}{2} = 4 \pi + \frac{\pi}{2} \][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[ \theta = \frac{\pi}{2} + 2 \cdot 3 \cdot \pi = \frac{\pi}{2} + 6\pi = \frac{\pi}{2} + \frac{12\pi}{2} = \frac{13\pi}{2} = 6 \pi + \frac{\pi}{2} \][/tex]
5. For [tex]\(k = 4\)[/tex]:
[tex]\[ \theta = \frac{\pi}{2} + 2 \cdot 4 \cdot \pi = \frac{\pi}{2} + 8\pi = \frac{\pi}{2} + \frac{16\pi}{2} = \frac{17\pi}{2} = 8 \pi + \frac{\pi}{2} \][/tex]
Next, let's list out these values:
[tex]\[ \theta = \frac{\pi}{2}, 2 \pi + \frac{\pi}{2}, 4 \pi + \frac{\pi}{2}, 6 \pi + \frac{\pi}{2}, 8 \pi + \frac{\pi}{2} \][/tex]
These are all the points where [tex]\(y = \sin \theta\)[/tex] equals 1 in the given interval [tex]\(0 \leq \theta \leq 8\pi\)[/tex].
Comparing with the given options, the correct option is the following:
[tex]\[ \theta = \frac{\pi}{2}, 2 \pi + \frac{\pi}{2}, 4 \pi + \frac{\pi}{2}, 6 \pi + \frac{\pi}{2}, 8 \pi + \frac{\pi}{2} \][/tex]
Therefore, the correct answer is:
[tex]\(\theta = \frac{\pi}{2}, 2 \pi + \frac{\pi}{2}, 4 \pi + \frac{\pi}{2}, 6 \pi + \frac{\pi}{2}, 8 \pi + \frac{\pi}{2}\)[/tex]
The sine function reaches its maximum value of 1 at specific points within each period. For the sine function, these points are given by:
[tex]\[ \theta = \frac{\pi}{2} + 2k\pi \][/tex]
where [tex]\(k\)[/tex] is any integer. Essentially, this formula tells us that the function [tex]\(y = \sin \theta\)[/tex] is equal to 1 at [tex]\(\frac{\pi}{2}\)[/tex] plus any integer multiple of the period of the sine function [tex]\(2\pi\)[/tex].
We need to find all such [tex]\(\theta\)[/tex] values within the interval [tex]\(0 \leq \theta \leq 8\pi\)[/tex]. Let's calculate these step-by-step by substituting values for [tex]\(k\)[/tex]:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[ \theta = \frac{\pi}{2} + 2 \cdot 0 \cdot \pi = \frac{\pi}{2} \][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{2} + 2 \cdot 1 \cdot \pi = \frac{\pi}{2} + 2\pi = \frac{\pi}{2} + \frac{4\pi}{2} = \frac{5\pi}{2} = 2 \pi + \frac{\pi}{2} \][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[ \theta = \frac{\pi}{2} + 2 \cdot 2 \cdot \pi = \frac{\pi}{2} + 4\pi = \frac{\pi}{2} + \frac{8\pi}{2} = \frac{9\pi}{2} = 4 \pi + \frac{\pi}{2} \][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[ \theta = \frac{\pi}{2} + 2 \cdot 3 \cdot \pi = \frac{\pi}{2} + 6\pi = \frac{\pi}{2} + \frac{12\pi}{2} = \frac{13\pi}{2} = 6 \pi + \frac{\pi}{2} \][/tex]
5. For [tex]\(k = 4\)[/tex]:
[tex]\[ \theta = \frac{\pi}{2} + 2 \cdot 4 \cdot \pi = \frac{\pi}{2} + 8\pi = \frac{\pi}{2} + \frac{16\pi}{2} = \frac{17\pi}{2} = 8 \pi + \frac{\pi}{2} \][/tex]
Next, let's list out these values:
[tex]\[ \theta = \frac{\pi}{2}, 2 \pi + \frac{\pi}{2}, 4 \pi + \frac{\pi}{2}, 6 \pi + \frac{\pi}{2}, 8 \pi + \frac{\pi}{2} \][/tex]
These are all the points where [tex]\(y = \sin \theta\)[/tex] equals 1 in the given interval [tex]\(0 \leq \theta \leq 8\pi\)[/tex].
Comparing with the given options, the correct option is the following:
[tex]\[ \theta = \frac{\pi}{2}, 2 \pi + \frac{\pi}{2}, 4 \pi + \frac{\pi}{2}, 6 \pi + \frac{\pi}{2}, 8 \pi + \frac{\pi}{2} \][/tex]
Therefore, the correct answer is:
[tex]\(\theta = \frac{\pi}{2}, 2 \pi + \frac{\pi}{2}, 4 \pi + \frac{\pi}{2}, 6 \pi + \frac{\pi}{2}, 8 \pi + \frac{\pi}{2}\)[/tex]
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