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Sagot :
To find [tex]\(\sin ^{-1} \left(\sin \frac{2\pi }{3}\right)\)[/tex], we need to understand the properties of the sine and inverse sine functions.
Step-by-Step Solution:
1. Understand the given argument of the sine function:
[tex]\[ \theta = \frac{2\pi }{3} \][/tex]
We start with this angle, which is in radians. This angle lies in the second quadrant of the unit circle, where sine is positive.
2. Calculate the sine of [tex]\(\theta\)[/tex]:
Let’s find the sine of [tex]\(\theta\)[/tex]:
[tex]\[ \sin \left( \frac{2\pi }{3} \right) = \sin \left( \pi - \frac{\pi }{3} \right) = \sin \left( \frac{\pi}{3} \right) \][/tex]
Knowing that [tex]\(\sin \left( \frac{\pi }{3} \right) = \frac{\sqrt{3}}{2}\)[/tex], we get:
[tex]\[ \sin \left( \frac{2\pi }{3} \right) = \frac{\sqrt{3}}{2} \][/tex]
3. Apply the inverse sine function:
Now, we need to find the inverse sine of [tex]\(\frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[ \sin ^{-1} \left( \sin \frac{2\pi }{3} \right) = \sin ^{-1} \left( \frac{\sqrt{3}}{2} \right) \][/tex]
The range of the inverse sine function, [tex]\(\sin ^{-1} (x)\)[/tex], is [tex]\([- \frac{\pi}{2}, \frac{\pi}{2}]\)[/tex]. We need to find an angle [tex]\(\phi\)[/tex] in this range such that:
[tex]\[ \sin (\phi) = \frac{\sqrt{3}}{2} \][/tex]
The angle [tex]\(\phi = \frac{\pi }{3}\)[/tex] satisfies this condition, as:
[tex]\[ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \][/tex]
Within the range of the inverse sine function, the angle [tex]\(\phi\)[/tex] is thus:
[tex]\[ \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} \][/tex]
4. Adjust for the principal range:
Since [tex]\(\frac{2\pi }{3}\)[/tex] is outside the principal range of [tex]\(\sin^{-1}(x)\)[/tex], we must adjust our result to fit within [tex]\([- \frac{\pi}{2}, \frac{\pi}{2}]\)[/tex]. While [tex]\(\sin (\frac{2\pi }{3})\)[/tex] and [tex]\(\sin (\frac{\pi}{3})\)[/tex] yield the same sine value, [tex]\(\frac{\pi}{3}\)[/tex] is within the range of inverse sine.
To account for this properly:
[tex]\[ \sin^{-1} \left( \sin \frac{2\pi }{3} \right) = \frac{\pi}{3} \][/tex]
In conclusion:
[tex]\[ \sin^{-1} \left( \sin \frac{2\pi }{3} \right) = \frac{\pi}{3} \][/tex]
Step-by-Step Solution:
1. Understand the given argument of the sine function:
[tex]\[ \theta = \frac{2\pi }{3} \][/tex]
We start with this angle, which is in radians. This angle lies in the second quadrant of the unit circle, where sine is positive.
2. Calculate the sine of [tex]\(\theta\)[/tex]:
Let’s find the sine of [tex]\(\theta\)[/tex]:
[tex]\[ \sin \left( \frac{2\pi }{3} \right) = \sin \left( \pi - \frac{\pi }{3} \right) = \sin \left( \frac{\pi}{3} \right) \][/tex]
Knowing that [tex]\(\sin \left( \frac{\pi }{3} \right) = \frac{\sqrt{3}}{2}\)[/tex], we get:
[tex]\[ \sin \left( \frac{2\pi }{3} \right) = \frac{\sqrt{3}}{2} \][/tex]
3. Apply the inverse sine function:
Now, we need to find the inverse sine of [tex]\(\frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[ \sin ^{-1} \left( \sin \frac{2\pi }{3} \right) = \sin ^{-1} \left( \frac{\sqrt{3}}{2} \right) \][/tex]
The range of the inverse sine function, [tex]\(\sin ^{-1} (x)\)[/tex], is [tex]\([- \frac{\pi}{2}, \frac{\pi}{2}]\)[/tex]. We need to find an angle [tex]\(\phi\)[/tex] in this range such that:
[tex]\[ \sin (\phi) = \frac{\sqrt{3}}{2} \][/tex]
The angle [tex]\(\phi = \frac{\pi }{3}\)[/tex] satisfies this condition, as:
[tex]\[ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \][/tex]
Within the range of the inverse sine function, the angle [tex]\(\phi\)[/tex] is thus:
[tex]\[ \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} \][/tex]
4. Adjust for the principal range:
Since [tex]\(\frac{2\pi }{3}\)[/tex] is outside the principal range of [tex]\(\sin^{-1}(x)\)[/tex], we must adjust our result to fit within [tex]\([- \frac{\pi}{2}, \frac{\pi}{2}]\)[/tex]. While [tex]\(\sin (\frac{2\pi }{3})\)[/tex] and [tex]\(\sin (\frac{\pi}{3})\)[/tex] yield the same sine value, [tex]\(\frac{\pi}{3}\)[/tex] is within the range of inverse sine.
To account for this properly:
[tex]\[ \sin^{-1} \left( \sin \frac{2\pi }{3} \right) = \frac{\pi}{3} \][/tex]
In conclusion:
[tex]\[ \sin^{-1} \left( \sin \frac{2\pi }{3} \right) = \frac{\pi}{3} \][/tex]
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