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To solve the problem [tex]\(\operatorname {sin}^{-1}\left( \operatorname {sin} \frac{3 \pi}{5}\right)\)[/tex], we can break it down into a few essential steps and understand the key trigonometric principles involved.
### Step 1: Calculate the Inside Angle
The given angle inside the sine function is [tex]\(\frac{3\pi}{5}\)[/tex]. Converting this to a numerical value in radians:
[tex]\[ \frac{3\pi}{5} \approx 1.8849555921538759 \][/tex]
### Step 2: Compute the Sine of the Angle
Next, we find the sine of the angle [tex]\(\frac{3\pi}{5}\)[/tex]. Using the sine trigonometric function:
[tex]\[ \sin\left(\frac{3\pi}{5}\right) \approx 0.9510565162951536 \][/tex]
### Step 3: Apply the Inverse Sine Function
Now, we need to find the angle whose sine is [tex]\(0.9510565162951536\)[/tex]. This is done using the inverse sine function, also known as [tex]\(\arcsin\)[/tex] or [tex]\(\sin^{-1}\)[/tex]:
[tex]\[ \sin^{-1}(0.9510565162951536) \approx 1.2566370614359175 \][/tex]
### Step 4: Consider the Principal Range of [tex]\(\arcsin\)[/tex]
The [tex]\(\arcsin\)[/tex] function provides an angle in the principal range [tex]\([- \frac{\pi}{2}, \frac{\pi}{2}]\)[/tex]. It's important to observe that the original angle [tex]\(\frac{3\pi}{5}\)[/tex] is outside this principal range, so [tex]\(\sin^{-1}\left( \sin \frac{3 \pi}{5}\right)\)[/tex] brings it back into the principal range:
[tex]\[ \frac{3 \pi}{5} \approx 1.8849555921538759 \quad \text{(outside)} \\ \arcsin(0.9510565162951536) \approx 1.2566370614359175 \quad \text{(inside \([- \frac{\pi}{2}, \frac{\pi}{2}])} \][/tex]
### Conclusion
Thus, the angle whose sine is equivalent to the sine of [tex]\(\frac{3\pi}{5}\)[/tex] and lies within the principal range of the inverse sine function is approximately:
[tex]\[ \boxed{1.2566370614359175} \][/tex]
### Step 1: Calculate the Inside Angle
The given angle inside the sine function is [tex]\(\frac{3\pi}{5}\)[/tex]. Converting this to a numerical value in radians:
[tex]\[ \frac{3\pi}{5} \approx 1.8849555921538759 \][/tex]
### Step 2: Compute the Sine of the Angle
Next, we find the sine of the angle [tex]\(\frac{3\pi}{5}\)[/tex]. Using the sine trigonometric function:
[tex]\[ \sin\left(\frac{3\pi}{5}\right) \approx 0.9510565162951536 \][/tex]
### Step 3: Apply the Inverse Sine Function
Now, we need to find the angle whose sine is [tex]\(0.9510565162951536\)[/tex]. This is done using the inverse sine function, also known as [tex]\(\arcsin\)[/tex] or [tex]\(\sin^{-1}\)[/tex]:
[tex]\[ \sin^{-1}(0.9510565162951536) \approx 1.2566370614359175 \][/tex]
### Step 4: Consider the Principal Range of [tex]\(\arcsin\)[/tex]
The [tex]\(\arcsin\)[/tex] function provides an angle in the principal range [tex]\([- \frac{\pi}{2}, \frac{\pi}{2}]\)[/tex]. It's important to observe that the original angle [tex]\(\frac{3\pi}{5}\)[/tex] is outside this principal range, so [tex]\(\sin^{-1}\left( \sin \frac{3 \pi}{5}\right)\)[/tex] brings it back into the principal range:
[tex]\[ \frac{3 \pi}{5} \approx 1.8849555921538759 \quad \text{(outside)} \\ \arcsin(0.9510565162951536) \approx 1.2566370614359175 \quad \text{(inside \([- \frac{\pi}{2}, \frac{\pi}{2}])} \][/tex]
### Conclusion
Thus, the angle whose sine is equivalent to the sine of [tex]\(\frac{3\pi}{5}\)[/tex] and lies within the principal range of the inverse sine function is approximately:
[tex]\[ \boxed{1.2566370614359175} \][/tex]
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