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Sagot :
Alright, let's resolve the given expression step-by-step. We are tasked with finding the inverse sine (also known as the arcsine) of the following expression:
[tex]\[ \sin^{-1}\left(\frac{a + b \cos n}{b + a \cos n}\right) \][/tex]
Here, [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(n\)[/tex] are variables, and it is given that [tex]\(b > a\)[/tex].
Step-by-Step Solution:
1. Identify the input of the arcsine function:
The expression inside the inverse sine function is:
[tex]\[ \frac{a + b \cos n}{b + a \cos n} \][/tex]
2. Understand the domain of the inverse sine function:
The function [tex]\(\sin^{-1}(x)\)[/tex] or [tex]\(\arcsin(x)\)[/tex] is defined for [tex]\(-1 \leq x \leq 1\)[/tex].
3. Validate the value falls within the domain:
For [tex]\(b > a\)[/tex], the expression [tex]\(\frac{a + b \cos n}{b + a \cos n}\)[/tex] should lie within the domain of the inverse sine function, as it calculates a ratio involving trigonometric functions which inherently produce values between -1 and 1 inclusive, assuming typical angles [tex]\(n\)[/tex].
4. Write the final expression:
Given the constraints and the expression, the arcsine will yield the angle [tex]\( \theta \)[/tex] such that:
[tex]\[ \theta = \sin^{-1}\left(\frac{a + b \cos n}{b + a \cos n}\right) \][/tex]
Therefore,
[tex]\[ \sin^{-1}\left(\frac{a + b \cos n}{b + a \cos n}\right) \][/tex]
Is the final simplified form of the expression, representing the angle whose sine value corresponds to the given ratio.
Thus, the detailed step-by-step solution ends with the expression:
[tex]\[ \boxed{\sin^{-1}\left(\frac{a + b \cos n}{b + a \cos n}\right)} \][/tex]
[tex]\[ \sin^{-1}\left(\frac{a + b \cos n}{b + a \cos n}\right) \][/tex]
Here, [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(n\)[/tex] are variables, and it is given that [tex]\(b > a\)[/tex].
Step-by-Step Solution:
1. Identify the input of the arcsine function:
The expression inside the inverse sine function is:
[tex]\[ \frac{a + b \cos n}{b + a \cos n} \][/tex]
2. Understand the domain of the inverse sine function:
The function [tex]\(\sin^{-1}(x)\)[/tex] or [tex]\(\arcsin(x)\)[/tex] is defined for [tex]\(-1 \leq x \leq 1\)[/tex].
3. Validate the value falls within the domain:
For [tex]\(b > a\)[/tex], the expression [tex]\(\frac{a + b \cos n}{b + a \cos n}\)[/tex] should lie within the domain of the inverse sine function, as it calculates a ratio involving trigonometric functions which inherently produce values between -1 and 1 inclusive, assuming typical angles [tex]\(n\)[/tex].
4. Write the final expression:
Given the constraints and the expression, the arcsine will yield the angle [tex]\( \theta \)[/tex] such that:
[tex]\[ \theta = \sin^{-1}\left(\frac{a + b \cos n}{b + a \cos n}\right) \][/tex]
Therefore,
[tex]\[ \sin^{-1}\left(\frac{a + b \cos n}{b + a \cos n}\right) \][/tex]
Is the final simplified form of the expression, representing the angle whose sine value corresponds to the given ratio.
Thus, the detailed step-by-step solution ends with the expression:
[tex]\[ \boxed{\sin^{-1}\left(\frac{a + b \cos n}{b + a \cos n}\right)} \][/tex]
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