The expression arccos(sin(x)) is interesting because it involves the inverse cosine function applied to the sine of x. To find its exact value, we can analyze this step by step:
1. **Understanding the Functions**: The sin(x) function outputs values within the range of -1 to 1. This means we can safely apply the arccos function because arccos is defined for inputs between -1 and 1. The arccos function itself returns an angle in the range of 0 to rac{ an^{-1}(1)}{1} rac{ an^{-1}(1)}{1} (or rac{ rac{ an^{-1}{1}}{1} rac{ an^{-1}{1}}{1} rac{ an^{-1}(1)}{1}}{1}) radians.
2. **Finding the Exact Value**: To express arccos(sin(x)), we first need to consider the relationship between sine and cosine. One key relationship is that:
sin(x) = cos(π/2 – x)
Thus, we can express everything in terms of cos:
arccos(sin(x)) = arccos(cos(π/2 - x))
3. **Applying Inverse Functions**: The property of inverse functions states that:
arccos(cos(θ)) = θ for angles θ in the range of [0, π].
So we need to determine whether (π/2 - x) falls within this range.
4. **Determining Range**: Depending on the value of x:
- If
0 ≤ x ≤ π/2, then0 ≤ π/2 - x ≤ π/2. In this case: arccos(sin(x)) = π/2 - x- If
π/2 < x ≤ π, then0 < π/2 - x < -π/2This results in: arccos(sin(x)) = π + (π/2 - x)- If
x > π, similarly, you would subtractπand adjust accordingly.
5. **Final Expression**: Therefore, we can summarize the results:
arccos(sin(x)) = π/2 - xfor0 ≤ x ≤ π/2arccos(sin(x)) = 3π/2 - xforπ/2 < x ≤ 3π/2
In conclusion, the exact value of arccos(sin(x)) varies depending on the range of x. Understanding this relationship is essential for solving problems involving trigonometric identities and inverse functions.