Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of 16((sec−1x)2+(csc−1x)2) is:
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Step-by-Step Solution
Step 1: Identify the domain and properties of inverse trigonometric functions
We know the identity for inverse secant and cosecant functions: sec−1x+csc−1x=2π. This identity holds true for all x in the domain where both functions are defined, which is ∣x∣≥1. This relationship will be crucial for simplifying the given expression.
Step 2: Express one inverse function in terms of the other
From the identity, we can express csc−1x in terms of sec−1x. This substitution will allow us to rewrite the given expression in terms of a single variable, making it easier to find its maximum and minimum values.
Step 3: Substitute into the given expression
Now, we substitute the expression for csc−1x into the given function. Let y=sec−1x. The expression becomes 16(y2+(2π−y)2).
Step 4: Simplify the expression
Expand the squared term and combine like terms. This simplifies the expression into a quadratic function of y.
Step 5: Determine the range of y=sec−1x
The principal value range for sec−1x is [0,π] excluding 2π. This means y can take values from 0 up to π, but not exactly 2π.
Step 6: Find the maximum and minimum values of the quadratic function
The function is f(y)=16(2y2−πy+4π2)=32y2−16πy+4π2. This is a parabola opening upwards. Its vertex is at y=−2(32)−16π=6416π=4π. Since 4π is within the domain of y, the minimum value occurs at y=4π. The minimum value is f(4π)=32(4π)2−16π(4π)+4π2=3216π2−4π2+4π2=2π2. The maximum value occurs at the endpoints of the domain, y=0 or y=π. f(0)=4π2 and f(π)=32π2−16π2+4π2=20π2. Thus, the maximum value is 20π2.
Step 7: Calculate the sum of maximum and minimum values
The sum of the maximum and minimum values is 20π2+2π2=22π2.