Math Problem Statement

Find the exact value of the real number y. Y= sin^-1 (1)

Solution

To find the exact value of y=sin1(1)y = \sin^{-1}(1), we need to determine the angle yy for which sin(y)=1\sin(y) = 1.

The sine of an angle is equal to 1 at y=π2y = \frac{\pi}{2} radians (or 90°) within the principal range of the inverse sine function, which is π2yπ2-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}.

Thus, the exact value of yy is:

y=π2y = \frac{\pi}{2}

Would you like more details on inverse trigonometric functions or their ranges?

Here are 5 related questions you might explore:

  1. What is the value of sin1(0)\sin^{-1}(0)?
  2. What is the range of the inverse sine function?
  3. How would you solve y=sin1(1)y = \sin^{-1}(-1)?
  4. Can you find the derivative of sin1(x)\sin^{-1}(x)?
  5. What are the values of sin1(12)\sin^{-1}(\frac{1}{2})?

Tip: Inverse trigonometric functions can help solve equations involving angles and their trigonometric values. Always remember the principal range when dealing with inverse functions.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Inverse Trigonometric Functions

Formulas

y = sin^-1(1)
sin(π/2) = 1

Theorems

Inverse Sine Function
Principal Range of Inverse Sine

Suitable Grade Level

Grades 9-11