Math Problem Statement

Sin^-1(sin7pi/6)

Solution

To solve sin1(sin(7π6))\sin^{-1}(\sin(\frac{7\pi}{6})), we first simplify sin(7π6)\sin(\frac{7\pi}{6}).

  1. 7π6\frac{7\pi}{6} is in the third quadrant, where sine is negative.
  2. The reference angle for 7π6\frac{7\pi}{6} is π6\frac{\pi}{6}, so: sin(7π6)=sin(π6)=12.\sin\left(\frac{7\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}.

Now, we need to find sin1(12)\sin^{-1}(-\frac{1}{2}). The principal value of the inverse sine function, sin1\sin^{-1}, returns values in the range [π2,π2][- \frac{\pi}{2}, \frac{\pi}{2}]. The angle in this range whose sine is 12-\frac{1}{2} is π6-\frac{\pi}{6}.

Thus: sin1(sin(7π6))=π6.\sin^{-1}(\sin(\frac{7\pi}{6})) = -\frac{\pi}{6}.

Would you like further details or have any questions?

Here are 5 related questions:

  1. How do you determine the reference angle for any given angle?
  2. What are the properties of the inverse sine function?
  3. How do you convert between radians and degrees for trigonometric functions?
  4. What are the sine values for standard angles in all four quadrants?
  5. How does the unit circle help in solving trigonometric equations?

Tip: Always remember the range of the inverse sine function is [π2,π2][- \frac{\pi}{2}, \frac{\pi}{2}], so it only gives values from the first and fourth quadrants.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Inverse Trigonometric Functions
Unit Circle
Reference Angle

Formulas

sin(θ) = opposite/hypotenuse
sin^-1(x) is the inverse sine function
Range of sin^-1: [-π/2, π/2]

Theorems

Reference Angle Theorem
Principal Value Theorem for Inverse Trigonometric Functions

Suitable Grade Level

Grades 10-12