# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 264: 25

$\sin^{-1}\sqrt{3}\ \ \$ is not defined.

#### Work Step by Step

Inverse Sine Function: $y=\sin^{-1}x$ or $y=$ arcsin $x$ means that $x=\sin y$, for $-\displaystyle \frac{\pi}{2} \leq y \leq \frac{\pi}{2}$. ------------------- Since $\sqrt{3} > 1$, there is no y such that $\sin y=\sqrt{3}$ $y=\sin^{-1}\sqrt{3}$ is not defined. ($\sin^{-1}$ is the inverse of $\sin,$ so its domain must be the range of sine. The range of sine is $[-1,1].$ $\sqrt{3}\not\in [-1,1] )$

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