# Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 21

$\frac{\sqrt{5}}{3}.$

#### Work Step by Step

Assume that $\sin\theta =\frac{2}{3}$, then by solving the triangle, we have: $$\cos \theta=\frac{\sqrt{5}}{3}.$$ Now, since $\theta =\sin^{-1}\frac{2}{3}$, we have $$\cos(\sin^{-1}\frac{2}{3})=\cos\theta=\frac{\sqrt{5}}{3}.$$

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