# Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 626: 54

$\dfrac{\sqrt 3}{2}$

#### Work Step by Step

Suppose $\theta =\sin^{-1} (\dfrac{1}{2})$ This gives: $\sin \theta=\dfrac{1}{2}$ Since, $r=\sqrt {x^2+y^2}$ This implies that $x=\sqrt {r^2-y^2}=\sqrt {2^2-(1)^2}=\sqrt 3$ Therefore, we have $\cos \theta =\cos [\sin^{-1} (\dfrac{1}{2})]=\dfrac{\sqrt 3}{2}$

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