## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.3 - Right Triangle Trigonometry - Exercise Set - Page 560: 14

#### Answer

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

#### Work Step by Step

Convert the angle measure to degrees by multiplying $\dfrac{180^o}{\pi}$ to obtain: $\dfrac{\pi}{3} \cdot \dfrac{180^o}{\pi} = 60^o$ RECALL: (1) $\sin{\theta} = \dfrac{\text{opposite side}}{\text{hypotenuse}}$ (2) $\tan{\theta} = \dfrac{\text{opposite side}}{\text{adjacent side}}$ (3) $\cos{\theta} = \dfrac{\text{adjacent side}}{\text{hypotenuse}}$ (4) $\sec{\theta} = \dfrac{\text{hypotenuse}}{\text{adjacent}}$ (5) $\csc{\theta} = \dfrac{\text{hypotenuse}}{\text{opposite side}}$ (6) $\cot{\theta}=\dfrac{\text{adjacent side}}{\text{opposite side}}$ Use formula (6) above. Use the 60-degree angle of the triangle on the right to obtain: $\cot{\frac{\pi}{3}} = \cot{60^o} = \dfrac{1}{\sqrt3}$ Rationalize the denominator by multiplying $\sqrt3$ to both the numerator and the denominator to obtain: $=\dfrac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} \\=\dfrac{\sqrt{3}}{3}$

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