Chapter 7.5, Problem 62E

To determine

To find: the exact value of the given trigonometric function.

Answer to Problem 62E

  $\sqrt{3-2\sqrt{2}}$

Explanation of Solution

Given information:

  $\cot 67.5°$

Calculation:

The value of $\cot 67.5°$ can be evaluated as follows:

  $\begin{array}{c}\cot 67.5°=\cot \left(\raisebox{1ex}{$135°$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\left[\text{Rewrite in the form of half angle}\right]\\ =\frac{1}{\tan \left(\raisebox{1ex}{$135°$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}\left[\text{Using reciprocal identity}\right]\\ =\frac{1}{\sqrt{\frac{1-\cos 135°}{1+\cos 135°}}}\left[\begin{array}{l}\text{Using half angle identity for tangent}\text{.}\\ \text{Tangent and cotangent are positive on quadrant I}\text{.}\end{array}\right]\\ =\sqrt{\frac{1+\cos 135°}{1-\cos 135°}}\\ =\sqrt{\frac{1+\left(\raisebox{1ex}{$-\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{1-\left(\raisebox{1ex}{$-\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}}\\ =\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}\left(\frac{2-\sqrt{2}}{2-\sqrt{2}}\right)}\left[\text{Multiply and divide by }2-\sqrt{2}\right]\\ =\sqrt{\frac{6-4\sqrt{2}}{2}}\\ =\sqrt{3-2\sqrt{2}}\end{array}$

Thus the value of $\cot 67.5°$is $\sqrt{3-2\sqrt{2}}$ .