Chapter 7.3, Problem 25E

To determine

To find: the exact value of the function by using sum or difference identities.

Answer to Problem 25E

  $2-\sqrt{3}$

Explanation of Solution

Given information:

  $\cot \frac{113\pi }{12}$

Calculation:

Simplify the function as follows:

  $\begin{array}{c}\cot \frac{113\pi }{12}=\cot \left(\frac{113\pi }{12}\left(\frac{180°}{\pi }\right)\right)\left[\text{Convert into degree measure}\right]\\ =\cot 1695°\\ =\cot \left(5\left(360°\right)-105°\right)\left[\text{Rewritting }\cot 1695°\right]\\ =-\cot 105°\\ =-\cot \left(45°+60°\right)\left[\text{Rewritting }\cot 105°\right]\\ =-\frac{1}{\tan \left(45°+60°\right)}\left[\text{Reciprocal identity}\right]\end{array}$

Now apply the sum identity for tangent in the above simplified function as shown below.

  $\cot \frac{113\pi }{12}=-\left[\frac{1}{\left(\frac{\tan 45°+\tan 60°}{1-\tan 45°\cdot \tan 60}\right)}\right]$

Put the values for the trigonometric function and evaluate $\cot \frac{113\pi }{12}$ as follows:

  $\begin{array}{c}\cot \frac{113\pi }{12}=-\frac{1}{\left(\frac{1+\sqrt{3}}{1-1\cdot \sqrt{3}}\right)}\\ =-\frac{1}{\left(\frac{1+\sqrt{3}}{1-\sqrt{3}}\right)}\\ =-\frac{1-\sqrt{3}}{1+\sqrt{3}}\left(\frac{1-\sqrt{3}}{1-\sqrt{3}}\right)\left[\text{Multiply and divide by }\left(1-\sqrt{3}\right)\right]\\ =-\frac{\left(1+3-2\sqrt{3}\right)}{1-3}\left[\begin{array}{l}\because {\left(a-b\right)}^2=a^2+b^2-2ab\\ \because \left(a-b\right)\left(a+b\right)=a^2-b^2\end{array}\right]\\ =\frac{4-2\sqrt{3}}{2}\\ =2-\sqrt{3}\end{array}$

Thus the exact value of the given function is $2-\sqrt{3}$ .