Chapter 7.3, Problem 24E

To determine

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

Answer to Problem 24E

  $\sqrt{6}-\sqrt{2}$

Explanation of Solution

Given information:

  $\csc \frac{5\pi }{12}$

Calculation:

Simplify the function as follows:

  $\begin{array}{c}\csc \frac{5\pi }{12}=\csc \left(\left(\frac{5\pi }{12}\right)\left(\frac{180°}{\pi }\right)\right)\left[\text{Convert into degree measure}\right]\end{array}$

  $\begin{array}{c}=\csc 75°\end{array}$

  $\begin{array}{c}=\csc \left(45°+30°\right)\left[\text{Rewritting csc75}°\right]\end{array}$

  $\begin{array}{c}=\frac{1}{\sin \left(45°+30°\right)}\left[\text{Reciprocal identity}\right]\end{array}$

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

  $\csc \frac{5\pi }{12}=\frac{1}{\sin 45°\cos 30°+\cos 45°\sin 30}$

Put the values for the trigonometric function and evaluate $\csc \frac{5\pi }{12}$ as follows:

  $\begin{array}{c}\csc \frac{5\pi }{12}=\frac{1}{\left(\frac{\sqrt{2}}{2}\times \frac{\sqrt{3}}{2}\right)+\left(\frac{\sqrt{2}}{2}\times \frac{1}{2}\right)}\end{array}$

  $\begin{array}{c}=\frac{1}{\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}}\end{array}$

  $\begin{array}{c}=\frac{1}{\frac{\sqrt{6}+\sqrt{2}}{4}}\end{array}$

  $\begin{array}{c}=\frac{4}{\sqrt{6}+\sqrt{2}}\end{array}$

  $\begin{array}{c}=\left(\frac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}\right)\left[\text{Multiply and divide by }\sqrt{6}-\sqrt{2}\right]\end{array}$

  $\begin{array}{c}=\frac{4\left(\sqrt{6}-\sqrt{2}\right)}{6-2}\left[\text{Using }\left(a-b\right)\left(a+b\right)=a^2-b^2\right]\end{array}$

  $\begin{array}{c}=\frac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}\end{array}$

  $\begin{array}{c}=\sqrt{6}-\sqrt{2}\end{array}$

Thus the exact value of the given function is $\sqrt{6}-\sqrt{2}$ .