Chapter 7.3, Problem 23E

To determine

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

Answer to Problem 23E

  $\sqrt{2}-\sqrt{6}$

Explanation of Solution

Given information:

  $\sec 1275°$

Calculation:

Simplify the function as follows:

  $\begin{array}{c}\sec 1275°=\sec \left(3\left(360°\right)-195°\right)\left[\text{Rewritting sec127}5°\right]\end{array}$

  $\begin{array}{c}=\sec 195°\end{array}$

  $\begin{array}{c}=\sec \left(180°+15°\right)\left[\text{Rewritting sec19}5°\right]\end{array}$

  $\begin{array}{c}=-\sec \left(15°\right)\end{array}$

  $\begin{array}{c}=-\sec \left(45°-30°\right)\left[\text{Rewritting sec15}°\right]\end{array}$

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

Now apply the difference identity for cosine in the above simplified function as shown below.

  $\sec 1275°=-\frac{1}{\cos 45°\cos 30°+\sin 45°\sin 30}$

Put the values for the trigonometric function and evaluate $1275°$ as follows:

  $\begin{array}{c}\sec 1275°=-\frac{1}{\left(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}\right)}\end{array}$

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

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

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

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

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

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

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

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

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