Chapter 7.3, Problem 7CFU

To determine

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

Answer to Problem 7CFU

  $\sqrt{6}+\sqrt{2}$

Explanation of Solution

Given information:

  $\sec 795°$

Calculation:

Simplify the function as follows:

  $\begin{array}{c}\sec 795°=\sec \left(2\left(360°\right)-75°\right)\left[\text{Rewritting }795°\right]\\ =\sec 75°\\ =\sec \left(45°+30°\right)\left[\text{Rewritting 75}°\right]\\ =\frac{1}{\cos \left(45°+30°\right)}\left[\text{Reciprocal identity}\right]\end{array}$

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

  $\sec 795°=\frac{1}{\cos 45°\cos 30°-\sin 45°\sin 30}\left[\begin{array}{l}\text{Use sum identity }\\ \text{of}\cos \left(\alpha +\beta \right)\end{array}\right]$

Put the values for the trigonometric function and evaluate $\sec 795°$ as follows:

  $\begin{array}{c}\sec 795°=\frac{1}{\left(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}\right)}\\ =\frac{1}{\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}}\\ =\frac{1}{\frac{\sqrt{6}-\sqrt{2}}{4}}\\ =\frac{4}{\sqrt{6}-\sqrt{2}}\\ =\left(\frac{4}{\sqrt{6}-\sqrt{2}}\right)\left(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}\right)\left[\text{Multiply and divide by }\sqrt{6}+\sqrt{2}\right]\\ =\frac{4\left(\sqrt{6}+\sqrt{2}\right)}{6-2}\left[\text{Using }\left(a-b\right)\left(a+b\right)=a^2-b^2\right]\\ =\frac{4\left(\sqrt{6}+\sqrt{2}\right)}{4}\\ =\sqrt{6}+\sqrt{2}\end{array}$

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