Chapter 5.4, Problem 45E

To determine

To calculate: The value of the trigonometric expression $\sec \left(v-u\right)$ provided that $\sin u=-\frac{3}{5}$ and $\cos v=\frac{15}{17}$ . Interval in which u and v lies are $\frac{3\pi }{2}<u<2\pi$ and $0<v<\frac{\pi }{2}$ .

Answer to Problem 45E

The value of the trigonometric expression $\sec \left(v-u\right)$ is $\frac{85}{36}$ .

Explanation of Solution

Given information:

The values,$\sin u=-\frac{3}{5}$ and $\cos v=\frac{15}{17}$ . Interval in which u and v lies are $\frac{3\pi }{2}<u<2\pi$ and $0<v<\frac{\pi }{2}$ .

Formula used:

The Pythagorean identity,${\sin }^{2}x+{\cos }^{2}x=1$ .

Conversion rule,$\sec \left(v-u\right)=\frac{1}{\cos \left(v-u\right)}$

Difference rule for cosine function $\cos \left(v-u\right)=\cos v\cos u+\sin v\sin u$ .

Calculation:

Consider the provided values,$\sin u=-\frac{3}{5}$ and $\cos v=\frac{15}{17}$ . Interval in which u and v lies are $\frac{3\pi }{2}<u<2\pi$ and $0<v<\frac{\pi }{2}$ .

Recall that sine function is negative and cosine function is positive in the interval $\left[\frac{3\pi }{2},2\pi \right]$ .

Recall that sine function is positive and cosine function is positive in the interval $\left[0,\frac{\pi }{2}\right]$ .

Recall that the Pythagorean identity,${\sin }^{2}x+{\cos }^{2}x=1$ .

Since,$\sin u=-\frac{3}{5}$ , so,$\cos u$ is evaluated as,

  $\begin{array}{c}{\sin }^{2}u+{\cos }^{2}u=1\\ {\left(\frac{-3}{5}\right)}^2+{\cos }^{2}u=1\\ {\cos }^{2}u=1-\frac{9}{25}\\ {\cos }^{2}u=\frac{16}{25}\end{array}$

Simplify it further as,

  $\begin{array}{c}\cos u=\pm \sqrt{\frac{16}{25}}\\ \cos u=\pm \frac{4}{5}\end{array}$

Since,$\frac{3\pi }{2}<u<2\pi$ and cosine function is positive in the interval so,$\cos u=\frac{4}{5}$ .

Since,$\cos v=\frac{15}{17}$ , so,$\sin v$ is evaluated as,

  $\begin{array}{c}{\sin }^{2}v+{\cos }^{2}v=1\\ {\sin }^{2}v+{\left(\frac{15}{17}\right)}^2=1\\ {\sin }^{2}v=1-\frac{225}{289}\\ {\sin }^{2}v=\frac{64}{289}\end{array}$

Simplify it further as,

  $\begin{array}{c}\sin v=\pm \sqrt{\frac{64}{289}}\\ \sin v=\pm \frac{8}{17}\end{array}$

Since,$0<v<\frac{\pi }{2}$ and sine function is positive in the interval so,$\sin v=\frac{8}{17}$ .

Apply the difference rule for cosine function $\cos \left(v-u\right)=\cos v\cos u+\sin v\sin u$ .

Substitute the values,$\sin v=\frac{8}{17}$ ,$\cos u=\frac{4}{5}$ ,$\sin u=-\frac{3}{5}$ and $\cos v=\frac{15}{17}$ .

  $\begin{array}{c}\cos \left(v-u\right)=\cos v\cos u+\sin v\sin u\\ =\left(\frac{15}{17}\right)\left(\frac{4}{5}\right)+\left(\frac{8}{17}\right)\left(-\frac{3}{5}\right)\\ =\frac{36}{85}\end{array}$

Recall that $\sec \left(v-u\right)=\frac{1}{\cos \left(v-u\right)}$ , so,

  $\begin{array}{c}\sec \left(v-u\right)=\frac{1}{\cos \left(v-u\right)}\\ =\frac{1}{\frac{36}{85}}\\ =\frac{85}{36}\end{array}$

Thus, the value of the trigonometric expression $\sec \left(v-u\right)$ is $\frac{85}{36}$ .