Chapter 7.3, Problem 28E

To determine

To find: the exact trigonometric value by using the given information.

Answer to Problem 28E

  $-\frac{36}{77}$

Explanation of Solution

Given information:

  $\begin{array}{l}Trignometric\text{ function: tan}\left(x-y\right)\end{array}$

  $\begin{array}{c}\sin x=\frac{8}{17},0<x<\frac{\pi }{2}\end{array}$

  $\begin{array}{c}\cos y=\frac{3}{5},0<y<\frac{\pi }{2}\end{array}$

Calculation:

Evaluate the value of $\cos x\text{ and }\sin y$ as shown below.

  $\begin{array}{c}\cos x=\sqrt{1-{\sin }^{2}x}\left[\text{Pythagorean identity}\right]\end{array}$

  $\begin{array}{c}=\sqrt{1-{\left(\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$17$}\right.\right)}^2}\left[\text{Put the value of }\sin x\right]\end{array}$

  $\begin{array}{c}=\sqrt{\raisebox{1ex}{$225$}\!\left/ \!\raisebox{-1ex}{$289$}\right.}\end{array}$

  $\begin{array}{c}=\frac{15}{17}\end{array}$

  $\begin{array}{c}\end{array}$

  $\begin{array}{c}\sin y=\sqrt{1-{\cos }^{2}y}\left[\text{Pythagorean identity}\right]\end{array}$

  $\begin{array}{c}=\sqrt{1-{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)}^2}\left[\text{Put the value of }\cos y\right]\end{array}$

  $\begin{array}{c}=\sqrt{\raisebox{1ex}{$16$}\!\left/ \!\raisebox{-1ex}{$25$}\right.}\end{array}$

  $\begin{array}{c}=\frac{4}{5}\end{array}$

Now evaluate $\tan x\text{ and}\tan y$ by applying quotient identity as shown below.

  $\begin{array}{c}\tan x=\frac{\sin x}{\cos x}\end{array}$

  $\begin{array}{c}=\frac{\left(\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$17$}\right.\right)}{\left(\raisebox{1ex}{$15$}\!\left/ \!\raisebox{-1ex}{$17$}\right.\right)}\left[\text{Put the value for }\sin x\text{ and }\cos x\right]\end{array}$

  $\begin{array}{c}=\frac{8}{15}\end{array}$

  $\begin{array}{c}\end{array}$

  $\begin{array}{c}\tan y=\frac{\sin y}{\cos y}\end{array}$

  $\begin{array}{c}=\frac{\left(\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)}{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)}\left[\text{Put the value for }\sin y\text{ and }\cos y\right]\end{array}$

  $\begin{array}{c}=\frac{4}{3}\end{array}$

Now solve $\text{tan}\left(x-y\right)$ by applying difference identity for tangent as shown below.

  $\begin{array}{c}\tan \left(x-y\right)=\frac{\tan x-\tan y}{1+\tan x\cdot \tan y}\end{array}$

  $\begin{array}{c}=\frac{\frac{8}{15}-\frac{4}{3}}{1+\frac{8}{15}\cdot \frac{4}{3}}\left[\text{Put the values for tangent functions}\right]\end{array}$

  $\begin{array}{c}=\frac{\frac{24-60}{45}}{\frac{45+24}{45}}\end{array}$

  $\begin{array}{c}=\frac{24-60}{45+32}\end{array}$

  $\begin{array}{c}=-\frac{36}{77}\end{array}$

Thus the value of given trigonometric function is $-\frac{36}{77}$ .