Chapter 10.6, Problem 15E

Solution

To Find: The value of $\sin (a-b)$ .

  $\sin (a-b)=\frac{18}{17}$ .

Given information:

  $\cos a=\frac{4}{5},0<a<\frac{\pi }{2}\&\sin b=-\frac{15}{17},\frac{3\pi }{2}<b<2\pi$ .

Concept and Formula Used: Use the trigonometric formula, $\sin (a-b)=\sin a\cos b-\cos a\sin b,{\cos }^{2}b=1-{\sin }^{2}b,{\sin }^{2}b=1-{\cos }^{2}b$ .

Calculation:

The given values are $\cos a=\frac{4}{5},0<a<\frac{\pi }{2}\&\sin b=-\frac{15}{17},\frac{3\pi }{2}<b<2\pi$ .

First find the values of $\sin a\&\cos b$ ,

  $\begin{array}{l}{\sin }^{2}a=1-{\cos }^{2}a\\ {\sin }^{2}a=1-{\left(\frac{4}{5}\right)}^2\\ {\sin }^{2}a=1-\frac{16}{25}\\ {\sin }^{2}a=\frac{25-16}{16}\\ {\sin }^{2}a=\frac{9}{16}\\ \sin a=\sqrt{\frac{9}{16}}\\ \sin a=\pm \frac{3}{4}\end{array}$

As ‘a’ lies in first quadrant therefore,

  $\sin a=\frac{3}{4}$ .

Now, for value of $\cos b$ use,

  $\begin{array}{l}{\cos }^{2}b=1-{\sin }^{2}b\\ {\cos }^{2}b=1-{\left(-\frac{15}{17}\right)}^2\\ {\cos }^{2}b=1-\frac{225}{289}\\ {\cos }^{2}b=\frac{289-225}{289}\\ {\cos }^{2}b=\frac{64}{289}\\ \cos b=\sqrt{\frac{64}{289}}\\ \cos b=\pm \frac{8}{17}\end{array}$

As ‘b’ lies in fourth quadrant therefore,

  $\cos b=\frac{8}{17}$ .

Now,

  $\begin{array}{l}\sin (a-b)=\sin a\cos b-\cos a\sin b\\ \sin (a-b)=\left(\frac{3}{4}\right)\left(\frac{8}{17}\right)-\left(\frac{4}{5}\right)\left(-\frac{15}{17}\right)\\ \sin (a-b)=\frac{6}{17}+\frac{12}{17}\\ \sin (a-b)=\frac{18}{17}\end{array}$

Conclusion:

  $\sin (a-b)=\frac{18}{17}$ .