Chapter C, Problem 27E

To determine

To calculate:

The value of following expression:

  $\sin \left(x+y\right)$

Answer to Problem 27E

The given expression $\sin \left(x+y\right)$ is evaluated as $\frac{4+6\sqrt{2}}{15}$ .

Explanation of Solution

Given information:

  $\begin{array}{l}\sin x=\frac{1}{3}\\ \sec y=\frac{5}{4}\\ 0\le x,y\le \frac{\pi }{2}\end{array}$

Formula Used:

Addition formula:

  $\sin \left(x+y\right)=\sin x\cos y+\cos x\cdot \sin y$

Calculation:

Know that:

  $\begin{array}{l}\sin x=\frac{opposite}{hypotenus}\\ \sin x=\frac{1}{3}\end{array}$

First need to sketch the triangle for $\sin x$ :

  

Fig. 1 Triangle for $\sin x$

So, hypotenuse is equal to $3$ and the length of the adjacent side is equal to $1$ . So, use the Pythagoras theorem:

  $\begin{array}{l}{r}^2=opposit{e}^2+adjacen{t}^2\\ 3^2=1^2+adjacen{t}^2\\ adjacen{t}^2=8\\ adjacent=2\sqrt{2}\end{array}$

Know that:

  $\begin{array}{l}\sec x=\frac{hypotenus}{adjecent}\\ \sec x=\frac{5}{4}\end{array}$

Need to sketch the triangle for $\sec x$ :

  

Fig. 2 Triangle for $\sec x$

So, hypotenuse is equal to $5$ and the length of the adjacent side is equal to $4$ . So, use the Pythagoras theorem:

  $\begin{array}{l}{r}^2=opposit{e}^2+a^2\\ 5^2=opposit{e}^2+4^2\\ opposit{e}^2=9\\ opposite=3\end{array}$

And the remaining side opposite to $\angle y$ is $3$ .

From first triangle:

  $\begin{array}{l}\sin x=\frac{1}{3}\\ \cos x=\frac{2\sqrt{2}}{3}\end{array}$

From second triangle:

  $\begin{array}{l}\sin y=\frac{3}{5}\\ \cos y=\frac{4}{5}\end{array}$

Now, evaluate the given expression as:

  $\begin{array}{l}\sin \left(x+y\right)=\sin x\cos y+\cos x\cdot \sin y\\ \sin \left(x+y\right)=\left(\frac{1}{3}\right)\left(\frac{4}{5}\right)+\left(\frac{2\sqrt{2}}{3}\right)\left(\frac{3}{5}\right)\\ \sin \left(x+y\right)=\frac{4}{15}+\frac{6\sqrt{2}}{15}\\ \sin \left(x+y\right)=\frac{4+6\sqrt{2}}{15}\end{array}$