Chapter 9.7, Problem 81E

To determine

To determine the values of $\sin (u+v)$ .

Answer to Problem 81E

The value of $\sin (u+v)=\frac{-7}{5\sqrt{2}}$ .

Explanation of Solution

Given information: $\sin u=-\frac{3}{5}\text{and}\cos v=\frac{1}{\sqrt{2}}\text{θ}\text{is}\text{in}\text{IV quadrant}$ .

Formula used: $\sin (A+B)=\sin A\cos B+\cos A\sin B$ .

Concept used: For finding the trigonometric functions, following concept is used as:

$\sin \theta$	$\cos \theta$	$\tan \theta$
P	B	P
H	H	B

Where, P, B and H be the perpendicular, base and hypotenuse of right angled triangle.

By Pythagoras theorem, $B^2+P^2=H^2$ .

Also, $\cot \theta =\frac{1}{\tan \theta }$ ; $\sec \theta =\frac{1}{\cos \theta }$ and $\cos ec\theta =\frac{1}{\sin \theta }$ .

Calculation:

Given,  $\sin u=\frac{P}{H}=\frac{3}{5}$ .

Using Pythagoras theorem, $B^2+P^2=H^2$ .

  $\Rightarrow B^2+3^2=5^2$ .

  $\Rightarrow B=4$ .

Thus, $\cos u=\frac{B}{H}=\frac{4}{5}$ .

Also, $\cos v=\frac{B}{H}=\frac{1}{\sqrt{2}}$ .

Using Pythagoras theorem, $B^2+P^2=H^2$ .

  $\Rightarrow 1^2+P^2={\left(\sqrt{2}\right)}^2$ .

  $\Rightarrow P=1$ .

Thus, $\sin v=\frac{P}{H}=-\frac{1}{\sqrt{2}}$ .

  $\begin{array}{l}\sin (u+v)=\sin u\cos v+\cos u\sin v\\ =\left(-\frac{3}{5}\right)\left(\frac{1}{\sqrt{2}}\right)+\left(\frac{4}{5}\right)\left(-\frac{1}{\sqrt{2}}\right)\\ \left(\frac{-4-3}{5\sqrt{2}}\right)=\frac{-7}{5\sqrt{2}}\end{array}$ .