Chapter 7.3, Problem 54E

To determine

To evaluate: given expression under the given conditions.

Answer to Problem 54E

  $\tan 2\theta =-\frac{24}{7}$

Explanation of Solution

Given information:

Given expression

  $\tan 2\theta ;\cos \theta =\frac{3}{5},$

  $\theta$ in Quadrant I.

Calculation:

First we Sketch the angle $\theta$ in standard position with terminal side in quadrant II as in figure shown below.

Since $\cos \theta =\frac{3}{5}$ , we can label a side and the hypotenuse of the triangle in the figure.

  

To find the remaining side, we use the Pythagorean theorem. We have

  $x^2+y^2=r^2$

On substituting $x=3,r=5$ , we get

  ${\left(3\right)}^2+{\left(y\right)}^2={\left(5\right)}^2$

Isolating $3^2$ on the right hand side, we get

  $y^2=25-9$

Solving for $y$ , we get

  $\begin{array}{c}y=\pm \sqrt{16}\\ =\pm 4\end{array}$

Now, $\sin \theta =\frac{y}{r}$ . Therefore, we get

  $\sin \theta =\frac{4}{5}$ , note that we have taken positive square root of $y$ because $\sin \theta$ is positive in quadrant I.

Also we know that $\tan \theta =\frac{\sin \theta }{\cos \theta }$ , therefore substituting the values $\sin \theta$ and $\cos \theta$ in $\tan \theta$ . We have,

  $\tan \theta =\frac{4}{3}$

Now, we use the double-angle formula for tangent,

  $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$ , to find value of $\tan 2\theta$ .

Substituting the value of $\tan \theta$ , we get

  $\tan 2\theta =\frac{2\left(\frac{4}{3}\right)}{1-{\left(\frac{4}{3}\right)}^2}$

Hence, we have

  $\tan 2\theta =-\frac{24}{7}$