Chapter 7.3, Problem 53E

To determine

To evaluate: given expression under the given conditions.

Answer to Problem 53E

  $\sin 2\theta =-\frac{8\sqrt{3}}{49}$

Explanation of Solution

Given information:

Given expression

  $\sin 2\theta ;\sin \theta =\frac{1}{7},$

  $\theta$ in Quadrant II.

Calculation:

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

  

Since $\sin \theta =\frac{y}{r}$ ,we have

  $y=1$ and $r=7$

The double angle formula for sine function is $\sin 2\theta =2\sin \theta \cos \theta$ .

Therefore, we need to evaluate $\cos \theta$ which is given by $\cos \theta =\frac{x}{r}$ .

Using the Pythagorean Theorem, we get

  $\begin{array}{l}{r}^2=y^2+x^2\\ 7^2=1^2+x^2\\ 49=1+x^2\end{array}$

Transposing the constant terms to the left hand side, we get

  $\begin{array}{l}48=x^2\\ x^2=48\end{array}$

Taking the square roots on both sides, we get

  $\begin{array}{c}x=\pm \sqrt{48}\\ =\pm 4\sqrt{3}\end{array}$

Since $x$ in II quadrant, we have

  $x=-4\sqrt{3}$

On substituting the value of $x$ in $\cos \theta =\frac{x}{r}$ , we get

  $\cos \theta =\frac{-4\sqrt{3}}{7}$

Using the double-angle formula for sine, we get

  $\begin{array}{c}\sin 2\theta =2\sin \theta \cos \theta \\ =2\left(\frac{1}{7}\right)\left(-\frac{4\sqrt{3}}{7}\right)\\ =-\frac{8\sqrt{3}}{49}\end{array}$

Hence, on evaluating, we get

  $\sin 2\theta =-\frac{8\sqrt{3}}{49}$