Chapter 7.3, Problem 109E

To determine

To prove: that $\sin 2\theta =2\sin \theta \cos \theta$

Explanation of Solution

Given information:

Given figure,

  

Calculation:

Consider the following sketch,

  

First find a relationship for the large right triangle using simple trigonometry

  $\begin{array}{c}\sin \theta =\frac{y}{2}\\ y=2\sin \theta \\ \cos \theta =\frac{x}{2}\\ x=2\cos \theta \end{array}$

Next find use the area formula for a right triangle

  $\begin{array}{c}A=\frac{1}{2}xy\\ =\frac{1}{2}\left(2\sin \theta \right)\left(2\cos \theta \right)\\ A=2\sin \theta \cos \theta \end{array}$

Now use the area of the two smaller triangles using the Law of sines and start with the lower left one.

  $\begin{array}{c}A=\frac{1}{2}ab\sin C\\ A_1=\frac{1}{2}\left(1\right)\left(1\right)\sin \left(180-2\theta \right)\\ A_1=\frac{1}{2}\left[\sin \left(180\right)\cos \left(-2\theta \right)+\sin \left(-2\theta \right)\cos \left(180\right)\right]\\ A_1=\frac{1}{2}\left[-\sin \left(-2\theta \right)\right]\\ A_1=\frac{1}{2}\left[\sin \left(2\theta \right)\right]\end{array}$

Now use the area of the two smaller triangles using the Law of sines and compute the are of the right bottom triangle.

  $\begin{array}{c}{A}_2=\frac{1}{2}ab\sin C\\ A_2=\frac{1}{2}\left(1\right)\left(1\right)\sin \left(2\theta \right)\\ A_2=\frac{1}{2}\left[\sin \left(2\theta \right)\right]\end{array}$

Add the two areas of the smaller triangles,

  $\begin{array}{c}{A}_1+A_2=\frac{1}{2}\left[\sin \left(2\theta \right)\right]+\frac{1}{2}\sin 2\theta \\ =\sin 2\theta \end{array}$

Set the total area equal to the area of the two smaller triangles of obtain,

  $\begin{array}{c}A=A_1+A_2\\ 2\sin \theta \cos \theta =\sin 2\theta \end{array}$

Hence, proved