Chapter 12.6, Problem 19WE

To determine

Prove

  $\cos 2\theta =1-2{\sin }^2\theta ;0°<\theta <90°$

Explanation of Solution

Calculation :

  

Consider the above figure.

By Cosine Formula ,

  $\begin{array}{l}{b}^2=1^2+1^2-2\times 1\times 1\cos 2\theta \\ \Rightarrow b^2=2-2\cos 2\theta \end{array}$

In the right half triangle,

  $\begin{array}{l}\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of }hypotenuse}=\frac{\left(\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{1}=\frac{b}{2}\\ \Rightarrow b=2\sin \theta \mathrm{..............}[\text{multiply each side by 2]}\\ \Rightarrow {\text{b}}^2=4{\sin }^2\theta \mathrm{..........}[\text{square each side]}\end{array}$

Equate both the values of

  $b^2$ :

  $\begin{array}{l}{b}^2=2-2\cos 2\theta =4{\sin }^2\theta \\ \Rightarrow 2(1-\cos 2\theta )=4{\sin }^2\theta \\ \Rightarrow 1-\cos 2\theta =2{\sin }^2\theta \mathrm{...............}[\text{divide each side by 2]}\\ \Rightarrow 1-\cos 2\theta -2{\sin }^2\theta =\mathrm{0........}[\text{subtract }2{\sin }^2\theta \text{ from each side]}\\ \Rightarrow 1-2{\sin }^2\theta =\cos 2\theta \mathrm{.............}[\text{add }\cos 2\theta \text{ each side]}\\ \Rightarrow \cos 2\theta =1-2{\sin }^2\theta \end{array}$

Hence proved.