Question

How to simplify ${\cos ^2}2\theta - {\sin ^2}2\theta $ ?

Answer 1

Hint: This is a trigonometric expression this can be simplified by using some of the trigonometric formulas. First of all add ${\sin ^2}\theta $ and subtract the same so that the expression remains the same. Then, put the value of $\sin 2\theta $ in the expression to get the required result.

Formula used: $\sin 2\theta = 2\sin \theta \cos \theta $
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $
${\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete step by step answer:
Given: we have to simplify ${\cos ^2}2\theta - {\sin ^2}2\theta $.
We have to add and subtract ${\sin ^2}2\theta $ to the given expression. Then, we get
$ = {\cos ^2}2\theta - {\sin ^2}2\theta $
$ = {\cos ^2}2\theta + {\sin ^2}2\theta - {\sin ^2}2\theta - {\sin ^2}2\theta$
We know a trigonometric formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$. By applying this formula, the above expression become
$ = 1 - 2{\sin ^2}2\theta $
Now, put the value of $\sin 2\theta = 2\sin \theta \cos \theta $ in this expression and then we get
$
   = 1 - 2{\left( {2\cos \theta \sin \theta } \right)^2} \\
   = 1 - 2 \times 4{\cos ^2}\theta {\sin ^2}\theta \\
   = 1 - 8{\cos ^2}\theta {\sin ^2}\theta
 $

Thus, the expression ${\cos ^2}2\theta - {\sin ^2}2\theta $ can be simplified to $1 - 8{\sin ^2}\theta {\cos ^2}\theta $.

Note: Alternatively, this expression can be simplified as firstly expand the expression using ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$. Expanding ${\cos ^2}2\theta - {\sin ^2}2\theta $ we get $\left( {\cos 2\theta + \sin 2\theta } \right)\left( {\cos 2\theta - \sin 2\theta } \right)$. Then by applying some trigonometric formula that is given above we have to expand this expression and then multiplying and manipulating the terms we simplify the above expression.
While solving the trigonometric equation we have to apply some other trigonometric formula that is
(1) ${\sec ^2}\theta - {\tan ^2}\theta = 1$
(2) $\cos e{c^2}\theta - {\cot ^2}\theta = 1$
Proof of the given formula $\sin 2\theta = 2\sin \theta \cos \theta $.
We have studied a trigonometric formula $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$.
We can write $2\theta $ as $\theta + \theta $. We have $A = \theta $ and $B = \theta $. So, by applying above formula we can write
$
   \Rightarrow \sin \left( {\theta + \theta } \right) = \sin \theta \cos \theta + \cos \theta \sin \theta \\
   \Rightarrow \sin 2\theta = 2\sin \theta \cos \theta
$
Hence, the above given formula is proved.
Similarly, we can prove $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $.
Proof of the given formula $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $.
We have studied a trigonometric formula $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$.
We can write $2\theta $ as $\theta + \theta $. We have $A = \theta $ and $B = \theta $. So, by applying above formula we can write
$
   \Rightarrow \cos \left( {\theta + \theta } \right) = \cos \theta \cos \theta - \sin \theta \sin \theta \\
   \Rightarrow \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \\
 $
Hence, the above given formula is proved.