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Question: How to simplify \({\cos ^2}2\theta - {\sin ^2}2\theta \) ?...

How to simplify cos⁡22θ−sin⁡22θ{\cos ^2}2\theta - {\sin ^2}2\theta ?

Explanation

Solution

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

Formula used: sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta
cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1

Complete step by step answer:
Given: we have to simplify cos⁡22θ−sin⁡22θ{\cos ^2}2\theta - {\sin ^2}2\theta .
We have to add and subtract sin⁡22θ{\sin ^2}2\theta to the given expression. Then, we get
=cos⁡22θ−sin⁡22θ= {\cos ^2}2\theta - {\sin ^2}2\theta
=cos⁡22θ+sin⁡22θ−sin⁡22θ−sin⁡22θ= {\cos ^2}2\theta + {\sin ^2}2\theta - {\sin ^2}2\theta - {\sin ^2}2\theta
We know a trigonometric formula sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1. By applying this formula, the above expression become
=1−2sin⁡22θ= 1 - 2{\sin ^2}2\theta
Now, put the value of sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta in this expression and then we get
=1−2(2cos⁡θsin⁡θ)2 =1−2×4cos⁡2θsin⁡2θ =1−8cos⁡2θsin⁡2θ = 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⁡22θ−sin⁡22θ{\cos ^2}2\theta - {\sin ^2}2\theta can be simplified to 1−8sin⁡2θcos⁡2θ1 - 8{\sin ^2}\theta {\cos ^2}\theta .

Note: Alternatively, this expression can be simplified as firstly expand the expression using a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right). Expanding cos⁡22θ−sin⁡22θ{\cos ^2}2\theta - {\sin ^2}2\theta we get (cos⁡2θ+sin⁡2θ)(cos⁡2θ−sin⁡2θ)\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⁡2θ−tan⁡2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1
(2) cos⁡ec2θ−cot⁡2θ=1\cos e{c^2}\theta - {\cot ^2}\theta = 1
Proof of the given formula sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta .
We have studied a trigonometric formula sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B.
We can write 2θ2\theta as θ+θ\theta + \theta . We have A=θA = \theta and B=θB = \theta . So, by applying above formula we can write
⇒sin⁡(θ+θ)=sin⁡θcos⁡θ+cos⁡θsin⁡θ ⇒sin⁡2θ=2sin⁡θcos⁡θ \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⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta .
Proof of the given formula cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta .
We have studied a trigonometric formula cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B.
We can write 2θ2\theta as θ+θ\theta + \theta . We have A=θA = \theta and B=θB = \theta . So, by applying above formula we can write
⇒cos⁡(θ+θ)=cos⁡θcos⁡θ−sin⁡θsin⁡θ ⇒cos⁡2θ=cos⁡2θ−sin⁡2θ  \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.