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Question: How do you prove \[\cos 2x = {\cos ^2}x - {\sin ^2}x\] using other trigonometric identities?...

How do you prove cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x using other trigonometric identities?

Explanation

Solution

Here we are given a trigonometric identity that is cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x and we are asked to prove this identity using other trigonometric identities. For approaching this question we need to know a basic trigonometric identity that is as follows cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta applying this and further simplification we could get through the asked proof easily.

Complete step-by-step answer:
Here we are given an equation that is cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x and we are asked the method to proof that can be done as taking the LHS of the given equation that is cos⁡2x\cos 2x and applying the trigonometric identity that is cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta on it and further simplifying it that is depicted as follows –
Taking the LHS of the given equation cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}xthat is cos⁡2x\cos 2x and applying the identity that is cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta -
Here cos⁡2x\cos 2x can be expressed as -
cos⁡2x=cos⁡(x+x)\cos 2x = \cos (x + x)
Now applying the identity as stated above that is of cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
Here the α=β=x\alpha = \beta = x so the RHS becomes –
cos⁡xcos⁡x−sin⁡xsin⁡x\cos x\cos x - \sin x\sin x
Now further simplifying the resultant as multiplying the like terms originated that comes out to be as –
cos⁡2x−sin⁡2x{\cos ^2}x - {\sin ^2}x
Which is equal to the RHS of the required proof that is cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x
Therefore LHS=RHS hence proved the required quantity that is -cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x.

Note: While solving such kind of the questions one should know the identities of the form cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta ,sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta which is used to express the angles and their trigonometric expressions in the desired way which would help to get the desired answer.