Question 326427
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A few things I want you to consider before I work on this problem.


You just put an equation out there without any explanation of what you wanted to do with it, what is giving you difficulty, or anything else.  This sort of behavior is:


*[tex \LARGE \ \ \ \cdot\ \ ]Rude


*[tex \LARGE \ \ \ \cdot\ \ ]Disrespectful


*[tex \LARGE \ \ \ \cdot\ \ ]Has a high probability of getting a less than helpful response

*[tex \LARGE \ \ \ \ \ --\ \ ] this is Algebra.com, not the Psychic Hotline.


My recommendation to you is to go back and <i>very carefully</i> re-read (or read in the first place) the instructions for posting questions on this site.  Then, after you are certain you understand the instructions, and this is the second part of a two-part recommendation, follow the instructions.


Now, as to your equation.  In fact, this is an identity, so I presume you want to demonstrate that it is an identity.


Let *[tex \Large \varphi\ =\ 5x].


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(10x)\ =\ \cos2(\varphi)\ =\ \cos^2(\varphi)\ -\ sin^2(\varphi)]


by the Cosine Double Angle Formula.


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(\varphi)\ -\ sin^2(\varphi)\ =\ \cos^2(5x)\ -\ sin^2(5x)]


by substitution.


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(10x)\ \equiv\ \cos^2(5x)\ -\ sin^2(5x)\ \forall\ x\ \in\ \mathbb{R}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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