Differentiating the identity to develop another identity

meeklobraca
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Homework Statement



Differentiate the identity sin2x = 2sinxcosx to develop the identity for cos2x, in terms on sin x and cos x

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The Attempt at a Solution



Im not sure where to start with this one. Should I find the derivative of both sides of the equation, and then where do I go from there? The right side of that equation is 2cos^2x + sin^2x but I am not sure how I can use that.

Your help is appreciated as always!
 
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I believe you differentiated the right-hand side incorrectly:

\frac{d}{dx}(2\sin(x)\cos(x)) = 2(\cos^{2}(x) + (-\sin^{2}(x))) \Rightarrow 2\cos(2x) = 2(\cos^{2}(x) - \sin^{2}(x))\Rightarrow \cos(2x) = \cos^{2}(x) - \sin^{2}(x)which is a well-known double-angle formula.
 
Actually, once you find the dertivative of both sides, you will get the identity instantly(barring some cancellations).

Find the derivative of the left and right side (what u have written is not correct), using product rule, and see what cancels, on both sides
 
Thank you very much. I mistakenly used part of the quotient rule instead of the product rule, meaning I subtracted instead add added which caused me to get the + sin ^2x

Cheers guys!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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