Calculus Question

Question

Differentiate the identity sin2x = 2sinxcosx to develop the identity for cos2x, in terms of sinx and cosx.

 

Help!?

Answer 1

First, let's differentiate:

 

d/dx (sin(2x)) = d/dx (2sin(x)cos(x) )

 

cos(2x) · 2 = 2cos(x) · cos(x) + 2sin(x) · -sin(x)   [using product rule]

 

cos(2x) = cos²(x) - sin²(x)     [dividing both sides by two & simplifying terms on right]

 

cos(2x) has three identities. Above is one. Substituting from the Pythagorean identity gives the other two.

 

sin²(x) + cos²(x) = 1

sin²(x) = 1 - cos²(x)

cos²(x) = 1 - sin²(x)

 

cos(2x) = cos²(x) - [1 - cos²(x)] = 2cos²(x) - 1

 

cos(2x) = [ 1 - sin²(x)] - sin²(x) = 1 - 2sin²(x)