this equation

2sin(x)cos(x)

 

the derivative of that is 2cos(2x)

but on my TI-89 it gives me 4(cos(x))^2-2

when plug in an integer, same output answer........

someone show me that they are equal

2sin(x)cos(x) = sin(2x) => sin(2x)' = 2cos(2x)

 

4cos²(x)-2 = 2(2cos²x-1) = 2cos(2x).

 

It's all good

2sin(x)cos(x) = sin(2x) => sin(2x)' = 2cos(2x)

 

4cos²(x)-2 = 2(2cos²x-1) = 2cos(2x).

 

It's all good

2cos²x-1=2cos(2x)

is that an identity or I am missing something, if it is identity then how do you make them equal?

Yes, that actually is an identity

 

[math]

cos(2 \theta )=cos^{2} ( \theta ) - sin^{2} ( \theta ) =2cos^{2} ( \theta) - 1 = 1 - 2sin^{2} ( \theta )

[/math]

 

LF