caseclosed Posted October 27, 2005 Share Posted October 27, 2005 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 Link to comment Share on other sites More sharing options...
TD Posted October 27, 2005 Share Posted October 27, 2005 2sin(x)cos(x) = sin(2x) => sin(2x)' = 2cos(2x) 4cos²(x)-2 = 2(2cos²x-1) = 2cos(2x). It's all good Link to comment Share on other sites More sharing options...
caseclosed Posted October 27, 2005 Author Share Posted October 27, 2005 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? Link to comment Share on other sites More sharing options...
LazerFazer Posted October 28, 2005 Share Posted October 28, 2005 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 Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now