If this is a homework assignment, please consider trying to answer it yourself first, otherwise the value of the reinforcement of the lesson offered by the assignment will be lost on you.
The deriviative d/dx cos2(x) can be evaluated using the product rule.
Recall that d/dx f(x)g(x) = g(x) d/dx f(x) + f(x) d/dx g(x)
In this case, both f(x) and g(x) are cos(x), and cos2x = cos(x)cos(x), so...
d/dx cos(x)cos(x) = -cos(x)sin(x) - cos(x)sin(x) = -2sin(x)cos(x)
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
d/dx(cos x) = -sinx
No, (sinx)^2 + (cosx)^2=1 is though
1. Anything divided by itself always equals 1.
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)
Sin squared, cos squared...you removed the x in the equation.
22
2 x cosine squared x -1 which also equals cos (2x)
use the double angle formula for cos(2x) which is: cos(2x)=2cos^2(x)-1 by this relation cos^2(x)=(cos(2x)+1)/2 now we'd integrate this instead this will give sin(2x)/4+x/2 =) hope this helps
d/dx(cos x) = -sinx
cos x
Cos^2 x = 1 - sin^2 x
No, (sinx)^2 + (cosx)^2=1 is though
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
cosx^2 differentiates too 2(cosx)^1 x the differential of cos which is -sin so u get -2sinxcosx use the chain rule!