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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)
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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)