d/dt of t + sin2t = 1 + 2cos2t tell me i'm not crazy?

Question

I have x(t) = t + sin 2t and y(t)= t-sin2t
this is what i get for derivatives
x'(t) =1 + 2cos2t and y'(t) = 1-2cos2t

the answer key i'm looking at gives
x'(t) =1 - 2cos2t and y'(t) = 1 + 2cos2t

is there a mistake on the key

Answer 1

You're not crazy -- the derivatives you have given are correct. Either there is an error on the key, or at some point during the problem you mixed up the functions x and y.

Answer 2

d/dt of t + sin2t = 1 + 2cos2t
d/dt of t - sin2t = 1 - 2cos2t

You're sane, it's the answer key that needs some help!

Answer 3

yes you are correct, since d/dt(sinat) = acosat