limx→0(sin(2x)−2sin(x))4(3+cos(2x)−4cos(x))3
without L'Hôpital.
I've tried using equivalences with (sin(2x)−2sin(x))4 and arrived at −x12 but I don't know how to handle (3+cos(2x)−4cos(x))3. Using cos(2x)=cos2(x)−sin2(x) hasn't helped, so any hint?
Answer
Hint: Note that
3+cos(2x)−4cos(x)=3+2cos2(x)−1−4cos(x)=2(cos(x)−1)2,
and that
sin(2x)−2sin(x)=2sin(x)cos(x)−2sin(x)=2sin(x)(cos(x)−1).
No comments:
Post a Comment