compute limit
lim cos(3x)-1
x->0 sin(x)sin(x)
Okay, it is
$\displaystyle \lim_{x \to 0} \ cos(3x)-1 = 1-1 = 0$
$\displaystyle \lim_{x \to 0} \ sin(x)sin(x) = 0$
Therefor we use L'Hospital
We know that $\displaystyle cos(3x) = 4cos(x)^3 - 3cos(x)$
Thus
$\displaystyle cos(3x)-1 = 4cos(x)^3 - 3cos(x) =:f(x) $
=> $\displaystyle f'(x) = - 8sin(x)*cos(x)^2 + 4*sin(x)^3 - sin(x)$
$\displaystyle g(x):= sin(x)sin(x)$
$\displaystyle g'(x) = 2sin(x)cos(x)$
=> $\displaystyle \lim_{x \to 0} \ \frac{cos(3x)-1}{sin(x)sin(x)}$
= $\displaystyle lim \ \frac{- 8sin(x)*cos(x)^2 + 4*sin(x)^3 - sin(x)}{2sin(x)cos(x)}$
= $\displaystyle lim \ \frac{-8cos^2(x)+4sin^2(x)-1}{2cos(x)} = \frac{-8+4*0-1}{2*1} = -9/2$
Any questions?
Yes, $\displaystyle cos(2x)= cos^2(x)- sin^2(x)$ and $\displaystyle sin(2x)= 2sin(x)cos(x)$
Write $\displaystyle cos(3x)= cos(2x+ x)$ and use the sum formula: cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) so that cos(3x)= cos(2x)cos(x)- sin(2x)sin(x). Now put the formulas form cos(2x) and sin(2x) into that.
Hi again.
I'm glad you ask, but what exactly is the question?
Click here:
List of trigonometric identities - Wikipedia, the free encyclopedia
Use this formula for n=3
???
Does it help?