Find Limit of (1-cos(x))/2sin^2(x) for x->0

[kholdstare121]

I don't know if this neccessarily belongs in the calculus section, but it's for my calculus class.
How would you go about finding the limit as x approaches 0 for
(1-cos(x))/2sin^2(x)

I know the answer is 1/4(unless I copied it down wrong :/ ) , but what first steps would you take to evaluating thing. Also this is for my review packet where we just know the basics of derivatives, so the only knowledge I should be going by are my trig identities and that the limit as x approaches 0 for sinx/x =1.
Thanks.

 

[arildno]

Expand your fraction by the factor (1+cos(x)).

 

[kholdstare121]

Thanks that helped out a lot, figured it out!
I have a test on this chapter Tuesday, and these trig limits might just kill me.
There are so many identities to use, so little time.

 

[arildno]

Most of those trig identities can be opportunely RE-derived from a few simple facts:
1. Sine is odd, cosine is even
2. cos(x-y)=cos(x)cos(y)+sin(x)sin(y)
3. cos^2(x)+sin^2(x)=1
4 tan(x)=sin(x)/cos(x)

Remembering those 4 is really all you need, along with skill at manipulating expressions.