Limit proof of trigonometric function

dustbin
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Homework Statement



Prove that limh→0[(cos(h-1))/h]=0.

Homework Equations



Half angle formula
From an example:

limh→0[(cos(h-1))/h] = limh→0[-(2sin2(h/2))/h]

They state that they use the half-angle formula in the following way:

cosh= 1-2sin2(h/2)

I'm not really sure how they are getting this from the half angle formula. Any pointer in the correct direction would be greatly appreciated. Thank you, in advance.
 
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Well known identity: cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

Write cos(h) = cos(h/2 + h/2) and use the identity above along with cos2θ + sin2θ = 1
 
Thank you, very much! Somehow I didn't notice to split cos(h) into cos(h/2 + h/2).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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