Limit proof of trigonometric function

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SUMMARY

The limit proof of the trigonometric function limh→0[(cos(h-1))/h]=0 is established using the half-angle formula. Specifically, cos(h) is expressed as cos(h/2 + h/2), allowing the application of the identity cos(a + b) = cos(a)cos(b) - sin(a)sin(b). This leads to the transformation of the limit into limh→0[-(2sin²(h/2))/h], which simplifies the evaluation of the limit as h approaches zero.

PREREQUISITES
  • Understanding of trigonometric identities, specifically the half-angle formula.
  • Familiarity with limits and continuity in calculus.
  • Knowledge of the sine and cosine functions and their properties.
  • Ability to manipulate algebraic expressions involving trigonometric functions.
NEXT STEPS
  • Study the half-angle formula in detail and its applications in limit proofs.
  • Learn about the derivation and use of trigonometric identities in calculus.
  • Practice evaluating limits involving trigonometric functions using various techniques.
  • Explore the relationship between sine and cosine functions through graphical representations.
USEFUL FOR

Students studying calculus, particularly those focusing on limits and trigonometric functions, as well as educators seeking to enhance their teaching methods in trigonometry.

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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).
 

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