Addition formulae proof for sin(x + y) and cos(x + y)

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SUMMARY

The forum discussion focuses on proving the addition formulae for sin(x + y) and cos(x + y) using the Mean Value Theorem. It defines the function f(x) as f(x) = sin(x+y) - sin(x)cos(y) - cos(x)sin(y) and introduces E(x) = f(x)^2 + f'(x)^2. The proof hinges on demonstrating that E(x) is identically zero, which implies that both f and its derivative f' are identically zero, thus confirming the addition formulae.

PREREQUISITES
  • Understanding of trigonometric functions and identities
  • Familiarity with calculus concepts, particularly derivatives
  • Knowledge of the Mean Value Theorem
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the Mean Value Theorem and its applications in calculus
  • Explore proofs of trigonometric identities, specifically addition formulas
  • Learn about the properties of derivatives and their implications
  • Investigate the relationship between functions and their squares in calculus
USEFUL FOR

Mathematics students, educators, and anyone interested in advanced calculus and trigonometric proofs will benefit from this discussion.

maximus101
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For a fixed [tex]y[/tex] [tex]\in[/tex] [tex]R[/tex] , if
[tex]f(x)[/tex] [tex]=[/tex] [tex][sin(x+y)][/tex][tex]-(sin x)[/tex] [tex](cos y)[/tex] − [tex](cos x)[/tex] [tex](sin y)[/tex]and we let [tex]E(x) =[/tex] [tex][f(x)]^2[/tex]+[tex][f'(x)]^2[/tex].

How do we prove the addition formulae for [tex]sin(x + y)[/tex]and [tex]cos(x + y)[/tex] by applying the Mean Value Theorem to [tex]E[/tex]
 
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maximus101 said:
For a fixed [tex]y[/tex] [tex]\in[/tex] [tex]R[/tex] , if
[tex]f(x)[/tex] [tex]=[/tex] [tex][sin(x+y)][/tex][tex]-(sin x)[/tex] [tex](cos y)[/tex] − [tex](cos x)[/tex] [tex](sin y)[/tex]and we let [tex]E(x) =[/tex] [tex][f(x)]^2[/tex]+[tex][f'(x)]^2[/tex].

How do we prove the addition formulae for [tex]sin(x + y)[/tex]and [tex]cos(x + y)[/tex] by applying the Mean Value Theorem to [tex]E[/tex]

Well, if you were to show E(x) ≡ 0 that would show both f and f' are identically 0.
 

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