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Question Number 127383 by Lordose last updated on 29/Dec/20

Is there any analytic proof of the result

\sin (x + y) = \sin (x) \cos (y) + \sin (y) \cos (x)

Answered by bemath last updated on 29/Dec/20

Answered by liberty last updated on 29/Dec/20

\sin α = \sin ∠ A C B = \frac{A B}{A C} = \frac{A B}{A E + E C} = \frac{\sin (α + β)}{\cos β + \frac{\sin β \cos α}{\sin α}}

s o w e h a v e t h e e q u a t i o n \frac{\sin (α + β)}{\cos β + \frac{\sin β \cos α}{\sin α}} = \sin α

t h e n \sin (α + β) = \sin α (\cos β + \frac{\sin β \cos α}{\sin α})

= \sin α \cos β + \sin β \cos α

Commented by liberty last updated on 29/Dec/20

Answered by MJS_new last updated on 29/Dec/20

\sin α = \frac{e^{i α} - e^{- i α}}{2 i} \land \cos α = \frac{e^{i α} + e^{- i α}}{2}

\sin (x + y) = \frac{e^{i (x + y)} - e^{- i (x + y)}}{2 i} = \frac{e^{i x} e^{i y} - e^{- i x} e^{- i y}}{2 i} =

[let e^{i x} = u \land e^{i y} = v]

= \frac{u v - \frac{1}{u v}}{2 i} = \frac{2 u v - \frac{2}{u v}}{4 i} =

= \frac{2 u v + \frac{u}{v} - \frac{u}{v} + \frac{v}{u} - \frac{v}{u} - \frac{2}{u v}}{4 i} =

= \frac{u v + \frac{u}{v} - \frac{v}{u} - \frac{1}{u v}}{4 i} + \frac{u v - \frac{u}{v} + \frac{v}{u} - \frac{1}{u v}}{4 i} =

\frac{(u - \frac{1}{u}) (v + \frac{1}{v})}{4 i} + \frac{(u + \frac{1}{u}) (v - \frac{1}{v})}{4 i} =

[u = e^{i x} \land v = e^{i y}]

= \frac{e^{i x} - e^{- i x}}{2 i} \times \frac{e^{i y} + e^{- i y}}{2} + \frac{e^{i x} + e^{- i x}}{2} \times \frac{e^{i y} - e^{- i y}}{2 i} =

= \sin x \cos y + \cos x \sin y

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