Beweis Additionstheoreme: TAN(x + y) = (TAN(x) + TAN(y)) / (1 - TAN(x)·TAN(y)) ?

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1 Antwort

tan(x+y)=sin(x+y)/cos(x+y)

Nun gilt for sin(x+y) :

sin(x+y) = sin(x)⋅* cos(y) + sin(y) * cos(x)
sin(x+y) = (cos(x) * cos(y)) * ( sin(x)⋅* cos(y) / (cos(x) * cos(y)) + sin(y) * cos(x) / (cos(x) * cos(y)))
sin(x+y) = (cos(x) * cos(y)) * ( tan(x)⋅ + tan(y) )

Nun gilt for cos(x+y) :

cos(x+y) = cos(x)⋅* cos(y) - sin(x) * sin(y)
cos(x+y) = (cos(x) * cos(y)) * ( cos(x)⋅* cos(y)/ (cos(x) * cos(y)) - sin(x) * sin(y)/ (cos(x) * cos(y)))
cos(x+y) = (cos(x) * cos(y)) * ( 1 - tan(x) * tan(y) )

d.h.

tan(x+y) = ( tan(x)⋅ + tan(y) ) / ( 1 - tan(x) * tan(y) )


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