Çözüldü Trigonometri

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sin(x)·cos(y) - sin(y)·cos(x) = cos(y) ⇒ tan(y) = [ sin(x) - 1 ] / cos(x), (x ≠ π / 2 + k·π, k ∈ Z)
tan(y) = { 2·sin(x / 2)·cos(x / 2) - [ sin(x / 2) ]^2 - [ cos(x / 2) ]^2 } / { [ cos(x / 2) - sin(x / 2) ]·[ cos(x / 2) + sin(x / 2) ] }
tan(y) = -[ sin(x / 2) - cos(x / 2) ]^2 / { [ cos(x / 2) - sin(x / 2) ]·[ cos(x / 2) + sin(x / 2) ]
tan(y) = -[ cos(x / 2) - sin(x / 2) ]^2 / { [ cos(x / 2) - sin(x / 2) ]·[ cos(x / 2) + sin(x / 2) ]
tan(y) = -[ cos(x / 2) - sin(x / 2) ] / [ cos(x / 2) + sin(x / 2) ]
tan(y) = [ sin(x / 2) - cos(x / 2) ] / [ sin(x / 2) + cos(x / 2) ]....(I)
tan[ (x / 2) - y ] = [ tan(x / 2) - tan(y) ] / { 1 + [ tan(x / 2) ]·tan(y)....(II)
(I) eşitliği (II)'deki yerine yazılınca tan[ (x / 2) - y] =
{tan(x / 2) - [ sin(x / 2) - cos(x / 2) ] / [sin(x / 2) + cos(x / 2) ]} / {1 + [tan(x / 2)]·[sin(x / 2) - cos(x / 2)] / [sin(x / 2) + cos(x / 2) ]}=
{sin(x / 2)/cos(x / 2)-[sin(x / 2)-cos(x / 2)]/[sin(x / 2)+cos(x / 2)]}/{[sin(x / 2)+cos(x / 2)]·cos(x / 2)+[sin(x / 2)]·[sin(x / 2)-cos(x / 2)]}=
{ [ sin(x / 2) ]^2 + (1 / 2)sin(x) - (1 / 2)sin(x) + [ cos(x / 2) ]^2 } / { (1 / 2)sin(x) + [ cos(x / 2) ]^2 + [ sin(x / 2) ]^2 - (1 / 2)sin(x) ] =
1 / 1 =
1.

WolframAlpha Kontrolu:
https://www.wolframalpha.com/input?i=sin(x-y)=cos(y),tan(x/2-y)=?
("Substitution" başlığı altında)