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Prove the equation
 
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[Solved] Prove the equation

 
Trigonometry 1 & 2
Last Post by Staff 4 years ago
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June 15, 2021 12:44 pm
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(@armor)
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June 15, 2021 12:44 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Since     \small acosB-bcosA=\frac{1}{c}(accosB-bccosA)\small =\frac{a^2+c^2-b^2}{2c}-\frac{b^2+c^2-a^2}{2c}=\frac{a^2-b^2}{c},   we have

\small \frac{acosB-bcosA}{sin(A-B)}=\frac{a^2-b^2}{csin(A-B)}=\frac{(\frac{c}{sinC}sinA)^2-(\frac{c}{sinC}sinB)^2}{csin(A-B)}

\small =\frac{c^2sin^2A-c^2sin^2B}{csin^2Csin(A-B)}=\frac{c(sinA-sinB)(sinA+sinB)}{sin^2Csin(A-B)}

\small =\frac{c2sin\frac{A-B}{2}cos\frac{A+B}{2}2sin\frac{A+B}{2}cos\frac{A-B}{2}}{sin^2Csin(A-B)}

\small =\frac{csin(A-B)sin(A+B)}{sin^2Csin(A-B)}=\frac{csin(A+B)}{sin^2C}=\frac{csinC}{sin^2C}=\frac{c}{sinC}.

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