Question

CosA + cosB - cosC = -1 + 4cosA/2 cosB/2 sinC/2

Answer 1

Answer:

Step-by-step explanation:

(cos A+ cos B)-cos C

$=2cos \frac{A+B}{2}cos \frac{A-B}{2}-cos C~~~...(1)\\A+B+C=180\\A+B=180-C\\\frac{A+B}{2}=90-\frac{C}{2}\\cos \frac{A+B}{2}=cos(90-\frac{C}{2})=sin \frac{C}{2}\\cos C=1-2sin^2\frac{C}{2}\\(1)=2 sin \frac{C}{2} cos \frac{A-B}{2}-1+2sin^2\frac{C}2}\\=2sin\frac{C}{2}[cos \frac{A-B}{2}+sin \frac{C}{2}]-1~~~...(2)\\\\now~again~A+B+C=180\\C=180-(A+B)\\sin\frac{C}{2}=sin(90-\frac{A+B}{2})=cos \frac{A+B}{2}\\(2)=2sin\frac {C}{2}[cos \frac{A-B}{2}+cos \frac{A+B}{2}]-1$

$=-1+2sin\frac{C}{2}*2cos \frac{\frac{A-B}{2} +\frac{A+B}{2} }{2} cos \frac{\frac{A-B}{2} -\frac{A+B}{2} }{2} \\=-1+4sin\frac{C}{2} cos \frac{A}{2} cos\frac{-B}{2} \\=-1+4 cos \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}\\(cos(-B)=cos B)$