tam giác ABC có : cosC.(sinA+sinB)=sinC.cos(A-B)
tính cosA+cosB
\Leftrightarrow [TEX](-2).cos(A+B).sin(\frac{A+B}{2}).cos(\frac{A-B}{2})=sin(A+B).cos(A-B)[/TEX]
\Leftrightarrow [TEX](-2).cos(A+B).sin(\frac{A+B}{2}).cos(\frac{A-B}{2})=2sin(\frac{A+B}{2}).cos(\frac{A+B}{2}).cos(A-B)[/TEX]
\Leftrightarrow [TEX](-1).cos(A+B)cos(\frac{A-B}{2})=cos(\frac{A+B}{2}).cos(A-B)[/TEX]
\Leftrightarrow [TEX](-\frac{1}{2})[cos(\frac{A+3B}{2})+cos(\frac{3A+B}{2})]=\frac{1}{2}[cos(\frac{3B-A}{2})+cos(\frac{3A-B}{2})][/TEX]
\Leftrightarrow [TEX]cos(\frac{3B-A}{2})+cos(\frac{3B+A}{2})=-[cos(\frac{3A+B}{2})+cos(\frac{3A-B}{2})][/TEX]
\Leftrightarrow [TEX]cos(\frac{3B}{2}).cos(\frac{A}{2})=-cos(\frac{3A}{2}).cos(\frac{B}{2})[/TEX]
từ CT nhân 3 có [TEX]cos(\frac{A}{2}).cos(\frac{B}{2}).[4cos^2(\frac{B}{2})-3]=-[4cos^2(\frac{A}{2})-3].cos(\frac{A}{2}).cos(\frac{B}{2})[/TEX]
\Leftrightarrow [TEX]2(1+cosB)-3=3-2(1+cosA)[/TEX]
\Leftrightarrow [TEX]cosA+cosB=1[/TEX] (đpcm)
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