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xuanquynh97

Ta có $1+4sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}$
$=1+4.\frac{1}{2}(cos\frac{A-B}{2}-cos\frac{A+B}{2}).sin\frac{C}{2}$
$=1+2(cos\frac{A-B}{2}.sin\frac{C}{2}-cos\frac{A+B}{2}.sin\frac{C}{2}$
$=1+2.\frac{1}{2}(sin\frac{C-A+B}{2}+sin\frac{A-B+C}{2}-sin\frac{C-A-B}{2}-sin\frac{A+B+C}{2})$
$=1+sin(90^o-A)+sin(90^o-B)-sin(C-90^o)$
$=cosA+cosB+cosC$
 
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