$sin^2a+sin^2b+sin^2c=1-2sina.sinb.sinc$$\Leftrightarrow sin^2a+sin^2b+sin^2c-1=-2sina.sinb.sinc$
Ta có $sin^2a+sin^2b+sin^2c-1=sin^2a+sin^2b-cos^2c$
$= sin^2a+sin^2b-sin^2(a+b)$ ( vì $a+b+c=\frac{\pi}{2})$
$= sin^2a+sin^2b-(sina.cosb+sinb.cosa)^2$
$= sin^2a+sin^2b-sin^2a.cos^2b-sin^2b.cos^2a-2sina.sinb.cosa.cosb$
$=sin^2a(1-cos^2b)+sin^2b(1-cos^2a)-2sina.sinb.cosa.cosb$
$=sin^2a.sin^2b+sin^2b.sin^2a-2sina.sinb.cosa.cosb$
$=2sina.sinb(sina.sinb-cosa.cosb)$
$=2sina.sinb.-cos(a+b)$
$=-2sina.sinb.sinc$
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