(1+sina+cosa)(sin(a/2)-cos(a/2))/根号(2+2cosa)化简 (0<a<π)-cosa
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(1+sina+cosa)(sin(a/2)-cos(a/2))/根号(2+2cosa)化简(0<a<π)-cosa(1+sina+cosa)(sin(a/2)-cos(a/2))/根号(2+2cos
(1+sina+cosa)(sin(a/2)-cos(a/2))/根号(2+2cosa)化简 (0<a<π)-cosa
(1+sina+cosa)(sin(a/2)-cos(a/2))/根号(2+2cosa)
化简 (0<a<π)
-cosa
(1+sina+cosa)(sin(a/2)-cos(a/2))/根号(2+2cosa)化简 (0<a<π)-cosa
因为 √(2+2cosa)
=√(2+2(2cos^2(a/2)-1)
=√(4cos^2(a/2))
=-2cos(a/2)
因为 1+sina+cosa
=1+2sin(a/2)cos(a/2)+2cos^2(a/2)-1
=2cosa/2*(sina/2+cosa/2)
(1+sina+cosa)[(sin(a/2-cos(a/2)]/√(2+2cosa)
=[2cosa/2*(sina/2+cosa/2)](sina/2-cosa/2)/(-2cosa/2)
=-(sin^2a/2-cos^2a/2)
=cosa
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