a+b+c=2π 证明sina+sinb+sinc=4sina/2sinb/2sinc/2
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(a+b)/2= π -c/2
sina+sinb+sinc
=2sin(a+b)cos(a-b)+2sin(c/2)cos(c/2)
=2sin[(a+b)/2]cos[(a-b)/2]-2sin[(a+b)/蔽察2]cos[(a+b)/2]
=2sin[(a+b)/2] [cos(a-b)/2-cos[(a+b)/2]
=2sinc/冲腔2 *[cos(a-b)/2-cos[(a+b)/散并衫2]
=4sina/2sinb/2sinc/2
得证.
sina+sinb+sinc
=2sin(a+b)cos(a-b)+2sin(c/2)cos(c/2)
=2sin[(a+b)/2]cos[(a-b)/2]-2sin[(a+b)/蔽察2]cos[(a+b)/2]
=2sin[(a+b)/2] [cos(a-b)/2-cos[(a+b)/2]
=2sinc/冲腔2 *[cos(a-b)/2-cos[(a+b)/散并衫2]
=4sina/2sinb/2sinc/2
得证.
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