若tan(a+ b)=3tana 求证2sin2b-sin2a=sin(2a+3b)
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若tan(a+ b)=3tana 求证2sin2b-sin2a=sin(2a+3b)
好像是2sin2b-sin2a=sin(2a+2b)
再问: 啊…失误!
再答: tan(a+b)=3tana
sin(a+b)cosa=3sina*cos(a+b)
sinb=2*sina*cos(a+b)
2cosb*sinb=2cosb*2sina*cos(a+b)
sin2b=4cosb*sina*cos(a+b)
=2{sin(a+b)+sin(a-b)}*cos(a+b)
=sin(2a+2b)+2*sin(a-b)*cos(a+b)
=sin(2a+2b)+sin2a-sin2b
移项,2sin2b-sin2a=sin(2a+2b)
再问: 啊…失误!
再答: tan(a+b)=3tana
sin(a+b)cosa=3sina*cos(a+b)
sinb=2*sina*cos(a+b)
2cosb*sinb=2cosb*2sina*cos(a+b)
sin2b=4cosb*sina*cos(a+b)
=2{sin(a+b)+sin(a-b)}*cos(a+b)
=sin(2a+2b)+2*sin(a-b)*cos(a+b)
=sin(2a+2b)+sin2a-sin2b
移项,2sin2b-sin2a=sin(2a+2b)
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