若0≤α<β<γ<π,且sinα+sinβ+sin=γ0,cosα+cosβ+cosγ=0,求β-α的值
问题描述:
若0≤α<β<γ<π,且sinα+sinβ+sin=γ0,cosα+cosβ+cosγ=0,求β-α的值
是sinα+sinβ+sinγ=0
答
sinv=-(sina+sinb)sin^2v=(sina+sinb)^2=sin^2a+sin^2b+2sinasinbcosv=-(cosa+cosb)cos^2v=(cosa+cosb)^2=cos^2a+cos^2b+2cosacosb于是 sin^2v+cos^2v=sin^2a+sin^2b+2sinasinb+cos^2a+cos^2b+2cosacosb=2+2(sinasin...