两角和与差的正弦1.已知5sinB=sin(2a+b),cosa≠0,cos(a+b)≠0,求证:tan(a+b)=3/2tana2.若将1.中的5改为m(m≠1),则tan(a+b)与tana之间有何关系?答案是tan(a+b)=(m+1/m-1)tana
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![两角和与差的正弦1.已知5sinB=sin(2a+b),cosa≠0,cos(a+b)≠0,求证:tan(a+b)=3/2tana2.若将1.中的5改为m(m≠1),则tan(a+b)与tana之间有何关系?答案是tan(a+b)=(m+1/m-1)tana](/uploads/image/z/10146951-63-1.jpg?t=%E4%B8%A4%E8%A7%92%E5%92%8C%E4%B8%8E%E5%B7%AE%E7%9A%84%E6%AD%A3%E5%BC%A61.%E5%B7%B2%E7%9F%A55sinB%3Dsin%282a%2Bb%29%2Ccosa%E2%89%A00%2Ccos%28a%2Bb%29%E2%89%A00%2C%E6%B1%82%E8%AF%81%3Atan%28a%2Bb%29%3D3%2F2tana2.%E8%8B%A5%E5%B0%861.%E4%B8%AD%E7%9A%845%E6%94%B9%E4%B8%BAm%28m%E2%89%A01%EF%BC%89%2C%E5%88%99tan%28a%2Bb%29%E4%B8%8Etana%E4%B9%8B%E9%97%B4%E6%9C%89%E4%BD%95%E5%85%B3%E7%B3%BB%3F%E7%AD%94%E6%A1%88%E6%98%AFtan%28a%2Bb%29%3D%28m%2B1%2Fm-1%29tana)
两角和与差的正弦1.已知5sinB=sin(2a+b),cosa≠0,cos(a+b)≠0,求证:tan(a+b)=3/2tana2.若将1.中的5改为m(m≠1),则tan(a+b)与tana之间有何关系?答案是tan(a+b)=(m+1/m-1)tana
两角和与差的正弦
1.已知5sinB=sin(2a+b),cosa≠0,cos(a+b)≠0,求证:tan(a+b)=3/2tana
2.若将1.中的5改为m(m≠1),则tan(a+b)与tana之间有何关系?
答案是tan(a+b)=(m+1/m-1)tana
两角和与差的正弦1.已知5sinB=sin(2a+b),cosa≠0,cos(a+b)≠0,求证:tan(a+b)=3/2tana2.若将1.中的5改为m(m≠1),则tan(a+b)与tana之间有何关系?答案是tan(a+b)=(m+1/m-1)tana
1.
5sinb=sin(2a+b)
5sin(a+b-a)=sin(a+b+a)
5[sin(a+b)cosa-cos(a+b)sina]=sin(a+b)cosa+cos(a+b)sina
4sin(a+b)cosa=6cos(a+b)sina
sin(a+b)/cos(a+b)=3/2sina/cosa
tan(a+b)=3/2tana
2.
msinb=sin(2a+b)
msin(a+b-a)=sin(a+b+a)
m[sin(a+b)cosa-cos(a+b)sina]=sin(a+b)cosa+cos(a+b)sina
(m-1)sin(a+b)cosa=(m+1)cos(a+b)sina
sin(a+b)/cos(a+b)=(m+1/m-1)sina/cosa
tan(a+b)=(m+1/m-1)tana