若tanA、tanB是方程x^2-2(log8 72 + log9 72)x-log8 72·log9 72=0的两个根
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若tanA、tanB是方程x^2-2(log8 72 + log9 72)x-log8 72·log9 72=0的两个根,求sinA·cosB+cosA·sinB+2sinA·sinB的值
![若tanA、tanB是方程x^2-2(log8 72 + log9 72)x-log8 72·log9 72=0的两个根](/uploads/image/z/6690245-5-5.jpg?t=%E8%8B%A5tanA%E3%80%81tanB%E6%98%AF%E6%96%B9%E7%A8%8Bx%5E2-2%28log8+72+%2B+log9+72%29x-log8+72%C2%B7log9+72%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9)
tanA+tanB=2(log8 72 + log9 72)
=2(log8 9+log8 8 + log9 8+log9 9)
=2[(2/3)log2 3 +(3/2)log3 2 +2]
所以(sinAcosB+cosAsinB)/cosAcosB=sin(A+B)/cosAcosB
=2[(2/3)log2 3 +(3/2)log3 2 +2]------------①
tanAtanB=-log8 72·log9 72
=-[(2/3)log2 3+1][(3/2)log3 2+1]
=-[(2/3)log2 3 +(3/2)log3 2 +2]
所以sinAsinB/cosAcosB
=-[(2/3)log2 3 +(3/2)log3 2 +2]---------②
由①式和②式得(sinAcosB+cosAsinB)/cosAcosB=-2sinAsinB/cosAcosB
所以sinAcosB+cosAsinB+2sinAsinB=0
=2(log8 9+log8 8 + log9 8+log9 9)
=2[(2/3)log2 3 +(3/2)log3 2 +2]
所以(sinAcosB+cosAsinB)/cosAcosB=sin(A+B)/cosAcosB
=2[(2/3)log2 3 +(3/2)log3 2 +2]------------①
tanAtanB=-log8 72·log9 72
=-[(2/3)log2 3+1][(3/2)log3 2+1]
=-[(2/3)log2 3 +(3/2)log3 2 +2]
所以sinAsinB/cosAcosB
=-[(2/3)log2 3 +(3/2)log3 2 +2]---------②
由①式和②式得(sinAcosB+cosAsinB)/cosAcosB=-2sinAsinB/cosAcosB
所以sinAcosB+cosAsinB+2sinAsinB=0
若tanA、tanB是方程x^2-2(log8 72 + log9 72)x-log8 72·log9 72=0的两个根
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