已知tan(a+45)=2则1+3sina·cosa-2cos^2=?这就是你那天答得题,
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已知tan(a+45)=2则1+3sina·cosa-2cos^2=?这就是你那天答得题,
/>tan(a+π/4)=2 = [1 + tana]/[1 - tana] ,
解得tana = 1/3
1+3sina*cosa-2*cos^2a
=3sina*cosa+(1-2*cos^2a )
= (3/2)·sin(2a) - cos(2a)
sin(2a) = 2tana/[1 + (tana)^2] = 3/5
cos(2a) = [1 - (tana)^2]/[1 + (tana)^2] = 4/5
故:1+3sina*cosa-2*cos^2a = (3/2)·(3/5) - 4/5 = 1/10
解得tana = 1/3
1+3sina*cosa-2*cos^2a
=3sina*cosa+(1-2*cos^2a )
= (3/2)·sin(2a) - cos(2a)
sin(2a) = 2tana/[1 + (tana)^2] = 3/5
cos(2a) = [1 - (tana)^2]/[1 + (tana)^2] = 4/5
故:1+3sina*cosa-2*cos^2a = (3/2)·(3/5) - 4/5 = 1/10
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