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已知a属于(0,二分之π),且2sin^2a -sinacosa-3cos^2a=0,求sin(a+π/4)/(sin2a+cos2a+1)
人气:182 ℃ 时间:2020-03-29 20:19:33
解答
2(sina)^2-sinacosa-3(cosa)^2
=(2sina-3cosa)(sina+cosa)=0
∵00
∴sina+cosa>0
∴2sina-3cosa=0
tana=3/2,
sina=3/√13,cosa=2/√13
sin(a+π/4)/(sin2a+cos2a+1)
=√2/2(sina+cosa)/(2sinacosa+2(cosa)^2)
=√2/(4cosa)
=√26/8
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