已知sina>0,tana>0,写出角a的集合,并指出a 2角所在象限
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sina+cosa=tanasina+cosa=sina/cosasina.cosa+cosa.cosa=sinasina.cosa+1-sina.sina=sinacosa=(sina.sina+s
(sina-cosa)²=(7/13)²展开整理得:1+2sinacosa=49/169解得:sinacosa=60/169(sina+cosa)²=(sina-cosa
(sina+cosa)/(tana+cota)=(sina+cosa)sinacosa=cos²a*sina(1+tana)0,tana再问:(sina+cosa)/(tana+cota)=
(sinA+cosA)^2=1/4sin^2A+cos^2A+2sinAcosA=1/42sinAcosA=-3/4sinAcosA=-3/8sinAcosA/(sin^2A+cos^2A)=-3/8
因为sina+cosa=√2sin(α+π/4)又因0
sina+cosa=√2/3两边平方可得sin²a+2sinacosa+cos²a=2/92sinacosa=-7/9<0.∴结合0<a<π可知,sina>0,cosa<0.∴si
CosA+sinA=1/5,A属于(0,派),单位圆三角函数线得90
原题是不是这样的(sin2a)^2+sin2acosa-cos2a=1如果是的话,解如下:(2SinaCosa)^2+2Sina(Cosa)^2-{[2(Cosa)^2]-1}=14[(Sina)^2
tana=sina+cosa=√2sin(a+π/4)∵0<a<π/2∴π/4<a+π/4<3π/4∴1<√2sin(a+π/4)≤√2∴1<tana≤√2∴π/4<a≤arctg√2再问:第一步是为
sina=±【根号下(1+m平方)】分之mcosa=±【根号下(1+m平方)】分之1TAT不会打你看得懂咩╮(╯_╰)╭过啦.
sina+cosa=-7/13①(sina+cosa)²=sin²a+cos²a+2sinacosa=49/169sin²a+cos²a=12sina
sina+cosa=1/5,且0<a<π由勾股数组3,4,5得:sina=4/5,cosa=-3/5所以,tana=-4/3
sina+cosa=1/5sin²a+cos²a=1解方程,sina=4/5,cosa=-3/5或sina=-3/5,cosa=4/5(舍,因为a属于(0,π)tana=sina/
在三角形中,有0
sina+cosa=tanasina+cosa=sina/cosasina.cosa+cosa.cosa=sinasina.cosa+1-sina.sina=sinacosa=(sina.sina+s
tana=sina+cosa=√2sin(a+π/4)∵0<a<π/2∴π/4<a+π/4<3π/4∴1<√2sin(a+π/4)≤√2∴1<tana≤√2∴π/4<a≤arctg√2
sinAcosA+cos^2A=sinAsinAcosA+1-sin^2A=sinAcosA=(sin^2A+sinA-1)/snA∵0<sinA<1,0<cosA<1∴sin^2A+sinA-1>0
sina+cosa=tana-1/tanasina+cosa=sina/cosa-cosa/sinasina+cosa=(sin²a-cos²a)/sinacosa1=(sina-
2(sinA)^2=1-cos2A=1-[1-(tanA)^2]/[1+(tanA)^2]=2(tanA)^2/[1+(tanA)^2]=2[(Sina-cosa)/(Sina+cosa)]^2/{1