www.fltk.net > 已知α为锐角,且有2tAn(π%α)<os(π2+β)+5=0,tAn(π+α)+6sin(π+...

已知α为锐角,且有2tAn(π%α)<os(π2+β)+5=0,tAn(π+α)+6sin(π+...

α为锐角,且有2tan(π-α)-3cos(π 2 +β)+5=0,可得-2tanα+3sinβ+5=0,即2tanα-3sinβ-5=0…①由tan(π+α)+6sin(π+β)-1=0,可得:tanα-6sinβ-1=0…②,①*2-②得:3tanα-9=0,∴tanα=3.tanα=sinα cosα =3,∵sin2α+cos2α=1,∴sin2α=9 10 ,又α为锐角,sinα>0,sinα=3 10 10 .故答案为:3 10 10 .

即-2tana+3sinb+5=0tana-6sinb-1=0所以sinb=1/3,tana=3sina/cosa=tana=3cosa=1/3*sinacosa=1/9*sina因为sia+cosa=1所以sina=9/10a是锐角则sina>0sina=3√10/10

∵ 2tan(π-α)+3cos( π 2 +β)+7=0 ,tan(π+α)+6sin(π+β)-1=0∴-2tanα-3sinβ+7=0…①tanα-6sinβ-1=0…②①*2+②得tanα=3,即 sinα cosα =3∵sin 2 α+cos 2 α=1∵α为锐角,∴sinα= 3 10 10 故答案为: 3 10 10

由2TAN(π-α)-3COSπβ1=0得2TANα-12SINβ+4=0 ①由TANπα6SINπβ2=0得TANα-6SINβ+2=0 ②由①②解得 SINβ=5/9 ③将③代入② 得 TANα=3/4∴(SINα)^2=(TANα)^2/(1+(TANα)^2)=16/25∴SINα=4/5

2tan(π-α)+3cos(π/2-β)+5=0-2tanα+3sinβ=-5tan(π+α)+6sin(π+β)-1=0tanα-6sinβ=1所以tanα=3sinα/cosα=tanα=3sinα=3cosα代入恒等式sinα+cosα=19cosα+cosα=1

∵2tan(πx)3cos(π/2+y)+5=0,tan(π+x)+6sin(π+y)-1=0∴-2tanx+3siny+5=0…①tanx-6siny-1=0…②①*2+②得tanx=3∵x为锐角,∴sinx=(3√10)/10

∵2tan(π?α)+3cos( π 2 +β)+7=0,tan(π+α)+6sin(π+β)-1=0∴-2tanα-3sinβ+7=0…①tanα-6sinβ-1=0…②①*2+②得tanα=3,即 sinα cosα =3∵sin2α+cos2α=1∵α为锐角,∴sinα= 3 10 10 故答案为: 3 10 10

cos(π/6-2a)=cos[π/2-2(a+π/6)]=sin[2(a+π/6)]=2sin(a+π/6)cos(a+π/6),a为锐角,∴sin(a+π/6)>0,∴原方程变为cos(a+π/6)=3/5,易知sin(a+π/6)=4/5,∴sina=sin(a+π/6-π/6)=(√3/2)sin(a+π/6)-(1/2)cos(a+π/6)=(4√3-3)/10.

由已知可得-2tan α+3sin β+5=0,tan α-6sin β=1,解得tan α=3,故sin α= .

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