2道三角函数问题化简{(SINA+COSA-1)(SINA-COSA+1)}/SIN2A已知sin^2 2a+sin2acosa-cos2a=1,A是第一象限求SINA和TAN A````

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2道三角函数问题化简{(SINA+COSA-1)(SINA-COSA+1)}/SIN2A已知sin^22a+sin2acosa-cos2a=1,A是第一象限求SINA和TANA````2道三角函数问题

2道三角函数问题化简{(SINA+COSA-1)(SINA-COSA+1)}/SIN2A已知sin^2 2a+sin2acosa-cos2a=1,A是第一象限求SINA和TAN A````
2道三角函数问题
化简
{(SINA+COSA-1)(SINA-COSA+1)}/SIN2A
已知sin^2 2a+sin2acosa-cos2a=1,A是第一象限
求SINA和TAN A````

2道三角函数问题化简{(SINA+COSA-1)(SINA-COSA+1)}/SIN2A已知sin^2 2a+sin2acosa-cos2a=1,A是第一象限求SINA和TAN A````
{(SINA+COSA-1)(SINA-COSA+1)}/SIN2A
={sinA+[cosA-1]}{sinA-[cosA-1]}/sin2A
=[sin^2A-cos^2A +2cosA-1]/sin2A
=[-2cos^2A +2cosA]/2sinAcosA
=1-cosA
-------
sinA
sin^2 2a+sin2acosa-cos2a=1
4sin^2a*cos^2a+2sinacos^2a+2sin^2a-1=1
2sin^2a*cos^2a+ sinacos^2a+ sin^2a=1
2sin^2a*cos^2a+ sinacos^2a+ sin^2a=sin^2a+cos
^2a
2sin^2a*cos^2a+ sinacos^2a+ = +cos
^2a
同除cos^2a
2sin^2a+ sina=1
因为A第一象限
所以sina=1/2
tana=三分之根号三