已知sin平方2a+sin2acos2a-cos2a=1,a在第一象限,求sina,tana
人气:432 ℃ 时间:2020-02-04 21:59:13
解答
sin平方2a+sin2acos2a-cos2a=1
sin平方2a-1+cos2a(2sinacosa-1)=0
-(cos2a)^2+cos2a(2sinacosa-1)=0
cos2a(2sinacosa-1-cos2a)=0
cos2a[2sinacosa-1-2(cosa)^2+1]=0
cos2a[2sinacosa-2(cosa)^2]=0
2cosacos2a(sina-cosa)=0
cosacos2a(sina-cosa)=0
因为a在第一象限,cosa不等于0
即cos2a(sina-cosa)=0
则cos2a=0,或sina-cosa=0
cos2a=0,
1-2(sina)^2=0
sina=根号2/2,tana=1
sina-cosa=0
sina=cosa
sina=根号2/2,tana=1
所以sina=根号2/2,tana=1
推荐
- 已知sin^2a+sin2acosa-cos2a=1,a∈(0,π/2),求sina,tana的值(详细过程)
- 已知sin^2 2a+sin2acosa-cos2a=1,a E(0,π /2) 求sina tana
- 已知sin方2a+sin2acosa-cos2a=1,a∝(0,90),求sina,tana得值
- sin^2a+sin2a.cosa-cos2a=1,求sina,tana的值 a∈(0,pai/2)
- 已知sin^2a+sin2acosa-cos2a=1 a(0,兀/2) 求sina,tana的值
- kalenjin women won all their events as well(同义句转换)
- 虚数的虚数次方:i^i唯一吗
- 一个圆锥与一个圆柱的底面积比是3:2,体积比是2:5,如果圆柱的高与圆锥高之和是36厘米,求圆锥的高是多少厘米.
猜你喜欢