作业帮(cosa+1)/(sina+3)=ma为[0°,180°]求m的取值范围
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(cosa+1)/(sina+3)=ma为[0°,180°]求m的取值范围
(cosa+1)/(sina+3)=m
a为[0°,180°]
求m的取值范围
(cosa+1)/(sina+3)=ma为[0°,180°]求m的取值范围
m=(sina+3-2)(sina+3)
=(sina+3)/(sina+3)-2/(sina+3)
=1-2/(sina+3)
0<=a<=180
0<=sina<=1
3<=sina+3<=4
所以1/4<=1/(sina+3)<=1/3
-2/3<=-2/(sina+3)<=-1/2
1-2/3<=1-2/(sina+3)<=1-1/2
所以1/3<=m<=1/2
y=cos²x+sinxcosx
=(1+cos2x)/2+1/2*sin2x
=(1/2)(sin2x+cos2x)+1/2
=(√2/2)(√2/2*sin2x+√2/2cos2x)+1/2
=(√2/2)(sin2xcosπ/4+cos2xsinπ/4)+1/2
=(√2/2)sin(2x+π/4)+1/2
T=2π/2=π
3sina=2cosa 则(cosa-sina/cosa+sina)+(cosa+sina/cosa-sina)=______
3sina=2cosa 则(cosa-sina/cosa+sina)+(cosa+sina/cosa-sina)=______
3sina-cosa/sina+3cosa=1,求tana
已知（2sinA+cosA)/(sinA-cosA)=-5 求1、(sinA+cosA)/(sinA-cosA) 2、3cos2A+sin2A
tana=-1/3 sina*sina+2sina*cosa-3cosa*cosa
证明=[(sina+cosa)+(sina+cosa)²]/(1+sina+cosa) =(sina+cosa)(1+sina+cosa)/(1+sina+cosa)
sina+sina=1 求cosa+cosa
求证：1+sina+cosa/1+sina-cosa+1-cosa+sina/1+cosa+sina=2/sina
求证：2sina+sin2a=2(sina)^3/(1-cosa)(sina)^3/(1-cosa)=sina*(1+cosa) 看不懂
(cosa+1)/(sina+3)=ma为[0°,180°]求m的取值范围
1+sina/cosa=?
证明：cosa/(1+sina)-sina/(1+cosa)=2(cosa-sina)/(1+sina+cosa)
求证cosa/1+sina-sina/1+cosa=2（cosa-sina）/1+sina+cosa
求证：1+sina+cosa+2sina cosa/1+sina+cosa=sina+cosa
tana=-1/2,则2*sina-3*sina*cosa-5*(cosa)*(cosa)的值是
证明：2(cosa-cosa)/(1+cosa+cosa)=cosa/(1+sina)-sina/(1+cosa).
证明 cosa/(1+sina0-sina/(1+cosa)=2(cosa-sina)/(1+sina+cos)cosa/(1+sina）-sina/(1+cosa)=2(cosa-sina)/(1+sina+cos)
已知3sina=-cosa,求值 （1）sina-3cosa/sina+3cosa （2）1+sinacosa