tan(a+π/4)=2,则1+3sina*cosa-2*cosa^2=?
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/10/01 07:34:38
tan(a+π/4)=2,则1+3sina*cosa-2*cosa^2=?
tan(a+π/4)=2 = [1 + tana]/[1 - tana] ,解得tana = 1/3
1+3sina*cosa-2*cosa^2 = (3/2)·sin(2a) - cos(2a)
sin(2a) = 2tana/[1 + (tana)^2] = 3/5
cos(2a) = [1 - (tana)^2]/[1 + (tana)^2] = 4/5
故:1+3sina*cosa-2*cosa^2 = (3/2)·(3/5) - 4/5 = 1/10
1+3sina*cosa-2*cosa^2 = (3/2)·sin(2a) - cos(2a)
sin(2a) = 2tana/[1 + (tana)^2] = 3/5
cos(2a) = [1 - (tana)^2]/[1 + (tana)^2] = 4/5
故:1+3sina*cosa-2*cosa^2 = (3/2)·(3/5) - 4/5 = 1/10
证明(1-cos^2a)/(sina+cosa)-(sina+cosa)/(tan^2a-1)=sina+cosa
证明(1-cos^2a)/(sina-cosa)-(sina+cosa)/(tan^2-1)=sina+cosa
已知tan(a+π/4)=-3,则2sina-cosa/sina+cosa的值是多少?怎么算?
tan(a+π/4)=-1/2,求2cosa(sina-cosa)/(1+ tana)
tan a/2=3 1-cosa-sina/1+cosa+sina
求证:1-cos^2a/sina-cosa - sina+cosa/tan^2a-1=sina+cosa
已知tan(π-a)=2分之1,则2sina-cosa分之sina+cosa=
已知(2sinA+cosA)/(sinA-cosA)=-5 求1、(sinA+cosA)/(sinA-cosA) 2、3
已知tan(π+a)=-1/2 求2cosa(sina-cosa)/1+tana
[(sina+cosa)^2-(sina-cosa)^2]/tana-sinacosa=4/tan^2a 求证
tan(a/2)=sina/(1+cosa)=(1-cosa)/sina
证明(1+sinA-cosA)/(1+sinA+cosA)=tan(A/2)