已知sin^2B=2sin^2A-1求证tan^2A=2tan^B+1
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 08:27:03
已知sin^2B=2sin^2A-1求证tan^2A=2tan^B+1
已知sin^2B=2sin^2A-1求证tan^2A=2tan^B+1
已知sin^2B=2sin^2A-1求证tan^2A=2tan^B+1
证明:sin^2B=2sin^2A-1得cos^2B=1-sin^2B=1-(2sin^2A-1)=2(1-sin^2A)=2cos^2A
于是tan^2A-2tan^B=sin^2A/cos^2A-2sin^2B/cos^2B=sin^2A/cos^2A-2sin^2B/(2cos^2A)
=sin^2A/cos^2A-sin^2B/cos^2A=(sin^2A-sin^2B)/cos^2A=[sin^2A-(2sin^2A-1)]/cos^2A
=(1-sin^2A)/cos^2A=1
故tan^2A=2tan^B+1
sin^2B=2sin^2A-1求证tan^2A=2tan^B+1sin^2B=2sin^2A-1求证tan^2A=2tan^B+1
求证:sin(a+b)sin(a-b)=sin^2a-sin^2b
3 在三角形ABC中,已知(a2+b2)sin(A-B)=(a2-b2)sin(A+B) 求证:ABC是等腰或直角三角形(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B) sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B)) sin^A*2c
已知3sin b=sin(2a+b) 求证tan(a+b)=2tana
已知 3sin A = sin (2B +A ),求证 : tan(A+B)= 2tanB
在△ABC中,求证;sin^(A/2)+sin^(B/2)+sin^(C/2)=1-2sin(A/2)sin(B/2)sin(C/2)
已知cos(a+B)+1=0,求证sin(2a+B)+sinB=0?(提示:sin(-a)=-sina).
求解 已知 sin(a+b)=1/2 ,sin(a_b)=1/3,求证:tan a=5tan b求解:已知 sin(a+b)=1/2 ,sin(a_b)=1/3,求证:tan a=5tan b
已知(tan^2)a=2(tan^2)B+1,求证:(sin^2)B=2(sin^2)A-1
已知tan^2A=2tan^B+1求证sin^2B=2sin^2A-1救急
已知tan^2a=2tan^2b+1,求证:sin^b=2sin^2a-1
已知tan^2A=2tan^B+1,求证sin^2B=2sin^2A-1
已知sin^2B=2sin^2A-1求证tan^2A=2tan^B+1
已知tan^2 a=2tan^ B +1,求证sin^2 b=sin^2 a -1
已知tan(a-r)/tana+sin^2b/sin^2a=1 ,求证tan^2b=tana*tanr..
已知tan²a=2tan²b+1,求证sin²b=2sin²a-1
已知tan²a=2tan²B+1,求证sin²B+1=2sin²a
已知tan²a=2tan²b+1,求证sin²b+1=2sin²a
已知sin(a)=2sin(b)求sin(a/2)/sin(b/2)的值.