急 ,sin2A/sin2B+cos2Acos2C=1,证明tan2A/tan2B=sin2C

来源：学生作业帮助网编辑：作业帮时间：2024/05/06 05:45:07

急 ,sin2A/sin2B+cos2Acos2C=1,证明tan2A/tan2B=sin2C
急 ,sin2A/sin2B+cos2Acos2C=1,证明tan2A/tan2B=sin2C

急 ,sin2A/sin2B+cos2Acos2C=1,证明tan2A/tan2B=sin2C
题中的2不是二倍,而是平方.
证明：
sin²A/sin²B+cos²Acos²C=1
cos²C= (1-sin²A/sin²B)/cos²A
= 1/cos²A - sin²A/cos²A/sin²B
= (sin²A+cos²A)/cos²A - tan²A/sin²B
= tan²A + 1 - tan²A/sin²B
= 1 + tan²A(1- 1/sin²B)
= 1 + tan²A(sin²B-1)/sin²B {sin²B-1 = -cos²B}
= 1 - tan²Acos²B/sin²B
= 1 - tan²A/tan²B
sin²C= 1-cos²C
= 1-(1 - tan²A/tan²B)
= tan²A/tan²B
证明完毕.

sin2a+sin2b-sin2a*sin2b+cos2a*cos2b=1求证求证明：sin2A+sin2B-sin2A*sin2B+cos2A*cos2B=1 sin2a+sin2b-sin2a*sin2b+cos2a*cos2b=1 2是平方若三角形中,A>B,如何证明sin2A>sin2B,和cos2A 证明(sin2a-sin2b)^2+(cos2a+cos2b)^2=4*(cos(a+b))^2 sin2a+cos2a sin2a-cos2a/cos2a-sin2a等于为什么sin2A(1+cos2B)+sin2B(1+cos2A) =4sinAcosA(cosB)^2+4sinBcosB(cosA)^2?怎么化出来的? (1-sin2a+cos2a)/(1+sin2a-cos2a 化简：(1+sin2a+cos2a)/(1+sin2a-cos2a) sin2A+cos2A(0° 化简sin2a-cos2a能化简吗 (1+sin2a+cos2a)/(1+sin2a-cos2a)+(1+sin2a-cos2a)/(1+sin2a+cos2a)= (1-cos2a)/sin2a化简化简 1-cos2a/sin2a 2sin2a*cos2a=? 急 ,sin2A/sin2B+cos2Acos2C=1,证明tan2A/tan2B=sin2C (y-sin2a)/(sin2b-sin2a)=(x-cos2a)/(cos2b-cos2a)如何化简得 xcos(a+b)+ysin(a+b)-cos(a-b)=0阿尔法、贝塔不会打,用a、b表示了.如题!