cosa=1/7,cos(a-b)=13/14且0

cosa=1/7,cos(a-b)=13/14且0 COSa=1/7,cos(a-b)=13/14,且0 三角函数 已知cosA=1/7 cos（A-B)=13/14 且0 已知cosa=1/7,cos(a-b)=13/14,0 19.已知cosa=1/7,cos(a-b)=13/14,0 已知cosa=1/7,cos(a-b )=13/14,且0 已知cosa=1/7,cos(a-b)=13/14,且0 已知cosA=1/7,cos(A-B)=13/14,且0 已知cosa=1/7,cos(a-b)=-13/14,且0 已知cosa=1/7,cos(a-b)=13/14且,0 cosA=1/7,cos(A+B)=13/14,o cosa=1/7,cos(a+b)=-13/14,o 若cosa*cosb=1,则cos(a+b)= cosa cosb=1,则cos(a-b)= 若cosa*cosb=1,则cos(a+b)= 求证cos^2A+cos^2B+cos^2C+2*cosA*cosB*cosC=1 有2/cosA=1/cos(A+B) + 1/cos(A-B) ,证 2^(1/2)cosA=cosB cos(a+b)=12/13,cos(2a+b)=3/5,求cosa