晌：三角函数求值!求值：(1)(2cos10-sin20)/sin70=?(10,20,70.指度数)(2)sinA+sinB+sinC=0,cosA+cosB+cosC=0.则cos(A-B)=?

Question

晌：三角函数 求值!
求值：(1) (2cos10-sin20)/sin70=?(10,20,70.指度数)
(2)sinA+sinB+sinC=0,cosA+cosB+cosC=0.则cos(A-B)=?

Answer 1

1,原式=[2cos(70-60)-cos70]/sin70
=(cos70+√3sin70-cos70)/sin70
=√3
2,(sinA+sinB)^2=(-sinC)^2...(1)
(cosA+cosB)^2=(-cosC)^2...(2)
(1)+(2):sin^A+cos^2A+sin^2B+cos^2B
+2(cosAcosB+sinAsinB)=1
cosAcosB+sinAsinB=-1/2
cos(A-B)=cosAcosB+sinAsinB=-1/2