∠A=60°,求sin(A+10°)[1-根号三tan(A-10°)]的值?_数学解答

∠A=60°,求sin(A+10°)[1-根号三tan(A-10°)]的值?

人气：280 ℃　时间：2020-05-11 10:21:49

解答

sin(A+10°)[1-√3tan(A-10°)]=sin(A+10°)[1-√3sin(A-10°）/cos(A-10°）]=sin(A+10°)[cos(A-10°）--√3sin(A-10°）]/cos(A-10°）=2sin(A+10°)[1/2cos(A-10°）--√3/2sin(A-10°）]/cos(A-10°）=2sin(A+1...