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

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

∠A=60°,求sin(A+10°)[1-根号三tan(A-10°)]的值?
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+10°)[sin30°cos(A-10°)--cos30°sin(A-10°)]/cos(A-10°)
=2sin(A+10°)sin(30°-A+10°)/cos(A-10°)
=2sin(A+10°)sin(40°-A)/cos(A-10°)
=2sin70°sin(-20°)/cos50°
=-2sin70°cos70°/cos50°
=-sin140°/cos50°
=-sin40°/cos50°
=-cos50°/cos50°
=-1