Question

求解sin40º(tan10º-√3)

Answer 1

解: sin40°(tan10°-√3)
=sin40°(sin10°-√3cos10°)/cos10°
=-2sin40°(√3/2·cos10°+1/2·sin10° ) /cos10°
=-2cos40°(sin30°cos10°-cos30°sin 10° /cos10°
=-2cos40°sin(30°+10°)/cos10°
=-2cos40°sin40°/cos10°
=-sin2·40°/cos10°
=-sin80°/sin80°
=-1

Answer 2

sin40º(tan10º-√3)
=sin40º(sin10º-√3cos10°)/cos10°
=-2sin40°cos40°/cos10°
=-sin80°/sin80°
=-1.