Question

高一数学必修4题目

tan70°cos10°（根号3xtan20°—1）x是乘号

Answer 1

tan70×cos10×(√3tan20-1)
=tan70×cos10×(tan60×tan20-1)
=tan70×cos10×[(sin60×sin20/cos60×cos20)-1]
=tan70×cos10×(sin60×sin20-cos60×cos20)/(cos60×cos20)
=tan70×cos10×[-cos(60+20)]/(cos60 ×cos20)
=-tan70×cos10×cos80/(cos60×cos20)
=-tan70×cos10×sin10/(cos60×cos20)
=-(sin70/cos70)×(1/2)×sin20/(cos60×cos20)
=-(cos20/sin20)×sin20/(2cos60×cos20)
=-1/(2cos60)
=-1

Answer 2

=cos10°（根号3·tan70°·tan20°—tan70°）
=cos10°（根号3—tan70°）
=cos10°（tan60°—tan70°）
=cos10°· tan(60°-70°)·(1+tan60°·tan70°)
=cos10°· tan(-10°)·(1+tan60°·tan70°)
=-sin10°·(cos60°·cos70°+sin60°·sin70°)/(cos60°·cos70°)
=-sin10°·cos(60°-70°)/(cos60°·cos70°)
=-sin10°·cos10°/(cos60°·cos70°)
=-2sin10°·cos10°/(2·cos60°·sin20°)
=-sin20°/(2·cos60°·sin20°)
=-1/(2·cos60°)
= -1

Answer 3

原式=sin70°／cos70°×cos10°×(√3×sin20／cos20－1)
=cos20°/sin20°×cos10°×(√3×sin20／cos20－1)
=cos20°／2sin10°×(√3×sin20／cos20－1)
=√3×sin20°／2sin10°－cos20°／2sin10°
=2[(√3／2)sin20°－(1／2)cos20°]／2sin10°
=sin(20°－30°)／sin10°
=－1