1. tan70°cos10(√3tan20°-1) = 2*tan70°cos10(cos30°sin20°-sin30°cos20°)/cos20° = 2*tan70°cos10[-sin(30°-20°)]/cos20° = -2*tan70°cos10sin10°/cos20° = -tan70°tan20° = -1 2. sin50°(1+√3tan10°） = 2*cos40°(sin30°cos10°+cos30°sin10°)/cos10° = 2*cos40°sin40°/cos10° = sin80°/cos10° = 1 3. sin＾4θ+cos＾4θ = [(1-cos2θ)/2]^2 +[(1+cos2θ)/2]^2 = [1 +(cos2θ)^2]/2 = [1 +(1+cos4θ)/2]/2 = (3 +cos4θ)/4
问：三角值cos20°cos10°+√3sin10°tan70°-2cos40°
答：因为 cot20=tan70 cot20cos10+(√3)sin10tan70-2cos40 =2cot20(1/2cos10+√3/2sin10)-2cos...详情>>
答：详情>>
