高一数学选择题图1
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一、1.解:答案为D
原式=1/2*(2(sin35°)^2-1)/(cos(10°+80°)+cos(10-80°))/2
=1/2*2*cos70°/cos70°
=1
2.解:答案为A
三角形面积=1/2*AB*AC*sinA
AC=2*三角形面积/(AB*sinA)=2x√3/2x1/2x2/√3=1
3.解:答案为A
cos2α=(cosα)^2-(sinα)^2=(cosα+sinα))(cosα-sinα)
cos(π/4-α)=cosπ/4cosα+sinπ/4sinα
=√2/2cosα+√2/2sinα
=√2/2(cosα+sinα)
∵ cos2α=cos(π/4-α)
∴ (cosα+sinα))(cosα-sinα)=√2/2(cosα+sinα)
∴ cosα-sinα=√2/2
又 ∵ (cosα-sinα)^2=(cosα)^2-2sinαcosα+(sinα)^2=1-sin2α=1/2
∴ 1-sin2α=1/2 sin2α=1-1/2=1/2
原式=1/2*(2(sin35°)^2-1)/(cos(10°+80°)+cos(10-80°))/2
=1/2*2*cos70°/cos70°
=1
2.解:答案为A
三角形面积=1/2*AB*AC*sinA
AC=2*三角形面积/(AB*sinA)=2x√3/2x1/2x2/√3=1
3.解:答案为A
cos2α=(cosα)^2-(sinα)^2=(cosα+sinα))(cosα-sinα)
cos(π/4-α)=cosπ/4cosα+sinπ/4sinα
=√2/2cosα+√2/2sinα
=√2/2(cosα+sinα)
∵ cos2α=cos(π/4-α)
∴ (cosα+sinα))(cosα-sinα)=√2/2(cosα+sinα)
∴ cosα-sinα=√2/2
又 ∵ (cosα-sinα)^2=(cosα)^2-2sinαcosα+(sinα)^2=1-sin2α=1/2
∴ 1-sin2α=1/2 sin2α=1-1/2=1/2
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