问题: 高一数学计算
1）sinc50°（1+√3tan10°)
2)(1+cosθ-sinθ)/(1-cosθ-sinθ)+(1-cosθ-sinθ)/(1+cosθ-sinθ)
3)求证sin^2@+cos@cos(π/3+a)-sin^2(π/6-@)的值是与@无关的定值。
O(∩_∩)O谢谢。
解答:
sin50°(1+√3tan10°)
=sin50°(cos10°+√3sin10°)/cos10°
=2sin50°cos50°/cos10°
=sin100°/cos10°
=sin80°/sin80°
=1
(1+cosθ-sinθ)/(1-cosθ-sinθ)+(1-cosθ-sinθ)/(1+cosθ-sinθ)
=(2cos^2 θ/2-2sinθ/2cosθ/2)/(2sin^2 θ/2-2sinθ/2cosθ/2)+(2sin^2 θ/2-2sinθ/2cosθ/2)/(2cos^2 θ/2-2sinθ/2cosθ/2)
=2cosθ/2(cosθ/2-sinθ/2) / 2sinθ/2(sinθ/2-cosθ/2)+2sinθ/2(sinθ/2-cosθ/2) / 2cosθ/2(cosθ/2-sinθ/2)
=-cosθ/2 / sinθ/2-sinθ/2 / cosθ/2
=-（cos^2 θ/2+sin^2 θ/2) / sinθ/2cosθ/2
= -2/sinθ
3)sin^2@+cos@cos(π/3+@)-sin^2(π/6-@)
=sin^2@+cos@[cos@cosπ/3 -sinπ/3sin@ -
[sinπ/6cos@ +cosπ/6sin@ ]^2
打开并整理
=1/4 [sin^2@+cos^2@)
=1/4是与@无关的定值
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