高中数学三角函数化简问题 高一数学三角函数的化简有什么技巧
\u9ad8\u4e2d\u6570\u5b66\u4e09\u89d2\u51fd\u6570\uff0c\u5316\u7b80\u6c42\u89d22sin²(c/2)+cos2c=1
\u22341-cosc+2cos²c-1=1
\u22342cos²c-cosc-1=(2cosc+1)(cosc-1)=0
\u2234cosc=1\u6216-1/2
\u2234c=2k\u03c0,\u6216c=2k\u03c0+2\u03c0/3,\u6216c=2k\u03c0+4\u03c0/3
(k\u2208z)
(\u5728\u4e09\u89d2\u5f62\u4e2dc\u53ea\u80fd\u53d62\u03c0/3)
\u5316\u7b80\u4e09\u6b65\u9aa4\uff0c\u9ad8\u6b21\u7684\u964d\u5e42\uff0c\u4e0d\u540c\u89d2\u7684\u5316\u540c\u89d2\uff0c\u90fd\u5316\u597d\u4e86\u5c31\u5316\u6210\u53ea\u6709\u4e00\u4e2asin \u5c31\u597d\u4e86
1sin²α+cosαcos(π/3+α)-sin²(π/6-α)
=sin²α+cosα(cosπ/3cosα-sinπ/3sina)-(sinπ/6cosα-cosπ/6sina)²
=sin²α+cosα(1/2cosα-√3/2sina)-(1/2cosa-√3/2sina)²
=sin²α+(1/2cosα-√3/2sina)(cosa-1/2cosa+3/2sina)
=sin²α+1/4cos²a-3/4son²a=1/4
2
sin50°（1+√3tan10°）-cos20°
=cos10°(sin50°（1+√3tan10°）-cos20°)/cos10°
=(sin50(cos10+√3sin10)-cos20cos10)/cos10
=(2sin50(1/2cos10+√3/2sin10)-cos20cos10)/cos10
=(2sin50cos(60-10)-cos20cos10)/cos10
=(2sin50cos(50)-cos20cos10)/cos10
=(sin100-cos20cos10)/cos10
=(cos10-cos20cos10)/cos10=1-cos20
所以原式=（1-cos20）/cos80°√1-cos20°=√(1-cos20）/cos80°
==√(1-(1-2sin10^2)）/cos80°=根号2sin10/cos80=根号2
1.化简sin²α+cosαcos(π/3+α)-sin²(π/6-α)
=sin²α+cosα(cosπ/3cosα-sinπ/3sina)-(sinπ/6cosα-cosπ/6sina)²
=sin²α+cosα(1/2cosα-√3/2sina)-(1/2cosa-√3/2sina)²
=sin²α+(1/2cosα-√3/2sina)(cosa-1/2cosa+3/2sina)
=sin²α+1/4cos²a-3/4son²a=1/4
.求值：[sin50°（1+√3tan10°）-cos20°]/ cos80°√1-cos20
=
①先计算sin(60°-θ)*sin(60°+θ)
=(sin60°cosθ-cos60°sinθ)*(sin60°cosθ+cos60°sinθ)
=sin²60°cos²θ-cos²60°sin²θ
=cos²θ-1／4
②再计算等式左边：
原式=4*(cos²θ-1／4)*sinθ
=4*[(1-sin²θ)-1／4]*sinθ
=3sinθ-4sin³θ
=sin3θ
∴原式=sin3θ
奇变偶不变,符号看象限
π/2是奇数
π是偶数
所以函数名不变
tan(-a)
符号看象限
假设-a是在第一象限,tan(-a-π)在第三象限
所以tan(-a-π)=tan(-a)
=-tan
a
至于下一题,看楼上好了,我俩是同一人(开玩笑)
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