Question

三角函数题目，高中的，谢谢大神

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Answer 1

sin(x+20°)=cos(90°-x-20°)=cos(70°-x)
cos(x+10°)+cos(x-10°)=2cosxcos10°
cos(70°-x)=2cosxcos10°
cos70°cosx+sin70°sinx=2cosxcos10°
cos70°+sin70°tanx=2cos10°
tanx=(2cos10°-cos70°)/sin70°
=[2cos(30°-20°)-sin20°]/cos20°
=(√3cos20°+sin20°- sin20°)/cos20°
=√3
(2)等式右边=sin(8π/15)/cos(22π/15)=tan(8π/15)=tan(π/5+π/3)
等式左边=分子分母同除以acos(π/5)=(tanπ/5+b/a)/(1-b/a·tanπ/5)
∴(tanπ/5+b/a)/(1-b/a·tanπ/5)=tan(π/5+π/3)
(tanπ/5+b/a)/(1-b/a·tanπ/5)=(tanπ/5+tanπ/3)/(1-tanπ/5tanπ/3)
(tanπ/5+b/a)/(1-b/a·tanπ/5)=(tanπ/5+√3)/(1-√3tanπ/5)
所以b/a=√3