求助！数学问题三角函数

显示全部楼层 · 2009-8-21 15:08:37

tan（-690°）= √3/3,tan29π/3= -√3,cot(-35π/3)=-√3/3,sin(-1071°)sin99°+sin(-171°)sin(-261°)=sin9°cos9°+sin9°cos9°=2sin18°sin347°cos148°+sin77°cos58°=-sin(-13°)cos(32°)+sin77°cos58°=cos77°sin58°+sin77°cos58°=sin135°=√2/2cosα=1/7,cos(α+β)=-11/14, α,β∈(0,π/2), cosβ=cos(α+β-α)=cos(α+β)cosα+sin(α+β)sinα=-11/14*1/7+5√3/14*4√3/...

千问 · 2009-8-21 15:08:37

tan（-690°）=tan30°=根号3/3tan29π/3=tan2p/3=-根号3/3sin(-17π/6)=sin7p/6=-1/2cot(-35π/3)=cotp/3=根号3/3 sin(-1071°)sin99°+sin(-171°)sin(-261°)=sin9°cos9°-sin9°cos9°=0sin347°cos148...

千问 · 2009-8-21 15:08:37

tan（-690°）= tan30=√3/3 tan29π/3=tan(-π/3)=-√3 sin(-17π/6)=sin(7π/6)=-1/2cot(-35π/3)=cotπ/3=√3/3sin(-1071°)sin99°+sin(-171°)sin(-261°)=sin9sin99-sin9sin99=0sin347°co...

千问 · 2009-8-21 15:08:37

tan（-690°）= tan（180°×4-690°）=tan30 °=√3/3tan29π/3=tan(29π/3-9π)=tan2π/3=-√3sin(-17π/6)=sin(-17π/6+4π)=sin(7π/6)=-1/2cot(-35π/3)=cot(-35π/3-12π)=cot(π/3)=√3/3sin(-1071°)si...

千问 · 2009-8-21 15:08:37

1 tan（-690°）=-tan690°=-tan330°=-tan(360°-30°)=-tan(-30°)=tan30°=√3/32 tan29π/3=tan(2π*5-π/3)=tan(-3π)=-√3/33 sin(-17π/6)=-sin(17π/6)=-sin(5π/6)=-1/24 cot(-35π/3)=1/(tan(-35π/...

求助！数学问题 三角函数

求助！数学问题三角函数