sin164ºsin224º+sin254ºsin314º.sin﹙α+β﹚cos﹙γ-β﹚-cos﹙β+α﹚sin﹙β-γ﹚
sin﹙α-β﹚sin﹙β-γ﹚-cos﹙α-β﹚﹙γ-β﹚.﹙tan5π/4+tan5π/12﹚/1-tan5π/12.sin﹙α+β﹚-2sinαcosβ/2sinα...
sin﹙α-β﹚sin﹙β-γ﹚-cos﹙α-β﹚﹙γ-β﹚.﹙tan5π/4+tan5π/12﹚/1-tan5π/12. sin﹙α+β﹚-2sinαcosβ /2sinαsinβ+cos﹙α+β﹚化简 帮帮忙thankyou
展开
3个回答
展开全部
sin164°sin224°+sin254°sin314°=-sin16°sin44°+cos16°cos44°=cos(44°+16°)=cos60°=1/2
sin(α+β)cos(γ-β)-cos(β+α)sin(β-γ)=sin(α+β)cos(β-γ)-cos(α+β)sin(β-γ)=sin(α+β-β+γ)=sin(α+γ)
(tan5π/4+tan5π/12)/(1-tan5π/12)=(tanπ/4+tan5π/12)/(1-tanπ/4tan5π/12)=tan(π/4+5π/12)=tan2π/3=-√3
[sin(α+β)-2sinαcosβ]/[2sinαsinβ+cos(α+β)]=(cosαsinβ-sinαcosβ)/(cosαcosβ+sinαsinβ)=sin(β-α)/cos(β-α)=tan(β-α)
sin(α+β)cos(γ-β)-cos(β+α)sin(β-γ)=sin(α+β)cos(β-γ)-cos(α+β)sin(β-γ)=sin(α+β-β+γ)=sin(α+γ)
(tan5π/4+tan5π/12)/(1-tan5π/12)=(tanπ/4+tan5π/12)/(1-tanπ/4tan5π/12)=tan(π/4+5π/12)=tan2π/3=-√3
[sin(α+β)-2sinαcosβ]/[2sinαsinβ+cos(α+β)]=(cosαsinβ-sinαcosβ)/(cosαcosβ+sinαsinβ)=sin(β-α)/cos(β-α)=tan(β-α)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询