sin25π/6+cos25π/3+tan25π/4等于多少?

duluabc2022-10-04 11:39:542条回答

已提交，审核后显示！提交回复

大山1971 共回答了23个问题 | 采纳率91.3%: sin(25π/6)+cos(25π/3)+tan(25π/4)
=sin(25π/6-4π)+cos(25π/3-8π)+tan(25π/4-6π)
=sin(π/6)+cos(π/3)+tan(π/4)
=(1/2)+(1/2)+(+1)
=2; 1年前

计算：sin25π/6+cos25π/4+tan(-25π/4)+sin(-26π/3) loveyisheng1年前2: 孤海蝴蝶共回答了15个问题 | 采纳率73.3%

原式=sin(4π+π/6)+cos(6π+π/4)-tan(6π+π/4)-sin(8π+2π/3)
=sinπ/6+cosπ/4-tanπ/4-sin2π/3
=1/2+√2/2-1-√3/2
=（√2-√3-1）/2

1年前查看全部

求值sin25π/6+cos25π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)-sin(- 求值sin25π/6+cos25π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)-sin(-5π/6)×cos(-5π/3) syywtyruci1年前1: gg我不怕终极PK 共回答了17个问题 | 采纳率76.5%

sin25π/6+cos25π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)-sin(-5π/6)×cos(-5π/3)
=sin(4π+π/6)+cos(8π+π/3)+tan(-25π/4+6π)+sin(-7π/3+2π)×cos(-13π/6+2π)-sin(π-π/6)×cos(-5π/3+2π)
=sinπ/6+cosπ/3+tan(-π/4)+sin(-π/3)×cos(-π/6)-sin(π/6)×cos(π/3)
=sinπ/6+cosπ/3-tanπ/4-sinπ/3×cosπ/6-sin(π/6)×cos(π/3)
=1/2+1/2-1-√3/2×√3/2-1/2×1/2
=-3/4-1/4
=-1

1年前查看全部