sin25兀/6 cos25兀/3 tan(-25兀/4)的解法
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发布时间：2024-10-01 03:22
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sin25兀/6 =sin(4π+π/6）=sinπ/6=1/2 cos25兀/3=cos(8π+π/3)=cosπ/3=1/2 tan(-25兀/4)=tan(-6π-π/4)=tan(-π/4)=-1
Sin25派/6+Cos25派/3+tan(-25派/4) 计算题 急!要过程 谢谢= sin(π/6+4π) + cos(π/3+8π) + tan(-π/4+6π)=1/2 + 1/2 -1 =0
sin25π6+cos25π3+tan(-25π4)=__sin25π6+cos25π3+tan（-25π4）=sinπ6+cosπ3-tanπ4=12+12-1=0故答案为0．
计算:sin25π/6+cos25π/4+tan(-25π/4)+sin(-26π/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
sin25/6+cos25/3+tan(-25/4) 化简原式 =sin(1/6)+cos(1/3)+tan(-1/4)=1/2+1/2-1 =0
(Ⅰ)计算:sin25π6+cos25π3+tan(?25π4);(Ⅱ)已知tanα=3,求4sin...（Ⅰ）sin25π6+cos25π3+tan(?25π4)＝sinπ6+cosπ3?tanπ4＝12+12?1＝0…（4分）（Ⅱ）显然cosα≠0∴4sinα?2cosα5cosα+3sinα＝4sinα?2cosαcosα5cosα+3sinαcosα＝4tanα?25+3tanα＝4×3?25+3×3＝57…（8分）
sin(25派/3)+cos(-25派/6)+tan(-29派/4)sin(25π/3)=sin(8π+π/3)=sinπ/3=√3/2 cos(-25π/6)=cos(-4π-π/6)=cos(-π/6)=cosπ/6=√3/2 tan(-29π/4)=-tan29π/4=-tan(7π+π/4)=-tanπ/4=-1 和=√3-1
sin六分之二十五派加cos三分之二十五派加tan(负四分之二十五派) 求值...=sin(4π+π/6)+cos(8π+π/3)+tan(6π+π/4)=sinπ/6+cosπ/3+tanπ/4=1/2+1/2+1=2
...sin六分之五π + cos 三分之二十五 π + tan 负四分之二十五π...利用诱导公式，得 sin5π／6+ cos 25π ／3+ tan （-25π／4）=sinπ／6+ cos π／3+ tan （-π／4）=0.5+0.5-1=0
急! 简单的高一数学题,解答详细的加悬赏=sin90=1 (2)因为正切的最小正周期是兀，原式=tan(3x180+135)+tan(4x180+45)-tan(-330+2x180)+tan(-690+4x180)=tan135+tan45-tan30+tan30=0 (3)原式=sin(4兀+1/6兀）+cos(8兀+1/3兀）+tan(-25/4兀+7兀）=sin1/6兀+cos1/3兀+tan3/4兀=1/2+1/2+（-1）=0 ...
