cos5π/12-根3sin5π/12过程

来源：学生作业帮助网 编辑：作业帮 时间：2024/11/05 23:28:23

cos5π/12-根3sin5π/12过程
cos5π/12-根3sin5π/12
过程

cos5π/12-根3sin5π/12过程
这个其实直接用辅助角公式就可以求出来了：
acosx-bsinx=√(a²+b²)*cos(x+φ) (其中tanφ=b/a)
cos5π/12-√3sin5π/12
=2cos(5π/12+π/3)
=2cos3π/4
=-√2

1 ,cos5π/12·sinπ/12 =cos(π/4+π/6)*sin(π/4-π/6) =((1-tanαtanβ)=(tanα+tanβ) 1+4/3tanβ=-4/3tanβ ,tanβ=-

cos5π/12-根3sin5π/12过程 sin5/3π= ,cos5/3π (sina5π/12+cos5π/12)(sin5π/12-cos5π/12) cos5π/12-tanπ/3sin5π/12的值是 （1）根号2cos3x+根号6sin3x； （2）cos5π/12-根号3sin5π/12 (sin5π/12+cos5π/12)(sinπ/12-cosπ/12) (sin5π/12-sinπ/12)(cos5π/12+cosπ/12)=? sin5/12π×cos5/12π怎么算?请给出详细过程 谢谢高手了.. sin5π和cos5π等于多少 cos5/6π=?sin5/6π=? cos5分之πcos10分3π-sin5分之πsin10分之3π度等于多少 已知sin(α+β)=3/5,sinβ=1/2,α+β,β均为锐角,求sinα的值.已知sin(α+β)=3/5,cos（α-β）=1/3,α+β,α-β均为锐角,求sin2α、cos2β.求[（sin5π/12）+（cos5π/12）]·[（sin5π/12）-（cos5π/12）]的值.求1+2cos²θ-cos2 (cot20°-√3)(cos5°+sin5°)(cos5°-sin5°) (cot30°-√3)(cos5°+sin5°)(cos5°-sin5°)求值 求值(cot20°-根号3)(cos5°+sin5°)(cos5°-sin5°),要详程 (cot20°-√3)(cos5°+sin5°)(cos5°-sin5°)计算 sin4π+cos3/2π+tan3π-sin5/2π+cos5π,帮算下求下化简过程 计算cos5分之π+cos5分之2π+cos5分之3π+cos5分之4π