计算sin10π-√2cos(-19π/4)-tan(-13π/3)=
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 16:32:06
计算sin10π-√2cos(-19π/4)-tan(-13π/3)=计算sin10π-√2cos(-19π/4)-tan(-13π/3)=计算sin10π-√2cos(-19π/4)-tan(-13
计算sin10π-√2cos(-19π/4)-tan(-13π/3)=
计算sin10π-√2cos(-19π/4)-tan(-13π/3)=
计算sin10π-√2cos(-19π/4)-tan(-13π/3)=
计算sin10π-√2cos(-19π/4)-tan(-13π/3)=?
计算sin10π-√2cos(-19π/4)-tan(-13π/3)=
sin10π-√2cos(-19π/4)+tan(-13π/3)=
求值sin10π/3-根号2cos(-19π/4)+tan(13π/3)
sin²π/8-cos²π/8= 1/sin10`-√3/cos10`=
求值:sin10兀/3-√2cos(-19兀/4)+tan(-13兀/3)
计算 2sin10+cos10+tan20sin10,
计算:(1/cos*2140°-3/sin*140°)·1/2sin10°
计算2cos40度-cot20度cos度-根号3sin10度tan70度.求解过程
式子sin10π/3-根号2cos(-19π/4)-1/2tan(-13π/3)的值是
计算2sin10°+cos10°+tan20°sin10°
化解√(1+2sin10ºcos170º)/[cos10º-√(1-cos²170º)?
sin-3π/4= cos11π/6= tan13π/4= sin7π/2= cos-3π/2= sin10π/3=
高一数学简单题1化简√(1+sin10)+√(1-sin10)具体做法2 设sin a-sinb=1/3 cos a +cos b=1/2则cos(a+b)=
计算:cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)
1) tan5π/12 2)1/sin10°-√3/cos10º
不查表计算sin^2(20°)+cos^2(80°)+√3cos20°cos80° 我化到了(4+√3)/4-√3/2*(sin10+cos10)
已知x∈[0,π/2],求函数y=cos(π/12 - x)-cos(5π/12 + x)的值域?已知函数f(x)=a*sinx+b*cosx,(ab≠0)的最大值是2,且f(π/6)=根号3.求f(π/3)化简:根号【1+sin10°】+根号【1-sin10°】