不理解 sin^2 π\12-cos^2 π\12=
sinπ/24*cosπ/24*sinπ/12=—— 答案是这么写的:1/2(2sinπ/24*cosπ/24)cosπ
sin²α+√3sinαcosα-2cos²α=0,α∈(π/6,5π/12),求:(1)sin(2
化简[cos(a-π)/sin(π-a)]sin(a-π/2)cos(π/2+a)=
为什么(sin&+cos&)^2=-1/5会变成sin&cos&=12/25?
已知(3sinα+cosα)/(3cosα-sinα)=2,则2-3sin(α-3π)sin(1.5π-α)-[cos(
不用计算器求值sinπ\8 cosπ\8=sin^2 π\12-cos^2 π\12=化简:(cosx+sinx)^2=
设函数f(x)=2+sin(3x+π/12)cos(x+π/6)+cos(3x+π/12)sin(x+π/6)
θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小
求证:sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos
sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin(π-α)·cos(π/2+α)/sin(π+α)=
已知cos(π/2+a)=2sin(a-π/2)求sin(π-a)+cos(a+π)/5cos(5π/2-a)+3sin
已知cos(π/2+α)=sin(α-π/2) 求sin(π-α)-cos(π+α)/cos(5π/2 -α)+2sin