(1)sin(-5π/12) (2)cos(-61π/12)
sin²α+√3sinαcosα-2cos²α=0,α∈(π/6,5π/12),求:(1)sin(2
sinπ/24*cosπ/24*sinπ/12=—— 答案是这么写的:1/2(2sinπ/24*cosπ/24)cosπ
为什么(sin&+cos&)^2=-1/5会变成sin&cos&=12/25?
【1】化简:sin(a-5π)/cos(3π-a)×cos(π/2-a)/sin(a-3π)×cos(8π-a)/sin
cos*π/12—sin*π/12
计算cosπ/12*sinπ/12
求值:sin(π/12)+cos(π/12)
已知 sinα+2cos(5π/2+α)/cos(π-α)-sin(π/2-α)=-1/4 求(sinα+cosα)平方
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)
已知α∈(π,2π) sinα+cosα=1/5 求sinα*cosα sinα-cosα
已知α=7/12π,那么cosα√(1-sinα)/(1+sinα)+sinα√(1-cosα)/(1-cosα)=
sinα+cosα=1/5,且π/2