cos(-π/4)*sin(5/6π)=?

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cos(-π/4)*sin(5/6π)=?
cos(-π/4)*sin(5/6π)=?

cos(-π/4)*sin(5/6π)=?
√2/4.

cos(-π/4)*sin(5/6π)=? 已知tan(α+π/4)=3,计算2sinαcosα+6cos方α-3/5cos方α-6sinαcosα-5sin方α sin(π+α)+cos(5π-α))/(sin(α-6π)+2cos(2π+α)=(-sinα-cosα)/(sinα+2cosα)为什么相等 sin(-19π/6)=?已知[3Sin(π+a)+Cos(π-a)]/4Sin(-a)-Sin(5π/2+a)=2,求tan a 三角函数求解!难题我采纳!10.化简.(1)【sin(2π-α)cos([π/2]+α)cos([π/2]-α)】/【sin(3π-α)sin(-π-α)sin([π/2]+α)】(2)【an(2π-θ)sin(2π-θ)cos(6π-θ)】/【(-cosθ)sin(5π+θ)】11.已知tanα=-2,求sin^2α-4sinαcos 已知sin(Π/6-α)=2cos(α-4Π),求cos(Π/2-α)+5sin(Π/2+α)分之2cos(Π/2+α)-sin(-α) 设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π+α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2(-sinα)(-cosα)+cosα/[1+sin² cos a=-4/5 第二象限 求sin a(a+、- π/6) cos(a+、-6/π) 为什么cos(β-α)=cos(α-β),已知sin(α- β)cosα-cos ( β- α )sin α =3/5,β是第三象限角,则sin ( β+5π/4)=因为cosα是偶函数,所以cos(β-α)=cos(α-β),则原式=sin(α-β)cosα-cos(β-α)sinα=sin(α-β)co 已知tanθ=(sin α-cos α)/(sin α+cos α) a,θ(0,π/2) 求证cos(3/2兀+ α) -sin(5π/2-α)=根号2sin(θ-4π) 已知6sin²α+sinαcosα-2cos²α=0,α∈(π/2,π),求值1).(sinα-3cosα)/(sinα-cosα);2).sinαcosα-sin²α;3).sin²α-3cosαsinα-2 已知sinα=4/5,且3π/2<α<2π,求sinα-cosα及(1/cos²α)-(1/sin²α)刚才有些错误,sinα+cosα=4/5 【三角函数】已知sin(α-π)=2cos(2π-α),求[sin(π-α)+5cos(2π-α)]/[3cos(π-α)-sin(-α)] sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=______ 已知tanα=3,分别求下列各式的值:(1)4sinα-2cosα/5cosα+3sinα(2)sinαcosα(3)(sinα+cosα)²(4)2sin²α+sinαcosα=3cos²α已知sin(π+α)=-1/2,分别求下列各式的值:(1)cos(2π-α)(2 若sin(α-π)=2cos(2π-α),求(sinα+5cosα)/(sinα-3cosα)的值, 已知α∈(π,2π) sinα+cosα=1/5 求sinα*cosα sinα-cosα 已知tan(π-α)=2,求sin²α-2sinαcosα-cos²α/4cos²α-3sin²α+1