已知sin(π 4 )=1 3 求2sin2α

tan(π-α)=-tanα=2tanα=-2sinα=-2cosα代入=4+4-1/4-12+1=-1

因为：tan(a+π/4)=[tana+tanπ/4]/[1-tanatanπ/4]=(tana+1)/(1-tana)=3,解得：tana=1/2(cosα+2sinα)/(cosα-sinα)=(

sin（a-3π）=2cos(a-4π）可化为sin（4π+a-3π）=2cos(4π+a-4π）即sin（π+a)=-sina=2cosa[sin（π-a)+5cos(2π-a)]/[2sin(3π

由和差化积公式：cosa-cosb=-2sin（a+b）/2sin（a-b）/2可知sin(π/4+2a)sin(π/4-2a）=1/2*（cos4a-cosπ/2）所以cos4a=1/2因为a∈（π

（1）由sin（π+α）=√3/2∴sinα=-√3/2,cosα=-1/2∴sin（3π/4-α）=sin3π/4cosα-cos3π/4sinα=（√2/2）×（-1/2）-（-√2/2）（-√3

因为cos(π/2+x)=-sinx,sin(x-π/2)=sin[π-(x-π/2)]=sin(π/2-x)=cosx,由cos(π/2+x)=sin(x-π/2),得：-sinx=cosx.所以[

sin(π/4+α)*sin(π/4-α)=sin(π/4+α)*cos(π/4-α)=1/2sin(π/2+2α)=1/2cos2α1/2cos2α=1/4cos2α=1/2α∈(π/4,π/2),

sin和cos周期都是2π所以sin(a-3π)=sin(a+π)=-sina2cos(a-4π)=2cosa所以sina=-2cosa[sin(π-a)+5cos((3π/2)-a)]/[2sin(

[sinα+2cos(5π/2+α)]/[cos(π-α)-sin(π/2-α)]=-1/4即(sinα-2sinα)/(-cosα-cosα)=-1/4,所以sinα/(2cosα)=-1/4,所以

∵2sin²α-sinαcosα-3cos²α=0,∴（2sinα-3cosα）(sinα+cosα)=0.∵α∈（0,π/2）,∴sinα>0,cosα>0,sinα+cosα>

(2sina-3cosa)(sina+cosa)=0a是锐角,sina>0,cosa>0所以sina+cosa不等于0所以2sina=3cosasina=3/2*cosa代入sina+cosa=1co

sin(θ+kπ)=-2cos(θ+kπ),可得tanQ=-24sinθ-2cosθ/5cosθ+3sinθ（分子分母同时除以cosQ）=10⑵(1/4)sin平方θ+(2/5)cos平方θ（分子分母

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

(1)展开,得sinAcosπ/4+cosAsinπ/4+sinAcosπ/4-cosAsinπ/4=2sinAcosπ/4=√2×sinA=√2/3所以,sinA=1/3.(2)原式=[sinAco

运用公式sin(α+β)=sinα*cosβ+cosα*sinβsin(α-π/4)=sinα*cos（-π/4）+cosα*sin（-π/4）=（sinα-cosα）/√2=5/13(1)sinα^

因为cosα=1.5sinα,所以原式＝0.5/2.5+2.5/0.5=1/5+5=5.2

,而sin^2a+cos^2a=1,得sin^2a=4/5f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-π/4).=sinx(cosx+sinx)+2sin(x+π/4)

(tanπ/4+tana)/(1-tanπ/4tana)=3tanπ/4=1所以tana=1/2sina/cosa=tana=1/2cosa=2sina带入恒等式sin²a+cos²

sinA+cosA=√2×sin(A+π)=-√2×sinAcosA=-(√2+1)×sinA由于|cosA|≤1,|sinA|≤1可得：|sinA|≤√2-1S/100=(10-9cosA)(10-

tan(α+π/4)=(tanα+tanπ/4)/(1-tanatanπ/4)=(tanα+1)/(1-tanα)=2所以tanα=1/32cosα-sinα/cosα+3sinα=(2-sina/c