已知ab属于(0,π 2)sinb=2cos(a b)sina,若tan

(1)sin(wx+π/6)=sinwxcosπ/6+coswxsinπ/6sin(wx-π/6)=sinwxcosπ/6-coswxsinπ/6f(x)=sin(wx+π/6)+sin(wx-π/6

根据sinα+cosα=2/3,两边平方,可得（sina）^2+（cosa）^2+2sina*cosa=4/9这样得到,2sina*cosa=-5/9；而（sina-cosa）^2=（sina）^2+

y=4sin(2x+π/4)+1单调区间为：单调增：2kπ-π/2

tanx=1/2tan2x=2tanx/(1-tan²x)=4/3sinx/cosx=tanx=1/2cosx=2sinxsin²x+cos²x=1所以sinx=√5/5

sinα-cosα=根号2两边同时平方得：sina^2+cosa^2-2sinα*cosα=2即：1-sin2a=2答案是-1

1、cosθ－sinθ=√10/5,－√2sin(θ－π/4)=√10/5,则2sin(θ－π/4)=－2√5/5；2、=sin(4π/3)cosθ＋cos(4π/3)sinθ=sin(4π/3＋θ)

2π/w=6π所以w=1/3x/3+φ=π/2+2kπ或x/3+φ=-π/2+2kπ（k属于z)φ=π/3+2kπ或φ=-5π/6+2kπ又-π

方便起见,用a,b来表示α,β由题意得：sina=√2sinb√3cosa=√2cosb两式平方相加得：sin²a+3cos²a=2即：1+2cos²a=2得：cos&#

解.2sin²a-sinacosa-3cos²a=(2sina-3cosa)(sina+cosa)=0∵a∈(0,π/2）∴2sina=3cosa即sina=3/√13,cosa=

a=(cos3/2x,sin3/2x),b=(cosx/2,-sinx/2),a^2=1b^2=1,ab=cos3/2xcosx/2-sin3/2xsinx/2=cos(3/2x_x/2)=cos2x

设tan(α/2)=asinα=2sin(α/2)cos(α/2)=2a*cos(α/2)*cos(α/2)（打不出平方,不好意思）=2a/(1+a*a)=4/5cosα=2cos(α/2)*cos(

1/sinβ=(cosαcosβ-sinαsinβ)sinα整理得:(1+cosα*cosα)sinβ=2sinαcosαcosβ所以,tanβ=sinαcosα/(1+1-2(sinα)^2)/2=

题目你写的不是很清楚啊,请补充完整.这种题目一点错误麻烦可不是点点哦还要注意用括号很好3sinθ-cos2θ/(-cosθ)*cosθ=3sinθ+cos2θ=-2sinθ平方+sinθ+1=1得si

（sin2α-cos2α+1）/（sinα+cosα）=[（sinα+cosα）^2+(（sinα^2-cosα^2）]/（sinα+cosα）=sinα=1/√10cosα=3/√10tan（2α+

tanα=2tan(α/2)/(1-(tan(α/2))^2)由已知4tan(α/2)=(1-(tan(α/2))^2)tanα=1/2,cosα=2*5^(1/2)/5,sin2=5^(1/2)/5

∵tan(π+α)=3==>tanα=3==>sinα=3cosα.(1)==>(3cosα)²+(cosα)²=1==>(cosα)²=1/10.(2)∴3sin

sinα+cosα=根号2sin(α+π/4)sinα>0sin(cosα)=cos(π/2-cosα)

(2sina-cosa)(3sina+2cosa)=0故2sina=cosa3sina=-2cosaa属于(π/2,π),第二象限sin和cos异号sina=2/根13cos=-3/根13sin(a+

6sin方a+sinacosa-2cos方a=0(2sina-cosa)(3sina+2cosa)=02sina-cosa=0即tana=1/2或3sina+2cosa=0即tana=-2/3因为a是

(1)sin(2A+π/6)+sin(2A-π/6)+2cos^2A>=2(sin2Acosπ/6+cos2Asinπ/6)+(sin2Acosπ/6-cos2Asinπ/6)+cos2A+1≥2(√