已知ab属于(0,π 2)sinb=2cos(a b)sina,若tan
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(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(√