若tan(π+α)=3,α∈(π,3/2π),求sin^2α+3sinαcosα+2cos^2α

2sin^2σ*sin^2β+2cos^2σcos^2β
=2(1-cos^2a)(1-cos^2b)+2cos^2acos^2b
=2-2(cos^2a+cos^2b)+4cos^2acos^2b
=-cos2a-cos2b+4(cos2a+1)(cos2b+1)/4
=1+cos2σ*cos2β

1) f(x)=cos(x+2π/3)+2cos²(x/2)=-(cosx)/2-(√3sinx)/2+1+cosx=1-[(√3sinx)/2-(cosx)/2]=1-[sin(x-π/6)]/2, ∴ 1/2≤f(x)≤3/2, 值域[1/2,3/2].由2kπ-π/2≤x-π/6≤2kπ+π/2,得2kπ-π/3≤x≤2kπ+2π/3,由2kπ+π/2≤x-π/6≤2kπ+3π/2,得2kπ+π/3≤x≤2kπ+5π/3, ∴ 减区间[2kπ-π/3,2kπ+2π/3];增区间[2kπ+π/32,kπ+5π/3]
(k∈Z)
(2) 利用你的结论:a²+c²-√3ac=1---->1≥(2-√3)ac---->
ac≤2+√3, ∵ S=0.5acsin30°=√3ac/4, ∴ 4S/√3≤2+√3,
∴ 0

原式=[sinacosa+2(cosa)^2-2(sina)^2]/[(sina)^2+(cosa)^2]
分式上下同除(cosa)^2,
则原式=[tana+2-2(tana)^2]/[(tana)^2+1]=-13/10

f(x)=cosx+1-0.5cosx-根号3/2sinx=cos(x+π/3)+1 f(B)=1 ，cos(B+π/3)=0 ， B=π/6 再由余弦定理 ，cosB=(3+a^2-1)/(2根号3*a) 整理得 a^2-3a+2=0 得到a=1或a=2

tan(θ+π/4)=3
(1+tanθ)/(1-tanθ)=3
tanθ=1/2
5sin^2θ-3sinθcosθ+2cos^2θ=(5sin^2θ-3sinθcosθ+2cos^2θ)/(sin^2θ+cos^2θ)
=(5tan^2θ-3tanθ+2)/(ta^2nθ+1)=(5/4-3/2+2)/(1/4+1)=7/5

(1)(cosa+sina)/(cosa-sina)=(1+tana)/(1-tana)将tana=根号2代入即可(2)sin^2a-sinacosa+2cos^2a=1-0.5sin2a+cos2a万能公式sin2a=2tana/(1+tan^2a);cos2a=(1-tan^2a)/(1+tan^2a)代入即可谢谢采纳

因为
f(B)=1=cos(B+2/3π)+2cos^2(B/2)
=cosB*cos(2/3π)-sinB*sin(2/3π)+1+cosB
=1/2*cosB-√3/2*sinB+1
=cos(B+π/3)+1
即 cos(B+π/3)=0 得
B+π/3=π/2 ==> B=π/6
又 b^2=a^2+c^2-2*a*c*cosB , b=1 , c=√3
所以 a^2-3a+2=0
得 a1=1, a2=2
即 a 的值为1或2

证明cosα+cos3α=2*cos((α-3*α)/2)cos((α+3*α)/2)=2cosα*cos(2α)
1+cos(2α)=2(cosα)^2
上面两个式子相加有1+cosα+cos2α+cos3α=2cosα（cos(2α)+cos(α））
这样原题等价于cos(2α)+cos(α）=cosα+2cos^2α
cos(2α)=2cos^2α
所以原题出的有点问题或者抄错题目了.

最后的乘积项用二倍角公式展开
原式＝2sin^2(x)sin^2(φ)+2cos^2(x)cos^2(φ)-[cos^2(x)-sin^2(x)][cos^2(φ)-sin^2(φ)]
=2sin^2(x)sin^2(φ)+2cos^2(x)cos^2(φ)-[cos^2(x)cos^2(φ)-cos^2(x)sin^2(φ)-sin^2(x)cos^2(φ)+
sin^2(x)sin^2(φ)]
合并同类项 =sin^2(x)sin^2(φ)+cos^2(x)cos^2(φ)+cos^2(x)sin^2(φ)+sin^2(x)cos^2(φ)
　　＝sin^2(x)[sin^2(φ)+cos^2(φ)]+cos^2(x)[cos^2(φ)+sin^2(φ)]
　　＝sin^2(x)+cos^2(x)
　　＝1

据:cos2(x)=cosx^2-sinx^2=2cosx^2-1得:2sin^2(x)sin^2(φ)+2cos^2(x)cos^2(φ)-cos2(x)cos^2(φ)=2sin^2(x)sin^2(φ)+cos^2(φ)=(2sin^2(x)-1)*sin^2(φ)+sin^2(φ)+cos^2(φ)=(sin^2(x)-cos^2(x))*sin^2(φ)+1=-co...

解
sin²a+2cos²a
=(sin²a+2cos²a)/(sin²a+cos²a)——除以sin²a+cos²a=1,值不变
=(tan²a+2)/(tan²a+1)——分子分母同时除以cos²a
=(1/4+2)/(1/4+1)
=9/4×4/5
=9/5
sinacosa+2cos²a
=(sinacosa+2cos²a)/(sin²a+cos²a)——与第一题同理
=(tana+2)/(tan²a+1)
=(1/2+2)/(1/4+1)
=5/2×4/5
=2

sin²2θ+2cos²θcos2θ
=(2sinθcosθ)²+2cos²θ(1-2sin²θ)
=4sin²θcos²θ+2cos²θ-4sin²θcos²θ
=2cos²θ

sina/cosa=3
sina=3cosa
sin²a=9cos²a
sin²a+cos²a=1
所以cos²a=1/10
sin²a=9/10
sin2a=2sinacosa
=6cos²a
=6/10
所以原式=23/10