cos2(π 4-a)-sin2(π 4 a)
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![cos2(π 4-a)-sin2(π 4 a)](/uploads/image/f/519887-47-7.jpg?t=cos2%28%CF%80+4-a%29-sin2%28%CF%80+4+a%29)
y=-cos(2x+π2)=sin2x,∵ω=2,∴T=π,∵sin(-2x)=-sin2x,则函数y为周期为π的奇函数.故选A
tan(α+π/4)=(1+tanα)/(1-tanα)=(sinα+cosα)/(cosα-sinα)=(sinα+cosα)^2/(cosα^2-sinα^2)=(1+sin2α)/cos2α=(
tan(π/4+θ)=(tanπ/4+tanθ)/(1-tanπ/4tanθ)=(1+tanθ)/(1-tanθ)=2由此可以解出tanθ=1/3.万能公式cos2θ=(1-tan^θ)/(1+tan
cos2(π/4-α)-sin2(π/4-α)=cos(π/2-2α)-sin(π/2-2α)=sin2α-cos2α=√2(√2/2sin2α-cos2α)=√2sin(2α-π/4)根据二倍角公式
题目写错了吧:应该是求证:sin2α+2cos2β=3证明:sin(π/4+α)=sinθ+cosθ(√2/2)(sina+cosa)=sinθ+cosθ两边同时平方得到:(1/2)(sin²
由两角和公式展开第一个等式得:根号2(sina+cosa)=sinθ+cosθ,再两边平方,得2(1+sin2a)=1+sin2θ得sin2a=(sin2θ-1)/2;cos2β=1-2sin^2β=
2sin(π/4+α)=√2(sina+cosa)√2(sina+cosa)=sinθ+cosθ将这个式子平方,得2(1+sin2a)=1+sin2θ2sin2β=sin2θ2(1+sin2a)=1+
(1)a·b=(cos2/3x,sin2/3x)*(cos2/x,-sin2/x)=cos2/3x*cos2/x-sin2/3x*sin2/x=cos(2/3x+2/x)=cos8/3x|a+b|=√
1f(x)=a·b+2λ|a+b|a·b=(cos(3x/2),sin(3x/2))·(cos(x/2),-sin(x/2))=cos(2x)|a+b|^2=|a|^2+|b|^2+2a·b=2+2c
由直角得:OA·OB=(a-b)(a+b)=a²-b²=0∴‖a‖=‖b‖由等腰得:‖OA‖=‖OB‖即‖a-b‖=‖a+b‖∴√(a-b)²=√(a+b)²∴
a=(cos3x/2,sin3x/2),b=(cosx/2,-sinx/2),(1)a*b=(cos3x/2,sin3x/2)*(cosx/2,-sinx/2)=cos(3x/2)*cos(x/2)-
sinπ=0cosπ=-1sin2π=0cos2π=1
∵tan(π4+α)=tanπ4+tanα1−tanπ4tanα=1+tanα1−tanα=12,∴3tanα=-1,解得:tanα=-13;∴sin2α−cos2α1+cos2α=2sinαcosα
题目是这样吧【tan(2α+π∕4)+6sin2α-cos2α】∕【3sin2α-2cos2α】方法是用辅助角公式展开tan2A=2tanA/(1-(tanA)^2);tan(A+B)=(tanA+t
原式=cos[2(π/4-a)]=sin(2a)公式:cos2x=(cosx)^2-(sinx)^2
tan(α+π/4)=(1+tanα)/(1-tanα)=(sinα+cosα)/(cosα-sinα)=(sinα+cosα)^2/(cosα^2-sinα^2)=(1+sin2α)/cos2α=(
设A=α+π/4,2A=2α+π/2tan(2A)=tan(2α+π/2)=-cot(2α)=-cos2α/sin2α=-b/atan(2A)=2tanA/(1-tanA^2)=-b/a解之得[2a+
tan[π/4+α]=(tanπ/4+tanα)/(1-tanπ/4tanα)=(1+sinα/cosα)/(1-sinα/cosα)=(cosα+sinα)/(cosα-sinα)=[2cos^2(
tan(θ+π/4)=sin(2θ+π/2)/[1+cos(2θ+π/2)]=cos(2θ)/[1-sin(2θ)]=b/[1-a]ortan(θ+π/4)=[1-cos(2θ+π/2)]/sin(2
tan(π/4-a)=[1-tana]/[1+tana]=3,则tana=-1/2.而sin2a-coa2a=[2sinacosa-cos²a+sin²a]/[sin²a