证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/24 23:11:40
证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°
证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2
证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2
证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2
证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2
左边=sin²a+sin²(a-60°)-2sinasin(a-60°)cos60°
=〔(sina-sin(a-60°)〕²+sinasin(a-60°)
=cos²(a-30°)-½〔cos(2a-60°)-cos60°〕
=½〔(cos2a-60°)+1〕-½cos(2a-60°) +½cos60°=¾=(sin60°)²=右边.
证毕.
步骤如下:
因为:sin(120°-α))与sin(120°-α)抵消则得出
(sinα)^2+sin-2sinα=(sin60°)^2
证明:sin(2α+β)/sinα - 2cos(α+β)=sinβ/sinα
证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2
证明:sin(360°-α)=-sinα
证明:1-cos2α/sinα=2sinα
证明cosα(cosα-cosβ)+sinα(sinα-sinβ)=2sin^2(α-β/2)第二个 证明sin(α+β)cosα-1/2[sin(2α+β)-sinβ]=sinβ
【证明】Sin A+sin B=2Sin 22
时间 证明sin(2α+β)/sinα-2cos(α+β)=sinβ/sin时间 证明sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
证明(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)=0
请证明在三角形ABC中:cosα=(sin²γ+sin²β-sin²α )/2sinγ*sinβ
在三角形ABC中,证明cosα=(sin²β+sin²γ-sin²α)/(2sinβ*sinγ)
三角函数证明(sinα+sinθ)*(sinα-sinθ)=sin(α+θ)*sin(α-θ)求证(sinα+sinθ)*(sinα-sinθ)=sin(α+θ)*sin(α-θ)
证明sin(α+β)sin(α-β)=sinα-sinβ
证明sin(a+b)sin(a-b)=sin^2 a-sin^2 b,
如何证明sin(sin(sinx))
证明:sin(-α)sin(丌-α)-tan(-α)cot(α-丌)-2cos^2(-α)+1=sin^2α
如何证明 sinα-sinβ=2cos((α+β)/2) ·sin ((α-β)/2)公式
求值:(tan10°-√3)×cos10°/sin50° 求证:[sin(2α+β)]/sinα- 2cos(α+β)=sinβ/cosα 化简cosθ+cos(θ+2π/3)+cos(θ+4π/3) 证明:sin(α+β)sin(α-β)=sin^2α-sin^2β
证明:(1+sinα-cosα)/(1+sinα+cosα)=tan(α/2)