(Cos²1°-cos²2°)/(sin3°•sin1°)

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(Cos²1°-cos²2°)/(sin3°•sin1°)
2cos³θ+sin²(360°-θ)+cos(360°-θ)-3/2+2cos²(180°-θ)+cos(-θ)怎样化

解题思路:同学,这个式子是不能化为cosθ-1,利用诱导公式就可以了,但要注意平方解题过程:

[2cos³θ+sin²(360°-θ)+cos(360°-θ)-3]/[2+2cos²(180°-θ)+cos(-θ)

解题思路:利用诱导公式和立方差公式来分解因式后约分化简。解题过程:

求证:sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos

sin^2/(sin-cos)-(sin+cos)/(tan^2-1)=sin^2/(sin-cos)-(sin+cos)/[(sin^2/cos^2)-1]=sin^2/(sin-cos)-(sin+cos)cos^2/(sin^2-co

化简:根号1+2sin(40°)cos(40°)

按照你的答案,你的题目输入有误,是化简:根号1-2sin(40°)cos(40°),原因是算术平方根是非负的.1-2sin(40°)cos(40°)=sin²40°+cos²40°-2sin(40°)cos(40°)=(

化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ

(sin²θ+2sinθcosθ+cos²θ+sinθ+cosθ)/(1+sinθ+cosθ)=[(sinθ+cosθ)²+(sinθ+cosθ)]/(1+sinθ+cosθ)=(sinθ+cosθ)(1+si

计算[3/(sin^2)20°-[1/(cos^2)20°]+64(sin^2)20°

原式=[3(cos20)^2-(sin20)^2]/(sin20)^2(cos20)^2+64(sin20)^2=[(√3cos20+sin20)(√3cos20-sin20)]/(sin20cos20)^2+64(sin20)^2=[2s

2cos x (sin x -cos x)+1

2cosx(sinx-cosx)+1=2sinxcosx-2cosx^2+1=sin2x+1-2cosx^2=sin2x-cos2x=√2sin(2x-π/4)

2sin@+cos@等于?

(2sina+cosa)=-5(sina-3cosa)7sina=14cosasina=2cosa

化简 (1-2sin@cos@)/(cos@^2-sin@^2)*(1+2sin@cos@)/(1-2sin@^2) 两

(sin@-cos@)^2/(cos@-sin@)(cos@+sin@)*(sin@+cos@)^2/(cos@-sin@)(cos@+sin@)=1

求2sin(-1110°)-sin960°+根号2cos(-225°)-cos(-210°)

2sin(-1110°)-sin960°+根号2cos(-225°)-cos(-210°)=2sin(-30-3*360)-sin(360*3-120)+根号2cos(135)-cos150=2*(-1/2)-(-(根号3)/2)+(根号2

化简1:cos(15°-a)sin15°-sin(165°+a)cos(-15°)

sin(165°+a)=sin[180`-(15°-a)]=sin(15°-a)cos(-15°)=cos15°1.原式=sin[15°-(15°-a)]=sina2、展开根号下{[(1-根号3)/2]*(cosa+sina)}=根号下{[

(cos^2 10°-cos^2 80°)+sin^2 20°

(cos^210°-cos^280°)²+sin^220°=(cos^210°-sin^210°)²+sin^220°=cos²20+sin²20°=1

2cos 30°-2sin 60°·cos 45°=?

2cos30°-2sin60°cos45°=2*(√3/2)-2*(√3/2)*(√2/2)=√3-√6/2

求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ

sinθ/(1+cosθ)+(1+cosθ)/sinθ=[sinθ^2+(1+cosθ)^2]/sinθ(1+cosθ)=(sinθ^2+1+2cosθ+cosθ^2)/sinθ(1+cosθ)=(2+2cosθ)/sinθ(1+cosθ)

sinα=-2cosα,求sin^2α-3sinαcosα+1

sina=-2cosatana=-2sin²a-3sinacosa+1=(sin²a-3sinacosa+sin²a+cos²a)/(sin²a+cos²a)=(tan²

求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ

左边=(2cos^2θ/2+cosθ/2)/2sinθ/2cosθ/2+sinθ/2=cosθ/2(2cosθ/2+1)/sinθ/2(2cosθ/2+1)=cosθ/2/sinθ/2=1/tanθ/2=1/(1-cosθ/sinθ)=si

证明Cos^A-Sin^A=1-2Sin^A=2Cos^A-1=cos^a-sin^a

根据余弦2倍角公式cos(a+b)=cosa*cosb-sina*sinbcos2a=cos(a+a)=cosa*cosa-sina*sina=cos²a-sin²a再根据三角函数的恒等式sin²a+cos&s

sinα^2+sinβ^2+sinγ^2=1,那么cosαcosβcosγ最大值等于

令x=cosα,y=cosβ,z=cosγ,则1=(sinα)^2+(sinβ)^2+(sinγ)^2=(1-x^2)+(1-y^2)+(1-z^2)=3-(x^2+y^2+z^2),所以x^2+y^2+z^2=2.由基本不等式,三次根号(

求证(1+sinα+cosα+2sincosα)/(1+sinα+cosα)=sinα+cosα

(1+sinα+cosα)*(sinα+cosα)=sinα+cosα+(sinα+cosα)^2=(sinα+cosα)+(sinα^2+2sincosα+cosα^2)=(1+sinα+cosα+2sincosα)两边除以(1+sinα

已知tan∝=2/3,求值:(1)(cos∝-sin∝/cos∝+sin∝)+(cos∝+sin

(cosα-sinα)/(cosα+sinα)+(cosα+sinα)/(cosα-sinα)=[(cosα-sinα)^2+(cosa+sinα)^2]/[(cosα)^2-(sinα)^2]=2[(cosα)^2+(sinα)^2]/[