化简 sin(α 180°)cos(-α)sin(-α-180°).

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.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3

cos(-α)=cosα,cos(180°+α)=-cosαsin(540°-α)=sinαsin(360°-α)=-sinα所以原式左边=(1/cosα-cosα)/(1/sinα-sinα)=(1

化简cos(90°-α)/sin(270°-α)*sin(360°-α)*sin(180°-α)

cos(90°-α)/sin(270°-α)*sin(360°-α)*sin(180°-α)=sinα/(-cosα)*(-sinα)*sinα=1/sinacosa

化简cos(540°-α)sin(α-360°)/sin(-α+180°)cos(180°+α)

cos(540°-α)sin(α-360°)/[sin(-α+180°)cos(180°+α)]=cos(180°-α)[-sin(360°-α)]/[sin(180°-α)cos(180°+α)]=

化简[ sec(-θ)+cos(-θ+180° ) ] / [ cos(180 °-θ)+sin(360 °-θ ) ]

[sec(-θ)+cos(-θ+180°)]/[cos(180°-θ)+sin(360°-θ)]=(1/cosθ-cosθ)/(cosθ-sinθ)=1+sinθ/cosθ-(cosθ)^2+sinθ

化简:sinαcosαcos2α

sinacosacos2a=1/2sin2acos2a=1/4sin4a再问:最后一步怎么得出来的啊??再答:=1/4(2sin2acos2a)=1/4sin(2x2a)=1/4sin4a

化简:sin(β+180°)cos(-β)sin(-β-180°)

=-sinβsinβcosβ=-1/2sinβsin2β

Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α)

这道题是考三角函数的性质cos(180+a)=-cosasin(a+360)=sinasin(-a-180)=-sin(a+180)=sinacos(-180-a)=cos(180+a)=-cosa所

化简:tanα*(cosα-sinα)+[sinα(sinα+tanα)/1+cosα]

是tanα(cosα-sinα)+[sinα(sinα+tanα)/(1+cosα)]吧?先看sinα(sinα+tanα)/(1+cosα),分子为sinα(sinαcosα/cosα+sinα/c

化简cos^2(-α)-[tan(360°+α)/sin(-α)]

cos^2(-α)-[tan(360°+α)/sin(-α)]=cos^2(α)+tanα/sinα=cos^2(α)+1/cosα这已经是够简的了,若还有问题可以直接HI我.

化简 (1)、sin(α+180度)cos(-α)sin(-α-180度) (2)、sin3(-α)cos(2π+α)t

(1)sin(a+180)cos(-a)sin(-a-180)=(-sina)cosa(-sin(a+180))=sin²acosa.(2)sin³(-a)cos(2π+a)=(-

化简:cos(90°-a)/sin(270°+a)·sin(180°-a)·cos(360°-a)=

原式=(sina/cosa)*sina*cos(-a)=[(sina)^2/cosa]cosa=(sina)^2再问:谢谢能不能再问道若f(sina)=3-cos(3π/2-a),则f(cosx)=A

化简:cos平方(θ+15°)+sin平方(θ-15°)+sin(θ+180°)cos(θ-180°)

cos平方(θ+15°)+sin平方(θ-15°)+sin(θ+180°)cos(θ-180°)=[1+cos(2θ+30°)+1-cos(2θ-30°)]/2+sin(θ)cos(θ)=1-sin(

化简cos(180°+x)*sin(x+360°)/[sin(-x-180°)*cos(-180°-x)]

cos(180°+x)*sin(x+360°)/[sin(-x-180°)*cos(-180°-x)]=-cosx×sinx/[-sin(180°+x)][cos(180°+x)]=-cosx×sin

化简,cos²(-α)-tan(360°+α)/sin(-α)

cos^2(-α)-[tan(360°+α)/sin(-α)]=cos^2(α)+tanα/sinα=cos^2(α)+1/cosα再问:为什么直接变+了再答:sin(-a)=-sina,把负号放在前

化简sin(α+180°)cos(-α)sin(-α-180°)

sin(α+180°)cos(-α)sin(-α-180°)=-sin(α)cos(α)sin(α)=-2sin(α)sin(2α)

化简cos(α+30°)cos(α-30°)-sin(α+30°)sin(α+150°),

sin(α+150°)=sin(α+180°-30°)=-sin(α-30°)原式=cos[(α+30°-(α-30°)]=cos60°=1/2再问:sin可以这样转换?不可以的吧?只可以2π加减,不

化简:(sin²αtanα+cos²α/tanα+2sinαcosα)sinαcosα

(sin²αtanα+cos²α/tanα+2sinαcosα)sinαcosα=(sinα)^4+(cosα)^4+2sin²αcos²α=(sin²

化简:tanα(cosα-sinα)+sinα(sinα+tanα)/1+cosα.

切化弦tanα(cosα-sinα)+sinα(sinα+tanα)/1+cosα.=sinα(cosα-sinα)/cosα+sinα(sinα+sinα/cosα)/(1+cosα)=sinα(c

化简:cos2α / (cosα-sinα) - sinα

cos2α=(cosα)^2-(sinα)^2=(cosα+sinα)(cosα-sinα)故答案为cosα