证明tan2分之塞塔-tan2分之塞塔分之1=-tan2分之塞塔分之2

看图.

tan2分之2α=tanα.sin4a=2sin2αcos2α=4sinαcosα[1-2(sinα)^2].

左边=cos²a/[(1+cosa)/sina]-[(1-cosa)/sina]=cos²a*sina/2*cosa=1/2sinacosa=1/4sin2a=右边即证!

方便起见,用a,b来表示：tan2a=tan[(a+b)+(a-b)]=[tan(a+b)+tan(a-b)]/[1-tan(a+b)tan(a-b)]=(3+5)/(1-15)=-4/7tan2b=

tan9=tan(9-2*3.14)>tan2

等式2边同时平方得：(sinα)^2-4sinαcosα+4(cosα)^2=5/21-2sin2α+3(cosα)^2=1+3/23(cosα)^2-2sin2α=3/2∵cos2α=2(cosα)

cos&=5分之4.tan&=4分之3.tan2&=7分之24.cos2&=25分之7

tan（a+π/4）+tan（a+3π/4）=tan（a+π/4）+tan（π/2+a+π/4）=tan（a+π/4）-cot（a+π/4）=sin(a+π/4)/cos(a+π/4)-cos(a+π

tan(a+4分之派)=2010,得(1+tana)/(1-tana)=2010,可得tana=2009/20111/cos2阿尔法=(sin^2a+cos^2a)/(cos^2a-sin^2a)=(

/>tan(π/4+α)=1/2∴[tan(π/4)+tanα]/[1-tan(π/4)tanα]=1/2∴(1+tanα)/(1-tanα)=1/2∴2+2tanα=1-tanα∴tanα=-1/3

已知A.B.C成等差数列则A+C=2B所以A+B+C=3B=180°故B=60°tan(B/2)=tan[90°-(A/2+C/2)]=cot(A/2+C/2)=1/tan(A/2+C/2)=[1-t

证明：由于A,B,C为△ABC中三个内角,则:tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=tanA/2*tanB/2+tanB/2*tan[pi/2-(A+B)

证：2sinβ/（cosα+cosβ）=[(sinα+sinβ)-(sinα-sinβ)]/(cosα+cosβ）=(sinα+sinβ)/（cosα+cosβ）-(sinα-sinβ)/(cosα+

(1)cosα=1/7,因为0＜α＜π/2,所以sinα=√(1-cosα)=√[1-(1/7)]=4√3/7所以tanα=sinα/cosα

证明：∵tan2θ＝2tanθ／(1－tan²θ)∴2tanθ／(tan²θ－1)＝－tan2θ∴(tan²θ－1)／2tanθ＝－1／tan2θ∴(tan²θ

tan（α+π/4）+tan（α+3π/4）=(tanα+tanπ/4)/(1-tanαtanπ/4)+(tanα+tan3π/4)/(1-tanα+tan3π/4)=(tanα+1)/(1-tanα

tan(2α-α)=(tan2α-tanα)/(1+tanαtan2α),tanαtan2α=(tan2α-tanα)/tanα-1tan2αtan3α=(tan3α-tan2α)/tanα-1┄┈┈

tan(π/4+α）-cot（π/4+α)=tan(π/4+α）-1/tan（π/4+α)=[tan²(π/4+α）-1】/tan(π/4+α）=-2/tan(π/2+2α)=-2cot(π

tan2α-sin2α=sin2α/cos2α-sin2α=sin2α-sin2αcos2α/cos2α=sin2α(1-cos2α)/cos2α=tan2αsin2α

tanα-1/tanα=sinα/cosα-cosα/sinα=(sin²α-cos²α)/(sinαcosα)=-cos2α/(1/2sin2α)=-2cos2α/sin2α=-