tan2分之π
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2cos^2Θ/2-sinΘ-1/√2sin(Θ+π/4)=(cosΘ-sinΘ)/(cosΘ+sinΘ){在分子和分母上同时除以cosΘ}=(1-tanΘ)/(1+tanΘ)tan2Θ=-2√2=2
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!
tan(α+四分之五π)=tan(α+四分之π)tan2/α=-2则从tanα=(2tan2/α)/(1-tan2/α的平方),且α是第三象限角所以tanα=4/3tan(α+四分之五π)=tan(α
答案如图片再问:根号2乘上sin(π/4+α)为什么等于sinα+cosα?再答:用公式展开
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
tan2分之α=2tana=2tana/2/(1-tan^2a/2)=-4/3⑴tan(α+4分之π)=(tana+1)/(1-tana)=(-1/3)/(7/3)=-1/7⑵(6sinα+cosα)
tan(α+2α)=tan(3α)=tan(3·π/9)=tan(π/3)=√3又tan(α+2α)=[tanα+tan(2α)]/[1-tanα·tan(2α)]因此[tanα+tan(2α)]/[
f(x)=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)∵f(θ+π/8)=√2/3∴f(θ+π/8)=√2sin[2^(θ+π/8)+π/4]=√2sin(2θ+