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tAnAtAn(π/4+A)sin(3π/2+A)

解:用诱导公式一个个简化,tan(π/4+α)=cot(π/4-α)=cos(π/4-α)/sin(π/4-α);sin(3π/2+α)=sin[2π-(π/2-α)]=-sin(π/2-α)=-cosα;cos(5π/2+α)=cos(2π+π/2+α)=cos(π/2+α)=-sinα;代入得:tanαtan(π/4+α)sin(3π/2+α)/[2cos(π/4-α)cos(5π/2+α)]=cos(π/4-α)/sin(π/4-α)/2cos(π/4-α)=2sin(π/4-α)cos(π/4-α)=sin(π/2-2α)=cos2α【cos2α=cosα-sinα=2cosα-1=1-2sinα】

sin(π/2+a)=cosa=-√5/5∴a∈(π/2,π)∴sina=2√5/5[cos(π/4+a/2)-cos(π/4-a/2)]/[sin(π-a)+cos(3π+a)]分子平方差公式=[cos(π/4+a/2)+cos(π/4-a/2)][cos(π/4+a/2)-cos(π/4-a/2)]/(sina-cosa)=√2cos(a/2)*(-√2sin(a/2))/(sina-cosa)=-2sin(a/2)cos(a/2)/(

第二个中间是不是漏了一项应该是sin(π-a)sin(π/2-a)1、(tana+tanπ/4)/(1-tanatanπ/4)=1/3(tana+1)/(1-tana)=1/33tana+3=1-tanatana=-1/22、sina/cosa=tana=-1/2cosa=-2sina代入sina+cosa=1sina=1/5cosa=4/5sinacosa=sina(-2sina)=-2sina=-2/5原式=2sina-sinacosa+cosa=2/5+2/5+4/5=8/5

〔sin(a-3π)+cos(π+a)〕/[sin(-a)-cos(π-a)]=(-sina-cosa)/(-sina+cosa)=(tana+1)/(tana-1) ∵tan(π+a)=2 ∴-tana=2即tana=-2∴原式=(-2+1)/(-2-1)=1/3

-2cosa+sina=0sina=2cosasina=4cosa因为sina+cosa=1所以cosa=1/5,sina=4/5a在第三象限所以cosa=-√5/5,sina=-2√5/5所以sin(a+π/4)=sinacosπ/4+cosasinπ/4=-3√10/10tana=sina/cosa=2所以tan(a+π/4)=(tana+tanπ/4)/(1-tanatanπ/4)=(2+1)/(1-2)=-3

1、(tana+tanπ/4)/(1-tanatanπ/4)=1/3(tana+1)/(1-tana)=1/33tana+3=1-tanatana=-1/22、sina/cosa=tana=-1/2cosa=-2sina代入sina+cosa=1sina=1/5cosa=4/5sinacosa=sina(-2sina)=-2sina=-2/5原式=2sina-sinacosa+cosa=2/5+2/5+4/5=8/5

cos(π+a)=-cosatan(π+a)=tana

因为y=tanx的周期是π 而tan(3π/2+a)=tan(π+π/2+a) 所以tan(3π/2+a)=tan(π/2+a) 龙者轻吟为您解惑,凤者轻舞闻您追问.如若满意,请点击[满意答案];如若您有不满意之处,请指出,我一定改正!希望还您一个正确答复!祝您学业进步!

sin(a+π/3)=1/4sinacosπ/3+cosasinπ/3=1/41/2*sina+√3/2*cosa=1/4√3cosa=1/2-sina两边平方3cosa=3(1-sina)=1/4-sina+sina4sina-sina-11/4=0sina=(1±3√5)/8a属于(0,π),sina>0所以sina=(1+3√5)/8

证明:∵tan(a+π/4)=(tana+tanπ/4)/(1-tana*tanπ/4)=(tana+1)/(1-tana)tan(a+3π/4)=(tana+tan3π/4)/(1-tana*tan3π/4)=(tana-1)/(1+tana)∴tan(a+π/4)+tan(a+3π/4)=(tana+1)/(1-tana)+(tana-1)/(1+tana)=[(tana+1)-(tana+1)]/(1-tana)=4tana/(1-tana)∵tan2a=2tana/(1-tana)∴tan(a+π/4)+tan(a+3π/4)=2tan2a

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