cosα=-1 根号下1 tanα--2
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tanα-1/tanα=√21,tan²α-2+1/tan²α=21;tan²α+2+1/tan²α=23;tanα+1/tanα=√23;sinαcosα=t
(1-tanα)/(1+tanα)=3+2√2(1-tanα)/(1+tanα)=2+2√2+1(1-tanα)/(1+tanα)=(√2+1)^2=(√2+1)/(√2-1)(1+tanα)(√2+
sin²α(1+1/tanα)+cos²α(1+tanα)=sin²α+sinacosa+cos²a+sinacosa=sin²α+2sinacosa
sinα+cosα=√2sin(π/4+α)=1π/4+α=π/2α=π/4tanα=1tanα+(1/tanα)=1+1/1=2
∵3π/2
∵(1+tanα)/(1-tanα)=3+2√2,∴(1+tanα)/[(1+tanα)+(1-tanα)]=(3+2√2)/[(3+2√2)+1],∴1+tanα=(3+2√2)/(2+√2),∴t
1、原式分子分母同除cosa得:原式=(tana+1)/(1-tana)把tana=√2代入=(√2+1)(1-√2)=-3-2√22、原式=(2sin²a-sinacosa+cos
根号下1+tan^2α=√(1/cosα)^2=|1/cosα|同理根号下1+cot^2α=|1/sinα|原式=cosα|cosα|+sinα|sinα|=-1而(cosα)^2+(sinα)^2所
tanα=-√2,(1)cos(π+α)=-cosa0,∴cosa=1/√[1+(tana)^2]=1/√3,sina=tana*cosa=-√(2/3),∴sinα+cosα=(√3-√6)/3.(
∵sinα-cosα=-√5/2==>(sinα-cosα)²=5/4==>sin²α+cos²α-2sinαcosα=5/4==>2sinαcosα=-1/4==>si
(1)原式=[(cosα+sinα)÷cosα]/[(cosα-sinα)÷cosα]=(1+tanα)/(1-tanα)=(1+根号2)/(1-根号2)=-3-2根号2(2)原式=(2sin平方α-
(1+tana)/(1-tana)=-1+2/(1-tana)=3+√21/(1-tana)=2+√21-tana=(2-√2)/2tana=√2/2sina/cosa=√2/2(sina)^2/(c
sinα+cosα=√2(sinα+cosα)²=2sin²α+cos²α+2sinαcosα=21+2sinαcosα=22sinαcosα=1sinαcosα=1/2
如果是(1+2sinαcosα)/(cos^2(α)-sin^2(α))可用处理齐次式的办法由(1-tanα)/(1+tanα)=3+2倍的根号下2可知tanα=1.5根号下2-2再将(1+2sinα
两边平方得4cosα^2=1+sinα+1-sinα-2根号(1-sinα^2)化简根号(1-sinα^2)=1-2cosα^2根号(1-sinα^2)=2sinα^2-1两边平方得1-sinα^2=
1/21+tana^2=1/cosa^2再问:答案是多少?
由题知sinα+cosα=(1-√3)/2.......①∴上式平方得,(sinα+cosα)^2=sin^2α+cos^2α+2sinαcosα=【(1-√3)/2】^2=1-√3/2又∵sin^2
第一题:sinαcosα=(sinαcosα)/[(sinα)^2+(cosα)^2]=tanα/[(tanα)^2+1]=-(3/2)/[(-3/2)^2+1]=-6/(9+4)=-6/13.第二题
化简为:根号下(1+1/tanα)=根号下(1+cosα/sinα)对吗?