已知tan(π/4+α)=3,求sin2α - 2(cosα)^2的值

已知tan(π/4+α)=3,求sin2α - 2(cosα)^2的值
已知tan(π/4+α)=3,求sin2α - 2(cosα)^2的值

已知tan(π/4+α)=3,求sin2α - 2(cosα)^2的值
tan(π/4+α)=3
(tanπ/4+tana)/(1-tanπ/4tana)=3
(1+tana)/(1-tana)=3
1+tana=3-3tana
tana=2/4=1/2
sin2a=2tana/(1+tan^2a)
=2*1/2 /(1+(1/2)^2)
=4/5
2(cosa)^2=cos2a+1
=(1-tan^2a)/(1+tan^2a) +1
=(1-(1/2)^2)/(1+(1/2)^2)+1
=3/5+1
=8/5
sin2a-2(cosα)^2
=4/5-8/5
=-4/5

答：
tan(π/4+a)=(tanπ/4+tana)/(1-tanπ/4tana)=(1+tana)/(1-tana)=3
即tana=1/2
sin2a-2cos²a (即分母为1)
=(2sinacosa-2cos²a)/1 (用sin²a+cos²a替换1)
=(2sinacosa-2cos²a)/(...

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答：
tan(π/4+a)=(tanπ/4+tana)/(1-tanπ/4tana)=(1+tana)/(1-tana)=3
即tana=1/2
sin2a-2cos²a (即分母为1)
=(2sinacosa-2cos²a)/1 (用sin²a+cos²a替换1)
=(2sinacosa-2cos²a)/(sin²a+cos²a) (分子分母同除以cos²a)
=(2tana-2)/(tan²a+1)
=(1-2)/(1/4+1)
=-4/5

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tan(π/4+α)=（tanα+tanπ/4)/(1-tanαtanπ/4)=(1+tanα)/(1-tanα)=3→tanα=1/2
sin2α-2cos²α=sin2α-(1+cos2α)=sin2α-cos2α-1
sin2α=2tanα/(1+tan²α)=4/5
cos2α=(1-tan²α)/(1+tan²α)=3/5
所以sin2α-2cos²α=sin2α-(1+cos2α)=sin2α-cos2α-1=4/5-3/5-1=-4/5

方法一
tan(π/4+α)=3
(tanπ/4+tana)/(1-tanπ/4tana)=3
(1+tana)/(1-tana)=3
1+tana=3-3tana
tana=2/4=1/2

sin2a=2tana/(1+tan^2a)
=2*1/2 /(1+(1/2)^2)
=4/5
2(cosa)^2=cos2...

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方法一
tan(π/4+α)=3
(tanπ/4+tana)/(1-tanπ/4tana)=3
(1+tana)/(1-tana)=3
1+tana=3-3tana
tana=2/4=1/2

sin2a=2tana/(1+tan^2a)
=2*1/2 /(1+(1/2)^2)
=4/5
2(cosa)^2=cos2a+1
=(1-tan^2a)/(1+tan^2a) +1
=(1-(1/2)^2)/(1+(1/2)^2)+1
=3/5+1
=8/5

sin2a-2(cosα)^2
=4/5-8/5
=-4/5
方法二
tan(π/4+a)=(tanπ/4+tana)/(1-tanπ/4tana)=(1+tana)/(1-tana)=3
即tana=1/2
sin2a-2cos²a (即分母为1)
=(2sinacosa-2cos²a)/1 (用sin²a+cos²a替换1)
=(2sinacosa-2cos²a)/(sin²a+cos²a) (分子分母同除以cos²a)
=(2tana-2)/(tan²a+1)
=(1-2)/(1/4+1)
=-4/5
方法三
tan(π/4+α)=（tanα+tanπ/4)/(1-tanαtanπ/4)=(1+tanα)/(1-tanα)=3→tanα=1/2
sin2α-2cos²α=sin2α-(1+cos2α)=sin2α-cos2α-1
sin2α=2tanα/(1+tan²α)=4/5
cos2α=(1-tan²α)/(1+tan²α)=3/5
所以sin2α-2cos²α=sin2α-(1+cos2α)=sin2α-cos2α-1=4/5-3/5-1=-4/5

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