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来源：学生作业帮助网 编辑：作业帮 时间：2024/07/28 22:39:27
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已知0<α<3π/4,cos(α+π/4)=3/5,则tana
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0<α<3π/4
∴π/4<α+π/4<π
cos(α+π/4)=3/5＞0
∴π/4<α+π/4<π/2
cos(α+π/4)=3/5
sin(α+π/4)=√{ 1-[cosα+π/4)]^2} = 4/5
tan(α+π/4) = sin(α+π/4)/cos(α+π/4) = 4/5 /(3/5) = 4/3
(tanα+tanπ/4)/(1-tanαtanπ/4) = 4/3
(tanα+1)/(1-tanα) = 4/3
3(tanα+1) = 4(1-tanα)
3tanα+3 = 4-4tanα
7tanα = 1
tanα = 1/7
0<α<3π/4
π/4<α+π/4<π
cos(α+π/4)=3/5
tan(α+π/4)=4/3
所以 4/3=tan(α+π/4)=[tanα+tan(π/4)]/[1-tanαtan(π/4)]
4/3=(1+tanα)/(1-tanα)
3(1+tanα)=4(1-tanα)
tanα=1/7
sinα+cosα=3根号5\\5，α属于（0，π\\4）, sin^2a+cos^2a=1 sinacos(2β)=-24/25,sin(2β)=7/25 sin2a=2sinacosa=4/5,tana=1/2
由cos(α+π/4)=3/5 知-π/2+2kπ<α+π/4<π/2+2kπ 0<α<3π/4 得到π/4<α<π/2
展开cos(α+π/4)=3/5 有cosa-sina=(3/5)X√2 (cosa-sina)2=18/25 sinacosa=7/50 sina+cosa>0 (sina+cosa)2=1+2sinacosa=32/25 ...
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由cos(α+π/4)=3/5 知-π/2+2kπ<α+π/4<π/2+2kπ 0<α<3π/4 得到π/4<α<π/2
展开cos(α+π/4)=3/5 有cosa-sina=(3/5)X√2 (cosa-sina)2=18/25 sinacosa=7/50 sina+cosa>0 (sina+cosa)2=1+2sinacosa=32/25 sina+cosa=4√2/5 cosa-sina/ sina+cosa=1-tana/tana+1=3/4 tana=1/7
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