(1-tanα)/(1+tanα)=3+2根号2 求(sinα+cosα)-1/(cotα-sinα乘cosα) 的值

问题描述:

(1-tanα)/(1+tanα)=3+2根号2 求(sinα+cosα)-1/(cotα-sinα乘cosα) 的值
1个回答 分类:数学 2014-12-15

问题解答:

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(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+1)=(1-tanα)(√2-1)
(√2+1)+(√2+1)tanα=(√2-1)-(√2-1)tanα
(√2+1+√2-1)tanα=√2-1-√2-1
2√2tanα=-2
tanα=-√2/2
(tanα)^2=(sinα/cosα)^2=(1-(cosα)^2)/(cosα)^2=1/(cosα)^2-1
cosα=±1/√(1+(tanα)^2)
(sinα+cosα)-1/(cotα-sinα*cosα)
=(tanα*cosα+cosα)-1/(cosα/sinα-sinα*cosα)
= ±(1+tan)/√(1+(tanα)^2)-sinα/(cosα-cosα(sinα)^2)
=±(1+tan)/√(1+(tanα)^2)-sinα/(cosα(1-(sinα)^2)
= ±(1+tan)/√(1+(tanα)^2)-tanα*1/(cosα)^2
=±(1+tan)/√(1+(tanα)^2)-tanα(1+(tanα)^2)
=±(1-√2/2)/√(1+(-√2/2)^2)+√2/2(1+(-√2/2)^2)
=±(1-√2/2)/√(3/2)+√2/2(3/2)
=±√(3/2)(1-√2/2)2/3+3√2/4
=±(√6-√3)/3+3√2/4
因为α的值不知是多少,cosα可正可负.所以cosα开方修改后加上±.
 
 
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