![設tanα=2,計算sinα+cosα/sinα-cosα,第1張 設tanα=2,計算sinα+cosα/sinα-cosα,第1張](data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==)
設tanα=2,計算sinα cosα/sinα-cosα網友廻答:
- (sinα cosα)/(sinα-cosα)
=(sinα cosα)(sinα-cosα)/(sinα-cosα)^2
=[(sinα)^2-(cosα)^2]/(1-2sinαcosα)
=-cos2α/(1-sin2α)
=-{[1-(tanα)^2]/[1 (tanα)^2]}/(1-sin2α)
=-{[1-2^2]/[1 2^2]}/(1-sin2α)
=(3/5)/(1-sin2α)
=(3/5)/{1-2tanα/[1 (tanα)^2]}
=(3/5)/{1-2*2/[1 2^2]}
=(3/5)/(1-4/5)
=(3/5)/(1/5)
=3
網友廻答:
- sinα cosα/sinα-cosα=(tanα 1)/(tanα-1)=(2 1)/(2-1)=3
就是等式上下都除以cosα,變成tanα的形式
網友廻答:
- 上下除以cosα
且sinα/cosα=tanα
原式=(tanα 1)/(tanα-1)=3/1=3
網友廻答:
- (sinα cosα)/(sinα-cosα)=(tanα 1)/(tanα-1)=3
注:tanα=sinα/cosα
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