已知銳角α,β,γ滿足cosα=tanβ；cosβ=tanγ；cosγ=tanα,求sinα的值

數(shù)學人氣：283 ℃時間：2020-01-25 20:43:32

優(yōu)質(zhì)解答

cosα=tanβ；cosβ=tanγ；兩式相乘得cosαtanγ=tanβ×cosβ=sinβ
所以（sinβ）^2=(cosαtanγ)^2; (cosβ)^2=(tanγ)^2
則1=（cosβ)^2+(sinβ)^2=(tanγ)^2[1+(cosα)^2]
=[(sinγ)^2/(cosγ)^2][1+(cosα)^2]
=[(1-(cosγ)^2)/(cosγ)^2][1+(cosα)^2]
=[(1-(tanα)^2)/(tanα)^2][1+(cosα)^2]
=[(1-2(sinα)^2)/(sinα)^2)][2-(sinα)^2]
所以[(1-2(sinα)^2)][2-(sinα)^2]=（sinα)^2
（sinα)^4-3（sinα)^2+1=0
（sinα)^2=（3-√5）/2
sinα=(√5-1)/2