已知α、β、γ是锐角,且满足联立方程{cosα=tanβ、cosβ=tanγ、cosγ=tanα,求sinα的值.
1个回答

利用公式：tan^2+1=1/cos^2,(1/tan^2)+1=1/sin^2

(cosα)^2+1=(tanβ)^2+1=1/(cosβ)^2=1/(tanγ)^2

=[1/(sinγ)^2]-1=(cosγ)^2/(sinγ)^2

=(cosγ)^2/[1-(cosγ)^2]=(tanα)^2/[1-(tanα)^2]

=(sinα)^2/[(cosα)^2-(sinα)^2]

设x=(sinα)^2,则(cosα)^2=1-x

所以可得：

2-x=x/(1-2x),即x^2-3x+1=0

所以解得：x=(3±√5)/2

因为x=(sinα)^2≤1,所以x=(3-√5)/2

所以

sinα=√[(3-√5)/2]

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