已知f(α)=2tanα-(2sin^2α/2-1)/(sinα/2cosα/2),求f(π/12)

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已知f(α)=2tanα-(2sin^2α/2-1)/(sinα/2cosα/2),求f(π/12)
已知f(α)=2tanα-(2sin^2α/2-1)/(sinα/2cosα/2),求f(π/12)

已知f(α)=2tanα-(2sin^2α/2-1)/(sinα/2cosα/2),求f(π/12)
f(a)
=2tana-(2sin^2a/2-1)/[(1/2)sina]
=2tana-(1-cosa-1)/[(1/2)sina]
=2tana+2cosa/sina
=2tana+2ctga
cos30 °= √3/2
cos15 °= √[(1 + cos30° )/2]
sin15° = √[(1 - cos30° )/2]
tan15° = sin15° /cos15°
= √[(1 - cos30° )/(1 + cos30° )]
= √[(2 - √3)/(2 + √3)]
= (2 - √3)
所以:
f(π/12)
=f(15°)
=tan15°+ctg15°
=(2 - √3)+1/(2 - √3)
=(2 - √3)+(2 + √3)
=4.