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)](/uploads/image/z/16104689-17-9.jpg?t=%E5%B7%B2%E7%9F%A5f%28%CE%B1%29%3D2tan%CE%B1-%282sin%5E2%CE%B1%2F2-1%29%2F%28sin%CE%B1%2F2cos%CE%B1%2F2%29%2C%E6%B1%82f%28%CF%80%2F12%29)
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.
=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.
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