20點 A.maths 2條題目

1)
tan2θ + sec2θ
=[sin2θ/cos2θ] + [1/cos2θ]
=[sin2θ+1] / cos2θ
=[2sinθcosθ+1] / [cos^2 θ-sin^2 θ]
=[2sinθcosθ+cos^2 θ+sin^2 θ] / [cos^2 θ-sin^2 θ]
=[cosθ+sinθ]^2 / [{cosθ+sinθ}{cosθ-sinθ}]
=[cosθ+sinθ] / [cosθ-sinθ]

2007-03-14 18:02:07 補充：
2a)tanθ cotθ=[sinθ / cosθ] [cosθ / sinθ]=[sin^2 θ cos^2 θ] / [cosθsinθ]=1 / cosθsinθ=1 / (1/2)[sin2θ sin0]=2 / sin2θ

2007-03-14 18:04:59 補充：
2a)tanθ 加 cotθ=[sinθ / cosθ] 加 [cosθ / sinθ]=[sin^2 θ 加 cos^2 θ] / [cosθsinθ]=1 / cosθsinθ=1 / (1/2)[sin2θ 加 sin0]=2 / sin2θ

1) prove that tan2θ + sec2θ = (cosθ + sinθ) / (cosθ - sinθ)

LHS

= tan2θ + sec2θ

= (sin2θ + 1)/(cos2θ)

= 2sinθcosθ + 1 / (cos^2θ - sin^2θ)

= (sinθ + cosθ)^2 / (cosθ + sinθ)(cosθ - sinθ)

= (cosθ + sinθ) / (cosθ - sinθ)

= RHS



2a) Prove that tanθ + cotθ = 2/2sinθ

tanθ + cotθ = 2/2sinθ

LHS

= tanθ + cotθ

= tanθ + 1/tanθ

= (tan^2θ + 1) / tanθ

= sec^2θ / tanθ

= 1/sinθcosθ

Please check your answer
圖片參考：http://hk.yimg.com/i/icon/16/7.gif






2b) if x = 2 + (√3) is a root of the equation x^2 - (tanθ + cotθ)x + 1 =0, fing the values of sin2θ and cos4θ

x^2 - (tanθ + cotθ)x + 1 =0

x^2 - x/sinθcosθ + 1 = 0

sinθcosθx^2 - x + sinθcosθ = 0

x = 2 + (√3) is a root of the equation sinθcosθx^2 - x + sinθcosθ = 0

Put x = 2 + (√3)

([2 + (√3)]^2 + 1)sinθcosθ - [2 + (√3)] = 0

(4 + 4√3 + 3 + 1)sinθcosθ = [2 + (√3)]

sinθcosθ = [2 + (√3)] / (8 + 4√3 )

sinθcosθ = 1/4

So,

sin2θ = 2sinθcosθ

= 1/2

cos4θ

= cos2(2θ)

= cos^2(2θ) - sin^2(2θ)

= [1 - sin^2(2θ)] - sin^2(2θ)

= 1 - 2sin^2(2θ)

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

= 1 - 1/2

= 1/2

1)




tan2θ + sec2θ




=[sin2θ/cos2θ] + [1/cos2θ]



=[sin2θ+1] / cos2θ




=[2sinθcosθ+1] / [cos^2 θ-sin^2 θ]



=[2sinθcosθ+cos^2 θ+sin^2 θ] / [cos^2 θ-sin^2 θ]



=[cosθ+sinθ]^2 / [{cosθ+sinθ}{cosθ-sinθ}]




=[cosθ+sinθ] / [cosθ-sinθ]



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