Question

已知tan(Θ+π/4）=3，求1+sin4Θ-cos4Θ/1+sin4Θ+cos4Θ的值。

Answer 1

先利用倍角公式化简 然后切换弦 求出正弦和余弦的值 代入就可以 类似的问题主要化简 公式要牢记

Answer 2

tan(Θ+π/4）=(1+tanΘ)/(1-tanΘ)=3 1+tanΘ=3-3tanΘ tanΘ=1/2
[1+sin4Θ-cos4Θ]/[1+sin4Θ+cos4Θ]
=[1+2sin2Θcos2Θ-(1-2sin^22Θ)]/[1+2sin2Θcos2Θ+(2cos^22Θ)]
=2sin2Θ(cos2Θ+-sin2Θ)/2cos2Θ(cos2Θ+-sin2Θ)
=sin2Θ/cos2Θ
=tan2Θ
=2tanΘ/(1-tan^2Θ) tanΘ=1/2
=1/(1-1/4)
=4/3

Answer 3

1+sin4Θ-cos4Θ=2sin2θcos2θ+2(sin2θ)^2=2sin2θ(sin2θ+cos2θ)
1+sin4Θ+cos4Θ=2sin2θcos2θ+2(cos2θ)^2=2cos2θ(sin2θ+cos2θ)
∴1+sin4Θ-cos4Θ/1+sin4Θ+cos4Θ
=tan2θ
= - cot2(θ+π/4)
= - (1-tan(θ+π/4)^2)/(2tan(θ+π/4)
=(3^2-1)/(2×3)
=4/3

Answer 4

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