中三數學 三角比問題

3cos^4θ-cos^2θ+3sin^2θcos^2θ
= 3cos^4 θ + 3(sin^2 θ)(cos^2 θ) - cos^2 θ
= 3cos4 θ + 3(1 - cos^2 θ)(cos^2 θ) - cos^2 θ
= 3cos^4 θ - 3 cos^4 θ + 3cos^2θ - cos^2 θ
= 2cos^2 θ

2010-06-10 19:43:19 補充：
第一步： 將 sin^2 θ 變成 1 - cos^2 θ
sin^2 θ + cos^2 θ = 1
sin^2 θ = 1 - cos^2 θ

2010-06-10 19:44:16 補充：
等二步： 將3(1 - cos^2 θ)(cos^2 θ)拆開
3(1-cos^2 θ)(cos^2 θ)
= (3 - 3cos^2 θ)(cos^2 θ)
= 3cos^2 θ - 3 cos^4 θ

2010-06-10 19:45:32 補充：
第三步：將(3cos^2 θ - 3 cos^4 θ)和(3cos^4 θ- cos^2 θ)依其指數排好
3cos^2 θ - 3 cos^4 θ + 3cos^4 θ- cos^2 θ
= 3cos^2 θ - cos^2 θ - 3cos^4 θ + 3cos^4 θ

2010-06-10 19:46:21 補充：
第四步：將- 3cos^4 θ + 3cos^4 θ互相抵消
- 3cos^4 θ + 3cos^4 θ = 0,

所以,3cos^2 θ - cos^2 θ - 3cos^4 θ + 3cos^4 θ
= 3cos^2 θ - cos^2 θ + 0

2010-06-10 19:47:02 補充：
第五步：抽因子
3cos^2 θ - cos^2 θ + 0
= 3cos^2 θ - cos^2 θ
= (3-1)cos^2 θ
= 2cos^2 θ

我覺得wy簡潔有力,匿名太複雜

3 cos^4 x - cos^2 x + 3 sin^2 x cos^2 x

= 3cos^2 x ( cos^2 x + sin^2 x) - cos^2 x

[因為cos^2 x + sin^2 x=1]

<---沒補這句話您發問者應該不懂吧

= 3cos^2 x - cos^2 x

= 2 cos^2 x.

3 cos^4 x - cos^2 x + 3 sin^2 x cos^2 x
= 3cos^2 x ( cos^2 x + sin^2 x) - cos^2 x
= 3cos^2 x - cos^2 x
= 2 cos^2 x.