1a 證明 sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
1b 由此, 證明 sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ
請問2條數點計
請問條數點計(10分 )
2007-03-10 10:18 pm
回答 (7)
2007-03-10 10:24 pm
✔ 最佳答案
1a)右方=1 - 2 cos ^2θ =sin^2 θ +cos^2 θ- 2 cos^2θ
= sin^2 θ - cos^2 θ=左方
所以sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
1b)因為sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
所以sin^4 θ - cos^4 θ ≡sin^2 θ - cos^2 θ
(sin^2 θ +cos^2 θ)(sin^2 θ - cos^2 θ)=sin^2 θ - cos^2 θ
(use the formular:x^2-y^2=(x+y)(x-y)
sin^2 θ +cos^2 θ=1
所以左方=右方，sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ
2007-03-10 10:35 pm
1a) sin^2 θ + cos^2 θ ≡ 1
sin^2 θ - cos^2 θ + 2cos^2 θ ≡ 1
sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
1 b) sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
(sin^2 θ - cos^2 θ) ‧(sin^2 θ + cos^2 θ)≡ (1 - 2 cos ^2 θ) ‧(sin^2 θ + cos^2 θ)
sin^4 θ - cos^4 θ ≡ sin^2 θ - 2 cos ^2 θ‧sin^2 θ + cos^2 θ - 2 cos ^2 θ
sin^4 θ - cos^4 θ ≡ (sin^2 θ + cos^2 θ) - ( 2 cos ^2 θ‧sin^2 θ + 2 cos ^2 θ)
sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ (sin^2 θ + cos ^2 θ)
sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ
sin^2 θ - cos^2 θ + 2cos^2 θ ≡ 1
sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
1 b) sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
(sin^2 θ - cos^2 θ) ‧(sin^2 θ + cos^2 θ)≡ (1 - 2 cos ^2 θ) ‧(sin^2 θ + cos^2 θ)
sin^4 θ - cos^4 θ ≡ sin^2 θ - 2 cos ^2 θ‧sin^2 θ + cos^2 θ - 2 cos ^2 θ
sin^4 θ - cos^4 θ ≡ (sin^2 θ + cos^2 θ) - ( 2 cos ^2 θ‧sin^2 θ + 2 cos ^2 θ)
sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ (sin^2 θ + cos ^2 θ)
sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ
2007-03-10 10:30 pm
sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
RHS=1-2cos^2θ
=sin^2θ+cos^2θ-2cos^2θ
=sin^2θ-cos^2θ
=LHS
∴sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ
RHS=1 - 2 cos ^2 θ
=sin^2θ-cos^2θ
LHS=sin^4 θ - cos^4 θ
=(sin^2θ+cos^2θ)(sin^2θ-cos^2θ) -------------------------->a^2-b^2=(a+b)(a-b)
=1(sin^2θ-cos^2θ)
=sin^2θ-cos^2θ
=RHS
∴sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ
RHS=1-2cos^2θ
=sin^2θ+cos^2θ-2cos^2θ
=sin^2θ-cos^2θ
=LHS
∴sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ
RHS=1 - 2 cos ^2 θ
=sin^2θ-cos^2θ
LHS=sin^4 θ - cos^4 θ
=(sin^2θ+cos^2θ)(sin^2θ-cos^2θ) -------------------------->a^2-b^2=(a+b)(a-b)
=1(sin^2θ-cos^2θ)
=sin^2θ-cos^2θ
=RHS
∴sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ
2007-03-10 10:28 pm
1a證明:
移項 sin^2 θ - cos^2 θ + 2 cos ^2 θ≡ 1
合并 sin^2 θ + cos^2 θ ≡ 1
根據公式 1=1
所以原式成立
1b證明:
用平方差公式 (sin^2 θ + cos^2 θ)(sin^2 θ - cos^2 θ)=1 - 2 cos ^2 θ
根據公式得 sin^2 θ - cos^2 θ=1 - 2 cos ^2 θ
接著與上題證法相同
移項 sin^2 θ - cos^2 θ + 2 cos ^2 θ≡ 1
合并 sin^2 θ + cos^2 θ ≡ 1
根據公式 1=1
所以原式成立
移項 sin^2 θ - cos^2 θ + 2 cos ^2 θ≡ 1
合并 sin^2 θ + cos^2 θ ≡ 1
根據公式 1=1
所以原式成立
1b證明:
用平方差公式 (sin^2 θ + cos^2 θ)(sin^2 θ - cos^2 θ)=1 - 2 cos ^2 θ
根據公式得 sin^2 θ - cos^2 θ=1 - 2 cos ^2 θ
接著與上題證法相同
移項 sin^2 θ - cos^2 θ + 2 cos ^2 θ≡ 1
合并 sin^2 θ + cos^2 θ ≡ 1
根據公式 1=1
所以原式成立
參考: me
2007-03-10 10:26 pm
(1a)
L.H.S = sin^2 θ - cos^2 θ
= (1 - cos^2 θ) - cos^2 θ
= 1 - 2 cos ^2 θ
= R.H.S.
Therefore, sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
(1b)
L.H.S. = sin^4 θ - cos^4 θ
= (sin^2 θ + cos^2 θ)(sin^2 θ - cos^2 θ)
= 1 * (sin^2 θ - cos^2 θ)
= 1 - 2 cos ^2 θ (from 1a)
= R.H.S.
Therefore, sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ
L.H.S = sin^2 θ - cos^2 θ
= (1 - cos^2 θ) - cos^2 θ
= 1 - 2 cos ^2 θ
= R.H.S.
Therefore, sin^2 θ - cos^2 θ ≡ 1 - 2 cos ^2 θ
(1b)
L.H.S. = sin^4 θ - cos^4 θ
= (sin^2 θ + cos^2 θ)(sin^2 θ - cos^2 θ)
= 1 * (sin^2 θ - cos^2 θ)
= 1 - 2 cos ^2 θ (from 1a)
= R.H.S.
Therefore, sin^4 θ - cos^4 θ ≡ 1 - 2 cos ^2 θ
2007-03-10 10:26 pm
1 (a).
左方
=sin2θ-cos2θ
=(1-cos2θ)-cos2θ
=1-2cos2θ
右方
=1-2cos2θ
因為左方=右方
所以sin2θ-cos2θ≡1-2cos2θ
1 (b).
左方
=sin4θ-cos4θ
=(sin2θ)2-(cos2θ)2
=(sin2θ+cos2θ)(sin2θ-cos2θ)
=(1)(1-2cos2θ)
=1-2cos2θ
右方
=1-2cos2θ
因為左方=右方
所以sin4θ-cos4θ≡1-2cos2θ
左方
=sin2θ-cos2θ
=(1-cos2θ)-cos2θ
=1-2cos2θ
右方
=1-2cos2θ
因為左方=右方
所以sin2θ-cos2θ≡1-2cos2θ
1 (b).
左方
=sin4θ-cos4θ
=(sin2θ)2-(cos2θ)2
=(sin2θ+cos2θ)(sin2θ-cos2θ)
=(1)(1-2cos2θ)
=1-2cos2θ
右方
=1-2cos2θ
因為左方=右方
所以sin4θ-cos4θ≡1-2cos2θ
2007-03-10 10:25 pm
1a)首先, 利用定理sin^2 θ + cos^2 θ = 1
sin^2 θ = 1-cos^2 θ
1b)利用定理(a+b)x(a-b)=a^2 - b^2
sin^4 θ - cos^4 θ = (sin^2 θ)^2 - (cos^2 θ)^2 = (sin^2 θ + cos^2 θ)x(sin^2 θ - cos^2 θ)
=1x(sin^2 θ - cos^2 θ)
利用1a結果,
sin^4 θ - cos^4 θ ≡ 1 - 2cos^2 θ
sin^2 θ = 1-cos^2 θ
1b)利用定理(a+b)x(a-b)=a^2 - b^2
sin^4 θ - cos^4 θ = (sin^2 θ)^2 - (cos^2 θ)^2 = (sin^2 θ + cos^2 θ)x(sin^2 θ - cos^2 θ)
=1x(sin^2 θ - cos^2 θ)
利用1a結果,
sin^4 θ - cos^4 θ ≡ 1 - 2cos^2 θ
參考: 自己
收錄日期: 2021-04-12 18:45:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070310000051KK02157