Prove the following trigonometric identities.
1. 1 + tan^2 Θ ≡ 1 / cos^2 Θ
2. sin Θ + cosΘ / tanΘ ≡ 1 / sinΘ
**請附上詳細解釋，總共2題，十二萬分感激 !
F. 3 數學問題 (Trigonmetric Relations) 急!!!
2007-04-09 3:15 am
回答 (3)
2007-04-09 3:22 am
✔ 最佳答案
1. 1 + tan^2 Θ = 1 / cos^2 ΘLHS
=1 + tan^2 Θ
=sec^2 Θ
=1/cos^2 Θ
=RHS
or
LHS
=1+ tan^2 Θ
=1+(sin^2 Θ/cos^2 Θ)
=(cos^2 Θ+sin^2 Θ)/cos^2 Θ
=1/cos^2 Θ
=RHS
-----------------------------------------------------
2. sin Θ + cosΘ/ tanΘ = 1 / sinΘ
LHS
=sin Θ + cosΘ / tanΘ
=sin Θ+cosΘ/(sin Θ/cos Θ)
=sin Θ+cos^2Θ/sin Θ
=(sin^2 Θ+cos^2Θ)/sin Θ
=1/sin Θ
=RHS
2007-04-09 3:25 am
1.
1 + tan^2 Θ
= 1 + sin^2Θ / cos^2Θ
= ( cos^2Θ + sin^2Θ ) / cos^2Θ
= 1 / cos^2 Θ ****cos^2Θ + sin^2Θ=1******
2.
sin Θ + cosΘ / tanΘ
= sin Θ + cos^2Θ / sin Θ
= ( sin^2Θ + cos^2Θ ) / sin Θ
= 1 / sinΘ ****cos^2Θ + sin^2Θ=1******
1 + tan^2 Θ
= 1 + sin^2Θ / cos^2Θ
= ( cos^2Θ + sin^2Θ ) / cos^2Θ
= 1 / cos^2 Θ ****cos^2Θ + sin^2Θ=1******
2.
sin Θ + cosΘ / tanΘ
= sin Θ + cos^2Θ / sin Θ
= ( sin^2Θ + cos^2Θ ) / sin Θ
= 1 / sinΘ ****cos^2Θ + sin^2Θ=1******
參考: 幫到你嗎？
2007-04-09 3:22 am
1)1 + tan^2 Θ = 1 + sin^2 Θ / cos^2 Θ
= (cos^2 Θ + sin^2 Θ) / cos^2 Θ ＜＝[通分母]
= 1 / cos^2 Θ // ＜＝[sin^2 Θ + cos^2 Θ) =1]
2)sin Θ + cos Θ / tan Θ = sin Θ + cos Θ / (sin Θ / cos Θ)
= sin Θ + cos^2 Θ / sin Θ ＜＝[除分母=乘分母倒數]
= (sin^2 Θ + cos^2 Θ) / sin Θ ＜＝[通分母]
= 1 / sin Θ // ＜＝[sin^2 Θ + cos^2 Θ) =1]
= (cos^2 Θ + sin^2 Θ) / cos^2 Θ ＜＝[通分母]
= 1 / cos^2 Θ // ＜＝[sin^2 Θ + cos^2 Θ) =1]
2)sin Θ + cos Θ / tan Θ = sin Θ + cos Θ / (sin Θ / cos Θ)
= sin Θ + cos^2 Θ / sin Θ ＜＝[除分母=乘分母倒數]
= (sin^2 Θ + cos^2 Θ) / sin Θ ＜＝[通分母]
= 1 / sin Θ // ＜＝[sin^2 Θ + cos^2 Θ) =1]
參考: by My Maths
收錄日期: 2021-04-13 00:43:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070408000051KK03769