Prove the Trigonometric identities
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Trigonometric identities
2012-05-18 4:35 am
回答 (4)
2012-05-18 4:50 am
✔ 最佳答案
tan(90 - Θ) - sinΘcosΘ= cosΘ/sinΘ - sinΘcosΘ
= (cosΘ - sin^2 ΘcosΘ)/sinΘ
= cosΘ(1 - sin^2Θ)/sinΘ
= cos^3 Θ/sinΘ
= cos^2 Θ/tanΘ
2012-05-20 1:01 am
L.H.S:tan(90 - Θ) - sinΘcosΘ
=tan Θ- sinΘcosΘ
= cosΘ/sinΘ - sinΘcosΘ
= (cosΘ - sin^2 ΘcosΘ)/sinΘ
= cosΘ(1 - sin^2Θ)/sinΘ
= cos^3 Θ/sinΘ
= cos^2 Θ/tanΘ
R.H.S=cos^2 Θ/tanΘ
(BECAUSE)L.H.S=R.H.S
(SO) It is an identity
2012-05-19 17:02:56 補充：
=1/tan Θ- sinΘcosΘ
=1÷sinΘ/cosΘ - sinΘcosΘ
sorry~
=tan Θ- sinΘcosΘ
= cosΘ/sinΘ - sinΘcosΘ
= (cosΘ - sin^2 ΘcosΘ)/sinΘ
= cosΘ(1 - sin^2Θ)/sinΘ
= cos^3 Θ/sinΘ
= cos^2 Θ/tanΘ
R.H.S=cos^2 Θ/tanΘ
(BECAUSE)L.H.S=R.H.S
(SO) It is an identity
2012-05-19 17:02:56 補充：
=1/tan Θ- sinΘcosΘ
=1÷sinΘ/cosΘ - sinΘcosΘ
sorry~
參考: me
2012-05-18 5:30 pm
tan (90 - Θ) - sin Θ cos Θ
= (1 / tan Θ) - sin Θ cos Θ
= (1 - sin Θ cos Θ tan Θ) / tan Θ)
= (1 - sin^2 Θ) / tan Θ)
= cos^2 Θ / tan Θ
= (1 / tan Θ) - sin Θ cos Θ
= (1 - sin Θ cos Θ tan Θ) / tan Θ)
= (1 - sin^2 Θ) / tan Θ)
= cos^2 Θ / tan Θ
參考: knowledge
2012-05-18 5:18 am
left=tan(90-Θ)-sinΘcosΘ
left=cosΘ/sinΘ-sinΘcosΘ
left=(cosΘ-sin^2ΘcosΘ)/sin^2Θ
left=[cosΘ(1-sin^2Θ)]/sinΘ
left=cos^3Θ/sinΘ
left=cos^2Θ/tanΘ
left=right
left=cosΘ/sinΘ-sinΘcosΘ
left=(cosΘ-sin^2ΘcosΘ)/sin^2Θ
left=[cosΘ(1-sin^2Θ)]/sinΘ
left=cos^3Θ/sinΘ
left=cos^2Θ/tanΘ
left=right
收錄日期: 2021-04-27 17:45:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120517000051KK00661