中3數學問題(3角學的應用cos sin tan)

2009-03-08 7:55 am
以下問題我有答案,但係我想要詳細的步驟,答案必須是我所提供的。

① (cosθ+sinθ)²+(cosθ-sinθ)²
answer : 2

② 1/1-sinθ + 1/1+sinθ
answer : 2/cos²θ

③ (1-cosθ/1+sinθ) / (1-sinθ/1+cosθ)
answer : tan²θ

④ cosθ4次方 - sinθ4次方
answer : 1-2sin²θ

⑤ 開方1-cosθ / 開方1+cosθ
answer : 1-cosθ/sinθ


(a) 已知cosθ=開方3 / 2
tanθ=1 / 開方3
sinθ=1/2
而且cosθ=0.1
求1+ (sinθtanθ/cosθ)的值。
answer : 100

(b)已知tanθ=2 , 求(3sinθ-4cosθ)/(3sinθ+4cosθ)的值。
answer : 1/5

回答 (3)

2009-03-08 8:47 am
✔ 最佳答案

(cosθ + sinθ)2 + (cosθ - sinθ)2
= (cos2θ + 2cosθsinθ + sin2θ) + (cos2θ - 2sinθcosθ + sin2θ)
= 2cos2θ + 2sin2θ
= 2(cos2θ + sin2θ)
= 2(1)
= 2


1/(1 - sinθ) + 1/(1 + sinθ)
= (1 + sinθ)/(1 - sinθ)(1 + sinθ) + (1 - sinθ)/(1 + sinθ)(1 - sinθ)
= (1 + sinθ)/(1 - sin2θ) + ( 1 - sinθ)/(1 - sin2θ)
= (1 + sinθ)/cos2θ + (1 - sinθ)/cos2θ
= [(1 + sinθ) + (1 - sinθ)]/cos2θ
= 2/cos2θ


[(1 - cosθ)/(1 + sinθ)] / [(1 - sinθ)/(1 + cosθ)]
= [(1 - cosθ)(1 + cosθ)] / [(1 - sinθ)(1 + sinθ)]
= [1 - cos2θ] / [1 - sin2θ]
= sin2θ / cos2θ
= tan2θ


(cosθ)4 - (sinθ)4
= (cos2θ)2 - (sin2θ)2
= (cos2θ + sin2θ)(cos2θ - sin2θ)
= (1)(cos2θ - sin2θ)
= cos2θ - sin2θ
= (1 - sin2θ) - sin2θ
= 1 - 2sin2θ

⑤ √(1 - cosθ) / √(1 + cosθ)
= √(1 - cosθ)(1 - cosθ) / √(1 + cosθ)(1 - cosθ)
= (1 - cosθ) / √(1 - cos2θ)
= (1 - cosθ) / √sin2θ
= (1 - cosθ) / sinθ


(a)
1 + [sinθtanθ/cosθ]
= 1 + [sinθ(sinθ/cosθ)/cosθ]
= 1 + [sin2θ/cos2θ]
= 1 + [(1 - cos2θ)/cos2θ]
= 1 + [(1 - 0.12)/0.12]
= 1 + [0.99/0.01]
= 1 + 99
= 100

(b)
[3sinθ - 4cosθ] / [3sinθ + 4cosθ]
= [(3sinθ - 4cosθ)/cosθ] / [(3sinθ + 4cosθ)/cosθ]
= [3tanθ - 4] / [3tanθ + 4]
= [3(2) - 4] / [3(2) + 4]
= 2/10
= 1/5
=
2009-03-08 8:35 am
③ (1-cosθ/1+sinθ) / (1-sinθ/1+cosθ)
=(1-cosθ)(1+cosθ) / (1+sinθ/ 1-sinθ)
=1-cos²θ/ 1-sin²θ
=sin²θ/cos²θ
=tan²θ

2009-03-08 00:38:44 補充:

cosθ=0.1
θ=84.3

跟住代入1+ (sinθtanθ/cosθ)掛...-----吾太肯定..


(b)同(a)差吾多..
參考: BY MYSELF
2009-03-08 8:16 am
1-sin²θ=cos²θ (a+/-b)²= a²+/- 2ab+b²
1-cos²θ= sin²θ
sinθ/cosθ=tanθ

① (cosθ+sinθ)²+(cosθ-sinθ)²
=(cos²θ+2sinθcosθ+ sin²θ)+(cos²θ- 2sinθcosθ+sin²θ)
= 2 (cos²θ+sin²θ) (上面公式換位)
answer : 2

② 1/1-sinθ + 1/1+sinθ
= 1/1-sinθ + 1/1+sinθ
= (1+sinθ+1-sinθ)/(1-sinθ ) (1+sinθ)
=2/(1+sin²θ)
answer : 2/cos²θ

2009-03-08 00:22:53 補充:
④ cosθ4次方 - sinθ4次方
= (cos²θ+sin²θ)(cos²θ-sin²θ)
=1(1-sin²θ-sin²θ)
answer : 1-2sin²θ

2009-03-08 00:35:24 補充:
a²-b²=(a+b)(a-b)
⑤ 開方1-cosθ / 開方1+cosθ
=(開方1-cosθ)² / (開方1+cosθ)(開方1-cosθ)
=(1-cosθ)/(1-cos²θ)
answer : 1-cosθ/sinθ

2009-03-08 01:04:08 補充:
⑥(a) 已知cosθ=開方3 / 2
tanθ=1 / 開方3
sinθ=1/2
而且cosθ=0.1
求1+ (sinθtanθ/cosθ)的值。
=1+tan²θ = 1+1/3 -->(1 / 開方3)²

ANS 有問題 = 1.33333333333333333333333333 [cosθ=0.1定(開方3 / 2 )ar]

answer : 100

(b)已知tanθ=2 , 求(3sinθ-4cosθ)/(3sinθ+4cosθ)的值。
(填番已知既ans 咪得lo)
answer : 1/5

2009-03-08 01:07:21 補充:
③ (1-cosθ/1+sinθ) / (1-sinθ/1+cosθ)
=(1-cosθ/1+sinθ) / (1+cosθ/1-sinθ)
=(1-cosθ)(1+cosθ)/(1+sinθ)(1-sinθ)
=(1²-cos²θ)/(1²-sin²θ)
=sin²θ/cos²θ

answer : tan²θ


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