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標題: Trigo. A.Maths. Test........o.... [打印本頁]

作者: 細薯 時間: 13/5/2005 05:56 PM
標題: Trigo. A.Maths. Test........o....
死傷嚴重.....
post D 問題上黎先

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Time allowed : 2 hours

Section A Questions X 10 @ 6 marks
Section B Questions X 5 @ 10 marks
note:Section B --the teacher will choose the best 4 questions to be the marks
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Section A
1. Find the general solution of the equation 10 sec^2 x - 3=17 tan x .
(我竟然忘了 >>sec^2 x = 1 - tan^2 x .........>< )

4.If sin 3x = p and sin^2 x = 3/4 - q ,prove that p^2 + 16 q^3 = 12 q^2 .
(唔識...)

6. Find the general solution of sin x - sin 2x = sin 4x - sin 3x .
(唔識...)

7. Prove that 4 tan^-1 1/5 - tan^-1 1/239 = π/4 without using calculator.
(唔識...)

Section B
11. In a triangle a= 4.832 , b= 2.186 , A-B=34∘24’ . Solve the triangle ABC .
(Hints : Use Sine formula and simplify a-b/a+b )
(唔識...)
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作者: oscar 時間: 13/5/2005 06:38 PM
4.If sin 3x = p and sin^2 x = 3/4 - q ,prove that p^2 + 16 q^3 = 12 q^2 .
sin 3x
= sin (2x+x)
= sin 2x cos x + cos 2x sin x
= (2 sin x cos x) cos x + (1 - 2 sin^2 x) sin x
= 2 sin x cos^2 x + sin x - 2 sin^3 x
= 2 sin x (1 - sin^2 x) + sin x - 2 sin^3 x
= 2 sin x - 2 sin^3 x + sin x - 2 sin^3 x
= 3 sin x - 4 sin^3 x
(由於下面的steps太難show出來，因此我用文字講講計法)
然後從「sin^2 x = 3/4 - q」計出sin x。
再把這term substitue到「3 sin x - 4 sin^3 x = p」，
再simplify便可以。

作者: 細薯 時間: 13/5/2005 06:45 PM
其實le題我後尾好似諗到點計...
但Times up......

作者: 抹茶可樂 時間: 13/5/2005 09:20 PM
Hints

sin x - sin 2x = sin 4x - sin 3x
sin x + sin 3x = sin 2x + sin 4x

4 tan^-1 1/5 - tan^-1 1/239 = π/4
Prove tan(4 tan^-1 1/5) = tan(tan^-1 1/239 +π/4)
(Use double angle formula on LHS)

a+b/a-b=k(sin A + sin B)/k(sin A - sin B)

作者: 細薯 時間: 14/5/2005 07:45 PM

Originally posted by 落雷 at 2005-5-13 09:20 PM:
Hints

sin x - sin 2x = sin 4x - sin 3x
sin x + sin 3x = sin 2x + sin 4x

4 tan^-1 1/5 - tan^-1 1/239 = π/4
Prove tan(4 tan^-1 1/5) = tan(tan^-1 1/239 +π/4)
(Use double angle formula on LH ...

Thx~~
(好厲害....羨慕......><)

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