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標題: [help~~~~]A.Maths. Queation of Trigonometic Function of Compound Angle [打印本頁]

作者: 細薯 時間: 3/2/2005 07:21 PM
標題: [help~~~~]A.Maths. Queation of Trigonometic Function of Compound Angle
In Fig. 1, AD prependicular to DE and AB=BC=CD=DE.Without using calculators,prove that a+b+c=90 degrees.

作者: 抹茶可樂 時間: 3/2/2005 09:38 PM
你用邊本書的？
====================

tanα=1/3
tanβ=1/2
tanγ=1

tan(α+β)
=(tanα+tanβ)/(1-tanαtanβ)
=(1/3+1/2)/(1-1/3．1/2)
=(5/6)/(5/6)
=1

tan(α+β+γ)
=[tan(α+β)+tanγ]/[1-tan(α+β)tanγ]
=2/0

tan(α+β+γ) is undefined,
it follows that α+β+γ=90°

[ Last edited by 落雷 on 2005-2-3 at 09:41 PM ]

作者: 細薯 時間: 4/2/2005 03:27 PM
my teacher has taught us in today's maths. lessons~~
but still thx u~~
my teacher use sin & cos to find the answer~~

my A.Maths. book is "New Trend"

作者: 細薯 時間: 4/2/2005 07:25 PM
New Trend Additional Mathematics P.154
no.27

Suppose y = sin^2x + sinxcosx.

(a)Express y in the form p+rsin(2x-a),where r>0 and a is acute.
[Hint:write p+rsin(2x-a) as an expression without double angles.]

(b)Hence find the maximum and minimum values of y.

作者: 抹茶可樂 時間: 4/2/2005 08:03 PM
y
=sin^2x + sinxcosx
=1/2．(1-cos2x) + 1/2．sin2x
=1/2．sin2x - 1/2．cos2x + 1/2
=√2/2．sin(2x-45°) + 1/2

-1 ≦ sin(2x-45°) ≦ 1
-√2/2 ≦ √2/2．sin(2x-45°) ≦ √2/2
(1-√2)/2 ≦ √2/2．sin(2x-45°) + 1/2 ≦ (1+√2)/2
(1-√2)/2 ≦ y ≦ (1+√2)/2

max=(1+√2)/2
min=(1-√2)/2

作者: dolphin_ice 時間: 4/2/2005 08:37 PM
英國D數真係冇得同香港比>.<"

我唔識做XD

作者: oscar 時間: 5/2/2005 12:04 AM

Originally posted by 紅唇娃 at 2005-2-4 20:37:
英國D數真係冇得同香港比>.<"

我唔識做XD

HK既數學係勁過其他國家嫁~

作者: 細薯 時間: 5/2/2005 10:59 AM
Re1 樓:
你去到 tan (a+b)=1
係咪已經prove到a+b=45 degrees??
因為tan (a+b)=1 ge solution只有 a+b=45degrees (both a & b are acute angles)
而 c= 45 degrees (base.angles,isos. triangle)

therefore,a+b+c=90 degrees

可ng可以??

Re4 樓:
睇唔明tim.......

[ Last edited by smallpotato226 on 2005-2-5 at 11:00 AM ]

作者: 抹茶可樂 時間: 5/2/2005 12:00 PM

Originally posted by smallpotato226 at 2005-2-5 10:59 AM:
Re1 樓:
你去到 tan (a+b)=1
係咪已經prove到a+b=45 degrees??
因為tan (a+b)=1 ge solution只有 a+b=45degrees (both a & b are acute angles)
而 c= 45 degrees (base.angles,isos. triangle)

ther ...

For Q1, it is ok.
But sometimes you may cope with tan(a+b+c)

Q2

cos2x=1-2sin^2x
2sin^2x=1-cos2x
∴sin^2x=1/2．(1-cos2x) <+strongly recommended to remember

sin2x - cos2x≡rsin(2x-a)
sin2x - cos2x≡rcosasin2x-rsinacos2x
rcosa=1
rsina=1
tana=1
a=45°
r=1

sin2x - cos2x≡sin(2x-45°)

作者: oscar 時間: 5/2/2005 10:52 PM

Originally posted by smallpotato226 at 2005-2-5 10:59:
Re1 樓:
你去到 tan (a+b)=1
係咪已經prove到a+b=45 degrees??
因為tan (a+b)=1 ge solution只有 a+b=45degrees (both a & b are acute angles)
而 c= 45 degrees (base.angles,isos. triangle)

ther ...

其實做數做到就得，方法唔只一個ge~

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