1. No category

Carmel Alison Lam Foundation Secondary School
F.4 Additional Mathematics
A Journey of Mathematical Thinking
Worksheet 1: How to solve it
Question:
A wire of length 100 cm is bent into a rectangle. Find the maximum area of the
rectangle.
I.
Understand the problem:
1. What is the unknown?
2. What are the data?
II. Devise a Plan:
3. What is the condition?
III. c a r r y o u t t h e p l a n :
IV. L o o k B a c k ! !
Checklist
l
l
l
l
l
the result
the argument
derive the solution sufficiently
see it a glance
use the result/method for some other
problem
Carmel Alison Lam Foundation Secondary School
F.4 Additional Mathematics
A Journey of Mathematical Thinking
Worksheet 2: Basic Skills of Problem Solving (1)
(A) Association (聯想)
Question 1:
Let A, B, C be any angles in a triangle ABC. Show that
tan A + tan B + tan C = tan A tan B tan C .
Think & associate:
(1) Any formula about the sum of three tangents? No!
(2)
Any formula about the sum of two tangents?
tan( A + B ) =
Yes!
tan A + tan B
.
1 − tan A tan B
Then, tan A + tan B = tan( A + B )(1 − tan A tan B ) .
What’s the meaning of three angles A, B, C to you?
Angle sum of triangle? Right!
A + B + C = 180o
(3)
Try this:
tan A + tan B + tan C
= tan( A + B)(1 − tan A tan B ) + tan C
= tan( A + B)(1 − tan A tan B ) + tan[180o − ( A + B)]
= tan( A + B)(1 − tan A tan B) − tan( A + B)
= − tan( A + B)tan A tan B
You’re in the right track!
Now, complete the proof yourself.
Solution:
Question 2:
Find the equation of the tangent to the curve y =
4x
which is parallel to
x −1
the line 4x + y = 5.
Think & associate:
(1)
What’s the meaning that the line you want to find is
parallel to the line 4x + y = 5?
Of course, the slope of this line is –4.
(2)
Can I use the technique of differentiation to solve
this problem? Yes! But wait a minute.
(3)
You’ve known the slope of the required line. You
only need to find one more thing, say, a point or an
y-intercept. Then the line can be solved.
Two-point form? Slope-intercept form?
(4)
What sort of information about a line being a
tangent to a curve can you remember?
For a curve y = ax 2 + bx + c , if the line y = mx + k is its
tangent, then the equation
ax 2 + (b − m) x + (c − k ) = 0
has only one root for x, i.e., ∆ = 0 .
So, let the required line be y = mx + k .
 y = 4x

Then, solve 
x −1 .
 y = −4 x + k
Caution: This method can only be applied to the
equation of degree 2.
Solution:
(B) Consideration of special cases (特殊化猜想)
Question 1:
The figure shows a quadrilateral ABCD and a circle whose centre O is on
the line AB and OA = OB. If AD, CD and BC are the tangents to the circle
at E, F and G respectively, show that AB ⋅ AB = 4 AD ⋅ BC .
Think & consider a special case:
(1)
Consider a special case for the quadrilateral ABCD:
If AB // CD, what happens?
Since OE = OG (radii),
OA = OB (given), and
∠AEO = ∠BGO = 90o
(radius
⊥ tangent),
∆AOE ≅ ∆BOG
(RHS)
Thus, ∠AOE = ∠BOG and ∠OAE = ∠OBG .
Therefore, ABCD is a trapezium and AD = BC.
(2)
Next, we have to find a relation between BC and
AB.
Since AD = BC, is possible that BC = AB/2 = OB and
AD = AB/2 = OA?
If so, we need to find a relation between BC and
OB.
Draw a line CH perpendicular to AB.
Clearly, CH = OF , and hence equal to OG (radii).
Then, ∆BHC ≅ ∆BGO (RHS).
Therefore, BC = OB =
Similarly, CD = OA =
(3)
AB
.
2
AB
. The result follows.
2
In this special case, we find that the key step to
solve the problem is to find a relation between BC
and AB. Similar argument can be applied to find a
relation between CD and AB.
Now, let’s go back to the problem.
In ∆OBC , how can you relate BC and OB by a
formula? Sine formula?
Try to solve the problem.
Solution:
Question 2:
(a) Prove, by mathematical induction, that
1× 2 + 2 × 3 + 3 ×4 +... + n(n + 1) =
n( n + 1)( n + 2)
.
3
(b) Hence, evaluate
1991× 1992 + 1992 × 1993 + 1993 × 1994 + ... + 2000 × 2001 .
Think & consider a special case:
(1)
The proof for the first part of question is standard.
You can find that in the very beginning of the proof,
we consider the case for n = 1 and check whether
the statement is true or not.
(2)
In the second part, you may try with small numbers,
say,
1× 2+ 2 × 3 + 3 × 4 + 4 × 5 + 5 ×6 =
5(5 + 1)(5 + 2)
= 70 .
3
If we want to find the sum 3 × 4 + 4 × 5 + 5 × 6 , we need
to subtract 1× 2 + 2 × 3 from the above sum, i.e.,
3× 4+ 4× 5 + 5 × 6
= (1× 2+ 2× 3+ 3× 4+ 4× 5 + 5 × 6)− (1× 2+ 2× 3)
= 70 −
2(2 + 1)(2 + 2)
= 70 − 8 = 62 .
3
Now, you are supposed to get the right
strategy.
Solution:
Carmel Alison Lam Foundation Secondary School
F.4 Additional Mathematics
A Journey of Mathematical Thinking
Worksheet 3: Basic Skills of Problem Solving (2)
(C) Search for hints (信息型分析法)
Question 1:
The figure shows a triangle ABC with C ≠ 90 o . Show that
a − c cos B sin B
=
.
b − c cos A sin A
Think & search for hints:
(1)
Suppose that the result is true, i.e.,
a − c cos B sin B
=
.
b − c cos A sin A
(1)
Then, by sine formula,
a − c cos B sin B b
=
= .
b − c cos A sin A a
(2)
We have,
a 2 − ac cos B = b 2 − bc cos A
a 2 − b2 = ac cos B − bc cos A
(3)
(4)
How can we get (4)? We need some hints.
By cosine formula,
a 2 = b2 + c2 − 2bc cos A and
b 2 = a2 + c2 − 2 ac cos B .
Hence, a 2 − b 2 = b 2 − a 2 + 2( ac cos B − bc cos A) .
Therefore, a 2 − b2 = ac cos B − bc cos A .
(2)
Now, reverse the steps: (4) → (3) → (2) → (1) and
complete the proof.
Solution:
Question 2:
Given y = xn + x1− n , where x ≠ 0 .
dy
Show that x + (n − 1) y = (2n − 1)x n .
dx
Think & search for hints:
(1)
Suppose that the result is true, i.e.,
x
dy
+ (n − 1) y = (2n − 1)x n .
dx
dy (2n −1) xn − ( n −1) y
=
.
dx
x
We have,
dy (2n − 1) x n − (n −1) x n − (n − 1) x1 −n
=
dx
x
n
1− n
nx − (n − 1)x
=
x
= nx n −1 − ( n − 1) x − n
Then
(2)
How can we get (3)? The hint is obvious.
(1)
(2)
(3)
dy
= nxn −1 + (1 − n) x −n .
dx
Now, reverse the steps: (3) → (2) → (1) and complete
the proof.
Solution:
(D) Sharpening the ideas (前進型分析法)
Question 1:
Let tanα and tan β be the unequal roots of equation
ax 2 + (2a − 3)x + (a − 2) = 0 , where a ≠ 0 .
Find the minimum value of tan(α + β ) .
Think & sharpen the ideas:
(1)
First we know that
tanα + tan β =
3 − 2a
a −2
and tanα tan β =
.
a
a
(2) Next, we recall the formula
tan(α + β ) =
tan α + tan β
.
1 − tan α tan β
Hence, we have tan(α + β ) =
3 − 2a 3
= −a .
2
2
(3) In order to find the minimum value of the
expression, we need to set an inequality.
(4) The quadratic equation has two unequal roots, hence,
its discriminant is non-negative, i.e.,
∆ ≥ 0 ⇒ (2a − 3) 2 − 4a (a − 2) ≥ 0 .
Solving, a ≤
9
9
3
3 9
3
⇒ −a ≥ − ⇒ − a ≥ − = − .
4
4
2
2 4
4
Question 2:
Given that tan A, tan B , tan C ≠ 0 , if
that
tan( A − B ) sin 2 C
+
= 1 , show
tan A
sin 2 A
tan2 C = tan A tan B .
Think & sharpen the ideas:
(1)
Notice that the result is all about the tangents.
We then have to modify the condition:
tan( A − B ) sin 2 C
+
=1
tan A
sin 2 A
⇒
tan( A − B) cosec 2 A
+
=1
tan A
cosec2 C
⇒
tan( A − B) 1 + cot 2 A
+
=1
tan A
1 + cot 2 C
⇒
tan( A − B ) tan 2 C (1 + tan 2 A)
+
=1
tan A
tan 2 A(1 + tan 2 C)
(2) In order to proceed, we need to expand tan(A – B).
tan( A − B ) tan 2 C (1+ tan 2 A)
+
=1
tan A
tan 2 A(1+ tan 2 C )
⇒
tan A − tan B
tan 2 C (1 + tan 2 A)
+
=1
tan A(1 + tan A tan B ) tan 2 A(1 + tan 2 C )
⇒
tan A − tan B
tan 2 C (1 + tan 2 A)
+
=1
tan A(1 + tan A tan B ) tan 2 A(1 + tan 2 C )
(3) It’s quite difficult to simplify the above equation.
However, we may make use of the constant on the
right hand side.
tan A − tan B
tan 2 C (1 + tan 2 A)
+
=1
tan A(1 + tan A tan B ) tan 2 A(1 + tan 2 C )
⇒
tan A − tan B
tan 2 C (1 + tan 2 A)
= 1−
tan A(1 + tan A tan B )
tan 2 A(1 + tan 2 C )
⇒
tan A − tan B
tan 2 A − tan 2 C
=
tan A(1 + tan A tan B ) tan 2 A (1 + tan 2 C )
Hence, tan A(tan A − tan B )(1+ tan2 C )
= (tan2 A − tan2 C )(1+ tan A tan B )
(4) Finally, we group the like terms together.
tan2 C (1 + tan 2 A) = tan A tan B (tan 2 A + 1)
⇒ tan 2 C = tan A tan B .
(E) Backward argument (追溯型分析法)
Question 1:
Given that sin B = k sin(2 A + B) , k ≠ 1 . Prove that
1+k
tan( A + B ) =
tan A .
1− k
Think & argue backwards:
(1)
The statement is true only if
tan( A + B ) 1 + k
=
.
tan A
1− k
(2) This is true only if
sin( A + B)cos A 1 + k
=
.
cos( A + B)sin A 1 − k
(3) Now, from the given condition,
sin(2 A + B) 1
sin( A + B )cos A + cos( A + B)sin B 1
= ⇒
=
sin B
k
sin B
k
(4) We study the expression in (2).
Recall that if
a c
a +b c + d
= , then
=
.
b d
a −b c −d
The expression in (2) is true only if
sin( A + B)cos A + cos( A + B)sin A 1 + k + 1 − k
1
=
=
sin( A + B)cos A − cos( A + B)sin A 1+ k − (1 − k ) k
It’s equivalent to
sin(2 A + B ) 1
= .
sin B
k
(5) Complete the proof by tracing the steps backwards.
Solution:
Question 2:
Given a > b , b > 0, a ≠ b . Show that
a +b
> ab .
2
Think & argue backwards:
(1)
Note that
(2) That means
a +b
a +b 
> ab only if 
 > ab .
2
 2 
2
a 2 + 2 ab + b 2
> ab .
4
(3) The inequality in (2) is true only if a 2 − 2ab + b2 > 0 .
(4) In fact, the last inequality is equivalent to
( a − b) 2 > 0 .
(5) It is always true since a > b as given.
Complete the proof by tracing the steps backwards.
Solution:
Carmel Alison Lam Foundation Secondary School
F.4 Additional Mathematics
A Journey of Mathematical Thinking
Worksheet 4: Basic Skills of Problem Solving (3)
(F) Two-way arguments (混合型分析法)
Question 1:
Given that the angles A, B, C and D are not equal to the integral multiple of
π . If cos( A + B )sin(C + D) = cos( A − B)sin(C − D) , show that
cot A cot B cot C = cot D .
Think & argue in two ways:
(1)
We try to solve the problem this way. On the one
hand, we make clear what we need in the expansion
of the given expression. On the other hand, we
argue backwards from the result. There should be
a connection between the arguments in two ways.
(2) We may design a form for such way of arguments.
The left hand side of a blank paper records the
“forward” arguments and the right hand side
records the “backward” arguments.
(3)
We start from the result.
cot A cot B cot C = cot D
⇔
cos A cos B cos C cos D
⋅
⋅
=
sin A sin B sin C sin D
⇔ cos A cos B cos C sin D =
sin A sin B sin Ccos D
The hint on the right tells us that we
have to expand the given expression:
(cos A cos B − sin A sin B) ⋅
(sin C cos D + cos C sin D)
= (cos A cos B + sin A sin B ) ⋅
(sin C cos D − cos C sin D)
Some simplification is needed here.
Solution:
Question 2:
In ∆ABC , BD = DC. Let ∠BAD = α , ∠DAC = β and ∠ADC = γ .
Show that cot α − cot β = 2cot γ .
Think & argue in two ways:
We start from the result.
cot α − cot β = 2cot γ
⇔ cot α − cot γ = cot β + cot γ
⇔
cos α cos γ cos β cos γ
−
=
+
sin α sin α sin β sin γ
⇔
sin γ cosα − cos γ sin α
=
sin α sin γ
sin β cos γ + cos β sin γ
sin γ sin β
⇔
Study (*) carefully.
sin( β + γ ) sin(180 o − C )
=
sin β
sin β
=
sin C AD
=
, and
sin β DC
sin(γ − α ) sin B
=
sin α
sin α
=
Hence,
sin B AD AD
=
=
.
sinα BD DC
sin( β + γ ) sin(γ −α )
=
.
sin β
sin α
Solution:
sin(γ − α ) sin( β + γ )
=
sin α sin γ sin β sin γ
(*)
(G) By analogy (類比推廣的訓練)
example:
We have learnt the nature of roots of a quadratic equation:
Let α and β be the roots of an equation ax 2 + bx + c = 0 .
b
c
Clearly, α + β = − and αβ = .
a
a
We can solve the problem of trigonometry by analogy:
If α and β are two distinct roots of the equation
5sin θ − 3cosθ = 2 , where α and β lie between −π and π ,
α
β
find the value of cot 2 + cot 2
without using calculators.
2
2
solution:
校本課程設計 － 數學思維之旅
〈中四試教一、二〉
教師：
對象：
課堂次數：
黃偉強
中四附加數學科學生
(I) 2/5/2001
(兩堂)
(II) 16/5/2001
(一堂)
A. 總目標：
1. 讓學生探索微積分的發展史；
2. 介紹解決數學難題的基本技巧，訓練學生的思維能力；
3. 培養學生學習數學的興趣。
B. 第一節及第二節 (2/5/2001)
1.
目標：
(a)
透過「數學思維之旅」光碟，介紹牛頓的生平。
(b) 介紹光碟內的一些小實驗，從而了解牛頓的工作。
(c)
利用 Java applets，鞏固學生對微分的認識。
２．教學準備：
(a) 上課前先將「數學思維之旅」光碟派發給所有同學，著他們細心觀看。
(b) 聯絡 IT 技術員，於課堂利用無線網絡上網。
３．教學程序：
(a) 播放「數學思維之旅」光碟，介紹牛頓的生平：少年時代、大學時代、神奇
年、逸事、挫折和時年時代，老師從旁補充說明。
(b) 學生發問時間：就牛頓的生平及其工作提出問題。
(c) 到以下網頁瀏覽：
http://cal.hkcampus.net/~cal-ltw/math.htm
並介紹其中有關微分的 Java applets.
４．課業：
學生須瀏覽「數學思維之旅」光碟，並閱讀內裏有關牛頓生平的文章。
B. 第三節 (16/5/2001)
１．目標：
(a) 介紹 G. Polya 所提出解決數學難題的基本步驟；
(b) 介紹「信息型分析法」和「前進型分析法」的解題技巧；
(a) 學生可以利用所學過的技巧去解決數學問題。
２．教學準備：
(a) 印備一份工作紙，引導學生找出那些 Java applets 的意義。
(b) 到以下網頁下載 G. Polya 所寫 How to solve it 的撮要：
http://www.malmo.lth.se/person/dw/prog0/polya.html
(c) 學生已學過的課題中選取例子，用以闡釋如何應用聯想和特殊化猜想去解
決數學問題。
３．教學程序：
(a) 老師簡介：數學思維方法。
(b) 派發 G. Polya: How to solve it 的撮要，並以下面的例子作闡釋，介紹解
決數學問題的四個步驟：
A wire of length 100 cm is bent into a rectangle. Find the maximum area of the
rectangle.
(c) 派發工作紙，介紹在數學思維過程中如何運用聯想能力，並列舉一些例子：
例一：Let A, B, C be any angles in a triangle ABC. Show that
tan A + tan B + tan C = tan A tan B tan C .
4x
例二：Find the equation of the tange nt to the curve y =
which is parallel to
x −1
the line 4x + y = 5.
(i) 老師依工作紙內容引導學生思考問題。
(ii) 老師發問：要解決這題，應按甚麼步驟來處理？
從這題目可以聯想到甚麼已學過的知識？
這個步驟可以想到甚麼方法來？
(d) 派發工作紙，介紹在數學思維過程中如何運用特殊化猜測，並列舉一些例子：
例一：The figure shows a quadrilateral ABCD and a circle whose centre O is on
the line AB and OA = OB. If AD, CD and BC are the tangents to the
circle at E, F and G respectively, show that AB ⋅ AB = 4 AD ⋅ BC .
例二：(a)
(b)
(i)
(ii)
Prove, by mathematical induction, that
n( n + 1)( n + 2)
1× 2 + 2 × 3 + 3 ×4 +... + n(n + 1) =
.
3
Hence, evaluate
1991× 1992 + 1992 × 1993 + 1993 × 1994 + ... + 2000 × 2001 .
老師依工作紙內容引導學生思考問題。
老師強調，在解決一些難題前，不妨自己擬出比較簡單的情況，看看
如何解決，然後才制定一套解題方案。
(e) 老師總結。
４．課業： Chapter 13 Technique of Differentiation
學生在思考習題時，盡量利用以上所學的技巧，並在下次附加數學堂講解習題
時，報告如何思考解題。
校本課程設計 － 數學思維之旅
〈中四試教三、四〉
教師：
對象：
課堂次數：
A.
1.
2.
3.
4.
黃偉強
中四附加數學科學生
(I) 23/5/2001
(兩堂)
(II) 30/5/2001
(一堂)
總目標：
介紹解決數學難題的基本技巧，訓練學生的思維能力；
培養學生學習數學的興趣；
應用資訊科技，鞏固學生微分的概念；
測試由老師制作的 Java Sketchpad 及 Shockwave 程式檔的教學成效。
B. 第一節及第二節 (23/5/2001)
１．目標：
(a) 重溫「聯想」和「特殊化猜測」的解題技巧；
(b) 介紹微分學中相對時間的變化率(rate of change over time)的意義及解決這類
問題的方法；
(c) 學生可以利用所學過的技巧去解決數學問題。
２．教學準備：
(a) 分析有關變化率問題的特性及類型。
(b) 準備 Java Sketchpad 及 Shockwave 程式檔，並上載於網頁中：
http://cal.hkcampus.net/~cal-mt/
(c) 教師預約流動 MMLC，並準備有關的器材及軟件。
(d) 與前一堂老師商量，使器材可以預先裝好。
３．教學程序：
ds
d  ds  d 2 s
及 a=  = 2 。
dt
dt  dt  dt
(b) 探討其他情況下的相對時間的變化率問題（只有一個相對時間的變數）。
例一：一個汽球在膨脹的過程中，它的半徑增長情況如何？
(i) 首先由學生用實物示範，並觀察其中半徑的變化。
(ii) 學生利用網上的 Shockwave 程式檔，細心觀察隨著時間的改變，半徑的
增長情況，然後選擇一幅合適的圖來表示。
事實上，半徑的增長與時間並不是成正比例的，時間愈長，半徑增加的
速度愈慢。
(iii) 研究有甚麼因素影響半徑的增長速度──如何用數學的方法去解決問
題。
4
－「汽球」聯想到「球體」，「球體」聯想到其公式： V = π r 3 。
3
dr
－「求半徑的相對時間的變化率」聯想到「求速度」的方法： 。
dt
dV 4
dr
dr
－將公式由左到右微分一次：
= π (3r 2 ) = 4π r 2 。
dt 3
dt
dt
dV
－假設氣體以固定速度吹進汽球裏，即
為常數，有
dt
dr
1 dV
k
=
= 2。
dt 4π dt r
－所以，半徑增加的速度與半徑的平方成反比。
(iv) 半徑增加的速度與半徑的關係圖為：
r
(a) 重溫有關速度及加速度的問題： v =
t
dr
d2 r
由學生以這圖來畫出 vs t 圖及 2 vs t 圖。
dt
dt
2
d r
從 2 vs t 圖可以發現，半徑的增加是以減速(decelerated)進行的。
dt
(v) 為更清楚了解這個情況，我們考慮一個特別的情況：
一個圓形增加其面積時，它的半徑增長情況如何？
－立即聯想到圓形面積： A = π r 2 。
dA
dr
dr
－將公式由左到右微分一次：
= π (2r ) = 2π r 。
dt
dt
dt
dr 1
－若面積增加的速度為一常數， ∝ 。
dt r
(c) 探討其他情況下的相對時間的變化率問題（有兩個相對時間的變數）。
例二：一個圓錐形狀的漏斗，若漏水的速度為常數，水位隨著時間的變化如
何？
1
－「圓錐體」聯想到其公式： V = π r 2 h 。
3
dV 1  2 dh
dr 
－將公式由左到右微分一次：
= π r
+ 2 rh  。
dt 3  dt
dt 
此處有兩個相對時間的變數。
－為使問題簡化，考慮將 h 和 r 拉上關係：
r
設立體角(solid angle)為θ ，有 tanθ = ⇒ r = h tan θ 。
h
2
π tan θ 3
這樣可將公式先行化簡： V =
h ，
3
dV
dh
然後微分，得
= (π tan2 θ )h 2
。
dt
dt
dV
假設漏水的速度是常數，則
為常數，有
dt
dh k
=
， 其中 k 為一常數。
dt h 2
(d) 老師引導學生得出結論：水位的下降速度與高度的平方成反比。
(e) 學生利用網上的 Shockwave 程式檔，細心觀察隨著時間的改變，高度的變
化情況。
(f) 堂課：Chapter 16 Ex 16B Q.3, 4, 5, 6.
dh
d2 h
(g) 家課：依 Shockwave 提供的 h vs t 圖，畫出
vs t 圖及 2 vs t 圖。
dt
dt
(h) 網上直播：「數學思維講座」內容。
老師示範如何在網頁上觀看講座片斷，並指出是次講座可以讓同學了解微
積分是如何發展的。
C. 第三節 (30/5/2001)
１．目標：
(a) 介紹「信息型分析法」、「前進型分析法」、「追溯型分析法」、「混合型分析
法」和「類比推廣」的解題技巧；
(b) 學生可以利用所學過的技巧去解決數學問題。
２．教學準備：
印制工作紙。
３．教學程序：
(a) 派發工作紙，介紹在數學思維過程中如何運用信息型分析法，並列舉一些例
子：
例一：The figure shows a triangle ABC with C ≠ 90 o . Show that
a − c cos B sin B
=
.
b − c cos A sin A
例二：Given y = xn + x1− n , where x ≠ 0 .
dy
Show that x + (n − 1) y = (2n − 1)x n .
dx
老師提出這種思考方法的特點：化簡－尋找信息－化簡。
(b) 介紹在數學思維過程中如何運用前進型分析法，並列舉一些例子：
例一：Let tanα and tan β be the unequal roots of equation
ax 2 + (2a − 3)x + (a − 2) = 0 , where a ≠ 0 .
Find the minimum value of tan(α + β ) .
例二：Given that tan A, tan B , tan C ≠ 0 , if
tan( A − B ) sin 2 C
+
= 1 , show
tan A
sin 2 A
that tan2 C = tan A tan B .
老師提出這種思考方法的特點：
(i) 逐次逼近分析；
(ii) 以這題不等式為例，從左邊出發，利用充分條件(sufficient condition)一
步步將問題轉化，尋找新的條件。
(c) 介紹在數學思維過程中如何運用追溯型分析法，並列舉一些例子：
例一：Given that sin B = k sin(2 A + B) , k ≠ 1 . Prove that
1+k
tan( A + B ) =
tan A .
1− k
a +b
例二：Given a > b , b > 0, a ≠ b . Show that
> ab .
2
老師引導學生從必要條件(necessary condition)出發，思考如何解決問題。
(d) 介紹在數學思維過程中如何運用混合型分析法，並列舉一些例子：
例一：Given that the angles A, B, C and D are not equal to the integral multiple
of π . If cos( A + B )sin(C + D) = cos( A − B)sin(C − D) , show that
cot A cot B cot C = cot D .
例二：In ∆ABC , BD = DC. Let ∠BAD = α , ∠DAC = β and ∠ADC = γ .
Show that cot α − cot β = 2cot γ .
所謂「混合型分析法」，就是將「前進型分析法」和「追溯型分析法」混合
使用。
(e) 介紹在數學思維過程中如何運用類比推廣來解決問題，並列舉一些例子：
We have learnt the nature of roots of a quadratic equation:
Let α and β be the roots of an equation ax 2 + bx + c = 0 .
b
c
Clearly, α + β = − and αβ = .
a
a
We can solve the problem of trigonometry by analogy:
If α and β are two distinct roots of the equation
5sin θ − 3cosθ = 2 , where α and β lie between −π and π ,
α
β
find the value of cot 2 + cot 2
without using calculators.
2
2
所謂類比推廣，就是將已學過的解題技巧應用在其他的問題上。
(f)
老師總結