[AM_20151130IAXS] A Motif for the Absolutely Absolute

Problem

Introduction

     This question would pose a challenge for many students, although theoretically it is within reach of a good Additional Mathematics student (~ grade 10).  Graphs of both  sin x  and  cos x  are waves that oscillate up and down.  There are many pairs of vertical bars, indicating the absolute values or modulus, and these seem confusing.

Strategy

     Let us graph the functions  y = |cos x|   and  y = |sin x|.   Note that  ||sin x| – |cos x|| = ||cos x| – |sin x||.   The absolute difference of  |cos x|   and  |sin x|  is the difference between them ignoring the negative sign (if any) of the result.  And this is just the difference between the higher value and the lower value. 

Can you see any repeating patterns?  [H04]  Can you visualise the required area?  How many times is that of the basic pattern (known as “motif” in art)?  [H09, H10, H11]

Solution

Remarks

     Our total area is made up of four congruent pieces.  When  0 < x < ^p/₄,  cos x  is higher than  sin x.  That allows us to strip away all the absolute signs and do the calculation.

Heuristics Used

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* challenge for GCE ‘O’ Additional Mathematics  * IB Mathematics SL HL

* GCE ‘A’ Level H2 Mathematics  * IB Mathematics HL

* AP Calculus AB & BC

* University / College calculus

* other syllabuses that involve integration and area

* whoever is game for a challenge in integration

[U_Calculus_STKC58] Exploiting symmetry for a Complicated Integral

Question

     This problem appears as problem 58 from a Facebook group and it is set by Kunihiko Chikaya.  Ordinary integration problems are already challenging, but this one is tough on steroids. 

The Standard Approach

Solution 1

With this result, I realised that there is a short cut.  We can make use of symmetry.  Note that  sin(p – x) = sin x   and   cos(p – x) = - cos x.

Solution 2

Heuristics Used

H04. Look for pattern(s):         “onions”, exploit symmetry

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* University / College Mathematics

* challenge for ‘A’ Level H2 Mathematics

* challenge for IB HL Mathematics

* other syllabuses that involve trigonometry and integration 

[AM_20151117] An “Unorthodox” Technique for Trigonometric Proof

Question

Introduction

     When proving trigonometrical identities, one usually starts from the “more difficult” side and try to work towards the other side.  The above identity looks like a tough nut to crack.  Both sides look equally complicated.  Where do we even begin?

     Here is one way to “cheat”.  Starting from, say, the LHS, we multiply the RHS expression and also multiply by its reciprocal i.e. dividing by the same amount.  It is like using a magic wand to create something out of nothing (无中生有 in Chinese), but doing so does not change the value of the LHS expression.  Now we do not touch the part that is equal to the RHS (highlighted in yellow below), but we try to find a way to cancel away the other stuff, as shown here.

Solution

Remarks

     In the third step, I had replaced  cos²A  with  1 – sin²A  and  sin²B  with  1 – cos²B.  This leads to the required cancellation and we are left with the yellow patch, which was never touched since the first step and it is the RHS.  This completes the proof.

     Just as in martial arts where deadly opponents require deadly strokes to counter them, evil questions require “unorthodox” techniques.  Even though most people would not have thought of it, all the steps presented above are actually legitimate.  This is because at every step, the equality is preserved.  All the “=” are really equal, and it’s legit (either work hard or you might as well quit), although my “can’t touch this” tactic seems a bit clairvoyant.

. 

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve proof of trigonometric identities

* anyone who loves mathematical challenges

[U_Complex20151117] Lagrange Identity for Complex Numbers

Question

Introduction

     I was revising my complex analysis just for fun (I had graduated almost 3 decades ago) and I came across this problem which is question 5 from page 9 of the book Complex Analysis by Ahlfors.  This book is pretty hard core for a first course in complex analysis, but this is not surprising since Lars Ahlfors was no less than a Field’s medallist.

     Intuitively, I knew that this identity looks like the vector identity

                                        |a · b|² = |a|² |b|² – |a ´ b|²

and if one strips away the |a|² |b|², it boils down to the Pythagorean Trigonometric Identity

                                         cos² q = 1 – sin² q

I thought if I could just define the appropriate dot and cross products (actually this can be done), I could solve it easily.  However, this itself requires proof.  I was begging the question.  In fact, it is precisely because of this Lagrange Identity (and the related Cauchy-Schwarz Inequality) that allows  cos q  and   sin q  to be meaningfully defined.

     OK, it looks like I have to do it the hard way.  The tricky part in the manipulation of those sums in sigma notation is to ensure, at each step, that I did not introduce any spurious terms, nor miss out any terms.  To simplify the notation, in what follows I shall assume that  i  and  j  are indices in the range of whole numbers  [n] = {1, ... , n}.

Solution

Remarks

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* University / College level Complex Analysis

* other syllabuses that involve Complex Analysis

* any independent learner game for a challenge

[AM_FTDA20150402] Proof of Triple Angle Sine Identity

Question

Solution

Discussion

     We are given a triple angle, and we want an expression in terms of  sin q  only, without any double or triple angle.  First, we split up 3q  into  2q +q.  see [1].  This allows us to use the compound angle formula, and then double angle formulas.  At every step, it is a good tactic is to compare what you have with what you want.  This problem solving heuristic is not in the official list, but from my experience it is a useful one.  So at [2], we have three choices for the cos 2q :  cos 2q  = cos²q  – sin²q ,  cos 2q  = 2cos²q  – 1,  and  cos 2q  = 1– 2sin²q.  Which one shall we choose?  Since everything needs to be expressed in terms of  sin q  only,  the best choice is the third formula.  At [3], we have a  cos²q  appearing in the first term.  Again we express that in terms of  sin q  via  cos²q  = 1 – sin²q.  After that, we simplify to get to the RHS.

Heuristics Used

H10. Simplify the problem

H11. Solve part of the problem

Hxx. compare what you have with what you want

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* revision for GCE ‘A’ Level H2 Mathematics

* revision for IB Mathematics HL / SL

* other syllabuses that teach further trigonometry