[AM_20151119] Anticipation and Bridging as Proof Tactics

Question

Introduction

     In a previous article, I have showed an “evil” tactic that can be used against “evil” questions.  Here is another “evil” trigonometric proof question.

     Many traditionalist school teachers insist on starting either from the LHS or the RHS, and working all the way to the other side.  Personally I do not mind any form of presentation as long as it is logical.  But not many students are able to do that.  People tend to fall into the trap of beginning a proof with the statement that they are supposed to prove in the first place.  This is called circular reasoning (or “begging the question” or petitio principii).  It is definitely a no-no.  So there is some advantage to sticking to the traditionalist mould.  The disadvantage is, of course, that it stifles creativity and this gives a misleading image of mathematics to the learner.

More “Evil” Tactics

     Now, if we do not want to “break the rules”, perhaps we can “bend the rules” a little.  On a piece of rough paper, or in your mind, secretly work from both sides and try to bridge them in the middle.  Let us compare

Think:  How are they similar?  How are they different?

As you can see, the LHS already has a preponderance of  cos 75°.  One of these  cos 75°  must somehow disappear.  The  RHS  has  4 sin 75°  which the LHS does not have.  So if we start from the LHS, we can use our magic wand [SV4] to create  4 sin 75°  out of thin air, remembering to divide by the same thing, so that the original value does not get changed.

Solution

Reflection

     Have you learned anything from this problem?  Let us review the heuristics used to help us solve this problem successfully.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Special Variants of Heuristics (Good for Trigonometric Proofs)

SV1   Work from both sides and try to bridge them in the middle [~ H04]

SV2   comparing (similarities/differences) what you have now with what you want [~ H05]

SV3   anticipate what will happen in the end and what you must do now [~ H05]

SV4   Magic Wand or create something out of nothing (无中生有) [~ H09]

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* IB SL & HL Mathematics (revision)

* other syllabuses that involve trigonometry

* just about anyone who is interested, really

[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_20150621TGCARF] Sum of sin and cos the cost of tan?

Question

Introduction

     This is a bonus Maths II (rough equivalent of Additional Mathematics syllabus) question from a test by Hwa Chong Institution (HCI) this year 2015.  Independent schools in Singapore like HCI are free to set their own curricula, but they usually end up covering slightly more than the mainstream curriculum, since their students also take the national examinations.  For their internal tests and exams, they can set bonus questions.  These are harder but optional questions that students can attempt and if they are successful, the bonus marks can be added to their normal marks.  It is also OK not to attempt the bonus questions.  That gives students the choice and opportunity to stretch their minds, but they are not penalised if they are unable to solve the bonus questions.     In this article, I present two approaches to tackling this question.  Let us review some important formulas first.

Some Useful Formulas

Solution 1  (via the Pythagorean Identity)

  

Solution 2  (via R-formula)

Reflections / Extension

     Here is another HCI question on trigonometry that involves sine, cosine and tangent.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

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 trigonometry

[AM_20150621TGSCQT] sin and cos Embroiled in some Quadratic

Question

Introduction

     This is a bonus question from another version of a test set by Hwa Chong Institution this year (2015).  It turns out that different students take the test at different dates, and the school took the trouble to set different versions of the test.  They have the manpower resources to do that!

     It is good to know what topics each question involves.  In this example, students’ knowledge of quadratic theory and trigonometry are being tested in a combined fashion.  Let us first review the relevant material.

Reminders

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

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 trigonometry

[JCH2CNEFTG_20150410] Exponential Half-Power Trick

Question

Introduction

     This is a complex-number question that appeared for ‘A’ Level Mathematics in November 1998.  Remember that mathematics is never out-dated.  Many Singapore schools keep this sort of questions in their question banks (for tutorial exercises, tests and examinations), in the hope that it becomes part of students’ repertoire.

     Students are expected to know the famous Euler’s Formula, one of the most beautiful formulas discovered by this visually-challenged but prolific mathematician.  It links the exponential function with trigonometry via the idea of angle rotation.  Adding e^if  with its reciprocal  e^-if (which is also its conjugate) gives a cosine expression, while subtracting gives a sine expression.

     The above question can be solved by rationalising the denominator and using heavy trigonometry and half-angle formulas.  There is nothing wrong with this approach.  I am going to illustrate a kewl approach, using what I call the exponential “half-power trick”.  Basically, whenever you see an expression like  1 ± e^{2kq i},  force out the factor  e^{kq i}.  This gives you either a sin or cos expression.  For example, e^{6q i} – 1 = e^{3q i} (e^{3q i} – e^{-3q i}) = i×2e^{3q i} sin q.

Solution

Observe that we have killed two birds with one stone.  At the last step, we just compare the real and imaginary parts to read off the answers.

Suitable Levels

* GCE ‘A’ Level H2 Mathematics (“Complex Numbers”)

* precocious students who love complex numbers

[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

[IB-HL_FTPR20150402] Periodicity cos a fortiori

     In this article, I give a rigorous proof of the period of cosine functions, using an a fortiori argument, signaled by the use of the words “in particular”.  When something is generally true for, say, all real numbers, it is “all the more” true for a particular case of a real number like zero.  I use this logic to nail down the answer.

Question

Solution

Suitable Levels

* revision for IB Mathematics HL / SL

* other syllabuses that teach trigonometry

[IB-HL H&H_Ex13e] pg 374 Q19 Angles with nice tans add up nicely

Question

This is taken from Haese and Harris textbook for IB Mathematics page 374.  It involves an interesting connection between certain angles whose tangents are nice fractions and the 45º degree angle i.e. ^p/₄.

Solution

Here is another look at the problem.

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

[AM_FTCA20150402] cos times cos causes cos plus cos

Question

This question involves formulas related to compound angles, specifically a product to sum formula and double angle formulas involving cosines.

Important Formulas

Solution

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