A few people felt that my experiment about solving the equation would have been more interesting if the equation had been more interesting. I’m not sure I agree with that: part of my intention was to get some evidence about what we do in a very simple situation. However, I now have another example that I want to know about, and it happens to be a harder one. Not much harder, but harder nevertheless.

One of the byproducts of its being a more interesting equation is that it will be quite a lot harder to encode all possible experiences that people might have and run a vote as I did last time. So I’ve made the vote a little bit vaguer, and I will be interested not just in the numbers but also in people’s descriptions of what they did (as I was first time round). My main aim is to get a fairly complete picture of all the ways that different mathematicians find it natural to think about the problem.

As last time, if you want to take part, then try to keep a close eye on all your thought processes: I’m interested not just in the ones that got you to the answer, but also in the ones that you might have thought about and dismissed. So try not to look at the problem until you are ready to keep track of what your reactions are. The problem will appear immediately after the fold. For the benefit of those who read this using RSS, I’ve also left a bit of space, and I’ve left space between the problem and my discussion of it.

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**Problem.** Find all such that

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**Some possible responses to the question.**

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*Approach 1.*

Step A. Spot that the left-hand side is equal to , so the equation holds if and only if

Step B. To solve this, observe (possibly after some thought) that the left-hand side can be written as

which equals .

Step C. It follows that is a solution if and only if is an odd multiple of

Step D. This is true if and only if is plus a multiple of .

*Approach 2.*

Step A. Spot that and rewrite the equation as

Step B. Recognise that the left-hand side is , so this can be rewritten .

Step C. It follows that is a solution if and only if is for some integer

Step D. This is true if and only if is plus a multiple of .

The main thing that interests me is, I think, clear. There are two patterns to exploit in this question. One is the polynomial , which we are trained to recognise as . The other is the identity . My initial instinct was that to use the latter identity would somehow be to split the expression up in an unnatural way, and it was only later that I realized that the appearance of the pattern as the middle term moved the goalposts a bit and not only made this splitting less unnatural than I thought, but actually led to an easier solution.

I deliberately didn’t say whether I expected people to solve this in their heads or to use paper. My guess is that those who used the first approach were more likely to need paper for it than those who used the second, but I’d be interested to know whether that is the case.

I would also guess that while some people abandoned the first approach for the second, not many people would have abandoned the second for the first — though some might have tried the first after completing the second. Also, my guess is that anyone who abandoned the first approach would have done so after carrying out Step A, since that is the point where one has to stop and think and possibly write things down.

For the vote, an expression such as 1A2D means, “I first tried Approach 1 and got as far as Step A. I then switched to Approach 2 and got to the end.” To give three more examples, 2D1D means you completed both approaches, starting with Approach 2, 1D means you completed Approach 1 and that was the end of what you did, and 2A1D means you spotted the identity but decided not to use it and went for Approach 1 instead. I have also put a “none of the above” option, since it’s quite likely that that will apply to some people.

As I have already said, if you want to give me a narrative account of your experiences, then that is great too. Also, if you think there is an option that I ought to add to the poll, then feel free to suggest it.

May 1, 2010 at 4:12 pm |

I first spotted the sin^2 + cos^2 identity and simplified to sin x cos x = 1/2, at which point I thought about the unit circle and decided you’d only get a rational answer from the angles at the eighth roots of unity. (Side note: not absolutely sure this is true, is it?) Brief testing revealed the answer.

May 1, 2010 at 4:14 pm

Extra note: I am meaning a rational answer when multiplying the sine and cosine together, not individually.

May 2, 2010 at 4:41 am

sin x cos x is a continuous function, so by the intermediate value theorem there are infinitely many x for which the product is rational.

May 2, 2010 at 5:29 pm

Ah, of course.

I’ve been thinking more about how I pattern matched sin^2 + cos^2 instantly but not 2 sin x cos x, and I think I have an explanation why:

I was doing the work in my head, but at least at the initial step, I was able to see sin^2 + cos^2 visually. At that point I “placed” the result in my head and worked from there, and somehow my access to my trig-pattern-matching bank was shut down. If I had done the process on paper I would have caught it.

May 1, 2010 at 4:28 pm |

I voted 1D, even though it is not exactly the truth. I first saw some cosines and sines and thought “I will be harder than previous time”… Then I did step 1A. But then I changed from your steps:

I studied the function x → cos x + sin x. It is 2π-periodic, and its derivative is cos x – sin x. It v

May 1, 2010 at 4:35 pm

Oops, I sent before I finished:

The derivative vanishes iff cos x = sin x, and it is well known that this happens iff x=π/4 (mod π).

Then I wanted to use the fact that the function takes the values ±sqrt(2) only twice between its maximal values (and the fact the I knew that π/4 was a candidate), but as the maximal values are for π/4 (mod π), then it is finished.

May 1, 2010 at 4:37 pm |

I voted 1A2D, but actually did something slightly different:

A) Recognize on first sight.

B) Stare at the whole expression to decide which flavor of trigonometric identity / binomial formula it is. (Also looking for and similar patterns). Decided that the binomial formula looked best.

C) Use binomial formula, recognize that can be expressed as a single sine/cosine. (That’s a trick I’ll never forget after not knowing it in a math competition.)

D) While now happy that I have pretty much solved the problem, I decide that I’m too lazy to perform the calculation and look for a different approach, in particular since there’s still the unused pattern.

E) Spot that and conclude . That’s the extremal value of the sine, done. I didn’t bother figuring out the actual form of the $x$ values, knowing that they are some shifted multiples of $\pi$ or something.

May 1, 2010 at 5:34 pm |

Yes, very interesting. 1A2D is very nearly what my thought process was. With regard to elementary-type problems such as this (involving quadratics or higher power terms), it seems we’ve been conditioned to factor first, and ask questions later.

May 1, 2010 at 5:35 pm |

The first thing I noticed was the sin(2x) identity, but at that stage it didn’t seem useful. After reaching 1A I decided to look for something simpler, and saw 2A. Then my first idea after reaching 2A was that I know both and $\latex \sin^2(x)+\cos^2(x)$ so I should be able to determine each separately. Only then did I recall my initial observation.

In retrospect, it seems I tend to follow unnecessarily convoluted lines.

May 1, 2010 at 5:40 pm |

I first spotted the sin(2x) identity. I then thought briefly that I would need to use the cos(2x) identity as well, before noticing that the other two terms summed to 1. Then it was just a matter of figuring out when sin(2x) = 1.

May 1, 2010 at 5:58 pm |

I spotted that 2sin(x)cos(x)=sin(2x), looked at the rest of the equation, saw \sin(x)^2 + \cos(x)^2 = 1, reduced to 2sin(x)cos(x)=1, converted to 2sin(2x)=1 and promptly got the wrong answer… Grrr! But I voted 2D

May 1, 2010 at 6:17 pm |

1. Oh Trigs! Hope there is an exact solution

2. Looks like a quadratic …

3. Trig identities might be useful here. Recall that $sin(x)^2 + cos(x)^2 =1$ and $sin(2x) = 2 cos(x)sin(x)$. Hope I don’t need anything else.

4. Tried to express it as a quadratic with $s = sin(x)$ Get $s^2 + 2s\sqrt(1-s^2) + (1-s^2) =2$.

5. Immediately notice the cancellation: $ 2s\sqrt(1-s^2) =1$

6. Notice that I could use $sin(2x) = 2 sin(x)cos(x)$ here but decide to press ahead with quadratic for now.

7. After little algebra get $s^2 – s^4=\frac{1}{4}$ and hence $s^2=\frac{1+\sqrt(2)}{2}$. Decide this isn’t very good.

8. Go back to Step 6 and use $sin(2x) = 2 sin(x)cos(x)$ and get $sin(2x) = 1$ and then easily work out x.

May 1, 2010 at 6:43 pm |

I put None of the Above, but it was pretty close to 1A2D. First I saw the quadratic could be factored, but before actually doing so I noted the term 2 sinx cosx could be rewritten. I then went ahead and factored it anyway, decided there was no easy way to solve sinx+cosx = +-sqrt 2, THEN noticed sin^2 x + cos^2 x = 1, and solved as in 2. (Actually I made a silly mistake and said it was pi(2k+1)/4 for integers k. I realized the mistake after reading the approaches.)

May 1, 2010 at 6:45 pm |

I put None of the Above, but it was pretty close to 1A2D. First I saw the quadratic could be factored, but before actually doing so I noted the term 2 sinx cosx could be rewritten. I then went ahead and factored it anyway, decided there was no easy way to solve sinx+cosx = +-sqrt 2, THEN noticed sin^2 x + cos^2 x = 1, and solved as in 2. (Actually I made a silly mistake and said it was pi(2k+1)/4 for integers k. I realized the error upon reading the approaches.)

[Sorry if I double-post, the first one didn't seem to go through.]

May 1, 2010 at 6:54 pm |

I was also a 1A2D: the pattern was more immediately visible than the one, but once the latter came into view, together with the pattern, it was clear that 2 was the way to go.

It would have been interesting (but beyond the capability of this blog) to do some randomised trials of, say, versus ; my guess is that with the latter the approach 2 would be much more prominent.

It may also unfortunately be the case that the reader pool here is now no longer representative for such experiments, having been conditioned by two such examples already…

May 1, 2010 at 7:25 pm

On that last point, I was worried about it even this time: perhaps one should say that there was an additional piece of input, namely that there was likely to be more than one way of doing the simplification. But there was a slight difference from the last equation, in that if you did the slower approach, you were less likely to think “Oh, silly me” on seeing the quicker one. Somehow the two approaches were “equally legitimate”, whatever that means.

May 1, 2010 at 7:02 pm |

Hi Tim,

I first noticed that the left hand side is equal to (sinx + cosx)^2, then I noticed that 2sinxcosx = sin2x, then I thought that I can probably use some trigonometric identities to simplify the expression, and then proceeded with your second route. So the best answer for me would be 1A2D, but with your steps A and B in the 2 tree interchanged:

1A – 2B – 2A – 2C – 2D.

Marko

May 1, 2010 at 7:46 pm |

1B2D. I was working in my head and first tried the factorization, then the rewrite (both habits burned into me from competitions like ARML), but was too lazy to figure out the resulting set of solutions, so then I noticed 2. It’s really interesting to me that I noticed 1 before 2; I think I was biased from the previous experiment.

May 1, 2010 at 7:48 pm

Actually, let me be precise. Initially I seem to have misread the problem slightly and I thought there was a sin 2x identity and a cos 2x identity rather than a sin 2x identity and a Pythagorean identity. I concluded that this would be difficult and turned my attention to the factorization before realizing that I had been wrong.

May 1, 2010 at 8:26 pm |

2A1A2D — I did the first steps of each before I decided which I could finish in my head, and saw the 2A first.

May 1, 2010 at 8:32 pm |

I thought about both the approaches suggested in the post. However, one difference w.r.t Approach 1 is the way I solved it. I treated sin(x) + cos(x) as a function and drew its graph to see where the function was +/- sqrt(2).

May 1, 2010 at 9:33 pm |

Think I’m 1A2D. Tried 1 as far as A, thought, “nah, too much bother to do in my head”, then looked again and spotted sin(2x), thought that was much simpler. I may have been conditioned by your previous post though.

May 1, 2010 at 9:51 pm |

I’m 1A2D too – spotted the factorisation – definitely had the last example in mind, realised this gave two cases and intuitively had a sense of what the answer would be, and also that there ought to be an easier way of getting there, so went directly through approach 2.

May 1, 2010 at 9:59 pm |

I’m roughly 1A2A?D, so I’m owning my “none of the above”. I recognized first the fact that you could factor the left into (sin(x)+cos(x))^2 = 2, so that sin(x)+cos(x)= ±√2. Then I realized that you could cancel sin(x)^2+cos(x)^2 = 1 from both sides, so that sin(x)cos(x) = 1/2. Together, these two equations determine sin(x) and cos(x) up to switching, by the quadratic formula, but by inspection a solution to sin(x)+cos(x)=√2, sin(x)cos(x) = 1/2 is sin(x)=cos(x)=1/√2, so the quadratic formula says these are the only values in that case. Hence the solution in the other case is sin(x)=cos(x)=-1/√2. These correspond together to x=π/4 + nπ.

May 1, 2010 at 10:01 pm |

2A1D – I’m not sure that knowing that this will be an experiment had any effect or not.

May 1, 2010 at 10:06 pm |

I saw approach 1 immediately and stuck with it. I got hung up on the addition identity— it took minutes to get it right. 95% of the time I spent on strategy 1 was in the 1B’s “possibly after some thought.” I wasn’t worried that I was wasting my time. I knew that there _was_ some identity like that, and once I could recall what it was, I could finish the problem.

Having done that, I looked at solution 2 and felt foolish. It would have been much faster—- I’m much more familiar with the identity for sin(2x) than I am with the addition formula. But it never occurred to me because I had the problem in a form I knew I could solve. In retrospect I felt like the mathematician in the jokes who is content to reduce something to a previously solved problem, even if it is more complicated than the situation requires. It is the feeling I get when I find some lemma I have spent a lot of time on done very simply in some old textbook— they got just where I wanted to go, more simply, because they had a different point of view.

This reminds me of a good reason, or perhaps the only reason, to go to research-level talks in fields far beyond your own field of expertise— where you have no chance of distinguishing the novelty in whatever the speaker has done, from the standard methods of the subject, or even from the basic consequences of the definitions in that subject. All of the details become a kind of white noise and you’re left with only an outline of the basic problem solving strategies that the person was using. Sometimes, exposure to this frees my brain from whatever patterns it has “locked onto”— where my instincts have led me to ignore something simple because I know something more complicated that will probably work.

May 1, 2010 at 11:08 pm |

I noticed first that s+c= + or – sqrt(2) and then that 2*s*c=1. This together leads to a pair quadratic equations (in c, say) each of which has a double root at c= + or – sqrt(2)/2. This is what Owen Biesel describes.

May 2, 2010 at 1:27 am |

I voted 1A2D but something like “1A-then-guess-1D-then-2D” is more accurate. After factoring, it was clear that the x’s for which sin(x) and cos(x) are both +sqrt(2)/2 or -sqrt(2)/2 were solutions, and it seemed reasonable to guess that these were the only solutions. However, it wasn’t instantly obvious how to prove this, so instead of reaching for a pencil I decided to look at the equation again, and then spotted approach 2 and worked that to the end.

May 2, 2010 at 2:31 am |

1A2D. To be more precise, I noticed the (a+b)^2 factorization first, and then almost immediately noticed that I could use sin^2 + cos^2 = 1, but I decided to go with (a+b)^2 until I reached step B which seemed hard and so I instead decided to use sin^2 + cos^2 = 1.

May 2, 2010 at 2:43 am |

I saw 2sin(x)cos(x)=sin(2x) first. I didn’t notice sin2(x)+cos2(x) right away, but instead kept staring at the equation, and started on approach 1. My thought process followed Jason Dyer’s, and I found π/4 by experiment … but because I had based it on experimentation and considered the argument inadequate (and therefore didn’t feel the approach required more effort right away), I went back to the original equation and combined observations to solve using B. So my thought process was:

2B-1A-(1D)-2D

May 2, 2010 at 4:13 am |

Sorry for the long and verbose post and also sorry for my bad english…But as I understand that the objective is to see how people really think I prefer to say more then less. I also think I got 2 ideas that are not already written above.

First I saw the perfect square and found sin(x)+cos(x) = sqrt 2. Here just seeing the sqrt(2) made me instantly think about 45 degrees (that is the principal angle related to sqrt(2) anyway) and testing to see that it worked. OK. (I returned to that line latter on…)

After that I caught my distraction and put “plus/minus sqrt(2)” and found the other solution (all other solutions are equivalent to these two anyway) as I got used not to care to much about writing down all solutions since it does not give anything new, but if something is not clear I make myself write all solutions down anyway, just to check things up.

After that I followed path B to the end. OK.

Not satisfied with using in path A recognition/trial of 45 degrees. I thought about isolating the sine in sin(x)+cos(x) = sqrt 2 and substituting that in sin^2+cos^2=1 seeing that would give a quadratic equation in cos, hence soluble by that means(not finding it particularly nice but workable anyway). Also because I got used to the fact that sin^2+cos^2=1 gives one more relation in trigonometric problems.

I don’t particularly like the follow-up that you mentioned in solution by path A. Probably because of my own experience in not seeing such kind of trick (putting sin(45) out of the blue) when first getting myself a problem of this kind. For me is hard to motivate such a trick. Perhaps by trying to think backwards?

Finally there is a small advantage in the more algebraic substitution since it would find something even if a nice angle was not “seeable”. I know that the method presented can be made to work in all cases (Ex. 3sinx+2cosx=1). But I see it more as some problem that really came from another set of ideas than something that would occur “naturally” to a student in this context.

I think my lines of reasoning are biased because I already have lots of experience with this kind of problem. But I am also used to make a conscious effort to emulate “reasonable” solutions since I also have experience with students in this level. But student experiences likely vary somewhat.

May 2, 2010 at 4:39 am |

I started with step 1A (although I forgot the +/- on the right until later), but then I found in my mind the specific fact that the maximum of sin x + cos x was sqrt{2}, and that it occurred when both sin x and cos x were sqrt{2}/2. I knew there was some trig identity like 1B out there, but instead I chose to verify this specific fact using calculus, solving for the zeros of the derivative which are where tan x = 1.

May 2, 2010 at 10:52 am |

1A2D: I noticed that this was (sin(x)+cos(x))^2, then thought for a bit to try to remember the standard way to solve this, and then noticed that I should instead cancel the sin(x)^2+cos(x)^2 part. My thinking was then very close to Tim’s sketch of method 2…

May 2, 2010 at 11:13 am |

1A2D.

May 2, 2010 at 11:23 am |

2D: took a pen, and decided to write things down before thinking about it – after having written sin(x)^2 down, I then looked at the screen again: the sin(x)^2+cos(x)^2 was then the only thing I saw. I haven’t noticed the factorization at all.

May 2, 2010 at 11:38 am |

I answered 1D, but my approach were slighly different. Recognizing the pattern, a^2+2ab+b^2=(a+b)^2, and remembering that sin(pi/4)+cos(pi/4)=\sqrt(2), the solution was immediate.

May 2, 2010 at 1:53 pm |

I simplified to sin x cos x = 1/2. Then I considered the portion of the unit circle in the upper right quadrant and noted that x = \pi/4 satisifies this. I then trusted my intuition that sin x cos x would be maximized at \pi/4. (This is easy enough to prove with the derivative, but when solving the problem, I just trusted it was true.) Then I thought about the other 4 corners of the unit circle and it was clear that only \pi + \pi/4 would satisfy the equation.

May 2, 2010 at 2:13 pm |

I reduced the equation to as in 1A, and then I solved this equation by looking at the unit circle: is a point on the unit circle, and the point should be on either the line or . I know that these lines are tangent to the unit circle, and it is easy to see that the intersections corespondents to and .

I answered “None of the above”.

May 2, 2010 at 2:18 pm |

My answer was “none of the above” but it was close to 1A2X:

I did 1A2A but then I applied again $1=\sin^2(x)+ \cos^2(x)$.

So I obtained $(\sin(x) – \cos(x))^2 = 0$ simplified to $\sin(x)=\cos(x)$

then reasonned mentally with the trigonometric circle in mind, giving

$\pi/4 + k \pi$.

It might be another option in the vote list.

May 2, 2010 at 5:02 pm |

I did not use paper and it took me about two minutes. At first, in analogy to the first experiment, I wanted to start with transforming the left-hand side to (sin+cos)^2, but did not immediately know the solutions of sin(x)+cos(x)=sqrt(2). So I started the second approach and got sin(x)cos(x)=1/2. For an instant, I wanted to plug in cos=sqrt(1-sin^2), but, having no paper at hand, looked for something less ugly. Then I thought about the unit circle and tried to guess a solution. I then thought: it would work if both factors equalled 1/sqrt(2) and came up with the solutions x=pi/4(+pi). I had the feeling that these were the only solutions. But how can one show this? Maximizing under the condition sin+cos=1 came to my mind, but recognized that the squares were missing. But I remembered my first approach had let me to sin(x)+cos(x)=sqrt(2) and I knew that under this condition, sin(x)cos(x) would be maximal if sin(x)=cos(x) leading exactly to the solutions I had guessed.

May 2, 2010 at 5:29 pm |

I voted 1D. More precisely, I got to $\cos(x) + \sin(x) = \sqrt{2}$, and remembered the left hand side should be something like $\sqrt{2} \sin(x \pm \pi/4)$. But I wasn’t sure I would get the sign right, so I just checked $k \pi/4$ for $0 \leq k \leq 7$ to figure out what the exact answer should be.

May 2, 2010 at 5:35 pm |

Answered 2D. But my thought process went like this:

1. Saw sin^2(x) and cos^2(x) first

2. Reduce 2 on RHS by 1=sin^2(x)+cos^2(x).

3. Saw middle term

4. Did sin(x)cos(x)=$\frac12$ so $\pm\frac{1}{sqrt{2}}$

5. Thought of pi/4

5. Suddenly formula for sin(2x) flashed in my head

6. Solved as 2B-2C-2D

May 2, 2010 at 11:32 pm |

Voted 2D.

I first spotted the Pythagorean identity and instantly got rid of the squared trig terms on the left and subtracted 1 from the right. I think my experience has trained me to always make that substitution when I see it.

Next I was tempted to divide both sides by 2 but I noticed the left hand was set up for another identity, yielding sin(2x) = 1. At that point I had to kind of imagine the unit circle and ask myself where sin is equal to 1, and then take half of that.

I’d say the whole process took 15-20 seconds.

May 3, 2010 at 3:43 am |

2A1A2D

Seeing I immediately looked for a . Since is the only trig identity I reliably remember off the top of my head, and I was working only in my head, after getting to I decided to start over. After recognizing the perfect square and factoring I decided I could come up with the right identity to handle faster than and went back to the original approach.

May 3, 2010 at 7:43 am |

I have voted 1A2D but this time it was not as obvious as last time:

* I am too tired to do this now *

I spot the square and factor

* Boring *

I spot sin^2 +cos^2 and cancel

* Interesting *

I make a mistake by thinking sin = 1/cos => no solution

I check the result by going back to 1 and realizing that the problem has a solution without actually computing it.

* embarrassment *

I go back to 2 and realize that I forgot the 2 in front of sin times cos.

* resignation *

I know some special values and “see” the solution

* frustration *

After reading the comments I realized that using a table rather than addition/multiplication theorems might actually qualify for a “None of the above” rather than 1A2D.

May 3, 2010 at 7:53 am |

2D.

I remember well a lot of the trigonometric identities, so right away I spotted the sin^2(x)+ cos^2(x) and the 2sin(x)cos(x) in the context of the identities and simplified to sin(2x)=1.

I never even saw the equation as the square of a sum until I read of approach A. I completely missed that. If the equation had been something like, say, tan^2(x) + 2tan(x)cos(x) + cos^2(x) = 2, devoid of familiar terms from the trig identities, then I surely would have gone the A route.

May 3, 2010 at 10:38 am |

Maybe this is totally wrong or somebody else posted something like this before, but it occurred to me that possibly one’s thought-processes are influenced by having to keep in mind how one solved the problem. I always tried to formulate how I had just proceeded while solving the problem. Perhaps when the problem at hand is quite easy, it just takes longer to solve it that way, but when it comes to more complex problems, reflecting upon what one has just done seems to have some impact on what road of attack one will choose next.

May 3, 2010 at 1:54 pm |

I voted 2D1D, but I feel it does not encapsulate my thought process with absolute accuracy. I looked at the equation and immediately recognized the left hand side as (sinx + cosx)^2. Almost simultaneously I substituted 1 in place of (sinx)^2 + (cosx)^2 in my head and went into approach 2. Out of curiosity, though, I redid the whole thing using approach 1 as well. So, as you can see, I couldn’t have voted 1A2D because I went only as much as half of step A in approach 1 the first time around. 2D1D came close to what I really did. An option like 1(0.5A)2D1D might help because I’m guessing that atleast a few others have gone through a thought process similar to mine!

May 3, 2010 at 7:57 pm |

On first look I saw the Sin[x]^2 + Cos[x]^2 = 1 identity, and realised that the centre term was an identity (I knew it was Sin[2x], or Cos[2x], or 1-Sin[2x] or something like that immediately. Recalling the exact one took only about a second.)

However, I then noticed it was a perfect square so I wrote it down as that on a bit of paper. I then realised that Sin[x] + Cos[x] = A is not too easy to solve, so I’ll go back to the other method which I knew would work.

So I decided 2x = Pi/2 + 2 Pi N and hence obtained the answer.

May 3, 2010 at 8:29 pm |

I did something entirely different than both options.

I noticed that the LHS is equal to (sin(x)+cos(x))^2, then saw that the RHS is 2 and thought “oh, that’s strange, isn’t the RHS too large? I think it might be a trick question, such that no x can give this”. So I thought what the obvious maximum of |sin(x)+cos(x)| is, and got 2. So it might be large enough. So I thought what x would maximize sin(x)+cos(x), and it made intuitive sense that would be x which makes cos(x) and sin(x) equal. Which x is this? Well, I know that sin(x)^2+cos(x)^2=1, so they are equal when they are both 1/sqrt(2). Then when you add them up you get exactly sqrt(2), and getting x from here is easy. This is not quite a proof but I was happy with it.

This is actually my usual way of doing math. Despite being an active theoretical computer scientist, I harbor great hate towards any kind of formula, so instead of solving them or manipulating them in sophisticated ways I try to understand what they say and then make sense of them by taking “toy examples” or using intuition to guess what’s actually going on.

May 3, 2010 at 9:38 pm |

2D.

I didn’t notice that the whole LHS factorises until after solving the equation (similarly, I also didn’t notice in the previous question that one could factorise). Otherwise – there was a moment where I thought to simplify

2sin(x)cos(x)=1 by dividing by 2, before noticing the trig identity.

July 19, 2010 at 5:52 pm

This was my process, taking less than a minute. I recently spent a week calculating light distributions on cones, using lots of trig substitutions, so these sorts of manipulations are fresh in my mind. I haven’t factored an equation for more than a year. Recent efforts definitely influenced my approach.

May 4, 2010 at 11:14 pm |

1A2D

What amused me is both in this example, and in the previous one, I very quickly started to draw pictures (in my head) to figure out the answer. I got as far as (sin x + cos x)^2 = 2 and then started to try and visualize this in terms of projections of points on the unit circle onto the x and y axes. However, no obvious picture emerged – I could see no obvious geometric interpretation of the left hand side – and so I gave up and went on to method 2.

I think this instinct towards visualization is partly out of a mistrust of my own algebraic skills, particularly in trig problems, where I might find it all to easy to forget about the negative square root, etc. It might also have something to do with my day job (physicist).

I’ve just read Sune Kristian Jakobsen’s elegant solution above – he saw the next step that I failed to see.

May 4, 2010 at 11:23 pm |

1A2A*D. Seeing the (x+y)^2 pattern and deciding that it is usually a good one to try, led to 1A. Only remembering the trigonometry basics at this point forced a retreat to a twist on approach 2. The geometric interpretation for sin and cos for angles smaller than pi/4 via a right-angled triangle with hypotenuse 1 gives 2A from first principles, which yields sin^2 x + cos^2 x = 2 sin x cos x when combined with the original equation. The quadratic form (sin x – cos x)^2 = 0 then leaps out. (Ah, so there was a quadratic form in there…) The triangle must therefore be isosceles and x must be pi/4. Inspection of the original gives x = pi/4 + k pi for some integer k. This approach is longer and more fragile than 2 but does not need identity 2B.

May 5, 2010 at 4:03 am |

I arrived at 1A, and happen to know that the sum of cos and sin is maximum when the the two are equal, i.e., at \pm\sqrt{2}/2 apiece, which meant \pi/4 + n\pi

May 5, 2010 at 11:09 am |

I did use a pencil and paper: I fetched them before looking at the question.

I did notice at first that as all the terms have degree 2, they can be rewritten as something like cos(2x) and sin(2x), but I didn’t get as far as actually rewriting it as 1 + sin(2x) because I noticed that it’s a perfect square. So I wrote (sin x + cos x)^2 = 2, so (sin x + cos x) = \pm sqrt(2) which I then solved visually by drawing a circle.

May 5, 2010 at 6:14 pm |

2D1B: (I just did it in my head.)

Approach 2: Spotting the identity was immediate, followed by cancellation of from both sides of the equation, and from there, followed, after which I used the fact that if , then , for all integers . In this case, . I can only attribute the last two statements to a lot of drilling I had on trigonometric identities during my high-school years.

Approach 1: Saw that allowed one to complete the square on the left hand side. And, then reached step B, after which I didn’t continue further partly due to laziness and partly because the next steps seemed similar to the ones done in Approach 2.

May 5, 2010 at 6:28 pm

I would just like to add that in Approach 1, when moving from step A to step B, I had in mind the fact that that I can say, for sure, was partly just a consequences of years of seeing it in so many places and partly a result of knowing that an expression of the form $E = latex a\sin(x) + b\cos(x)$ can be rewritten as for some suitable $\latex \beta$. The last expression is embedded in my head due to the fact that one one can use it to find the upper and lower bounds of , an expression which one encounters in high-school physics problems a lot of times.

May 6, 2010 at 6:15 am |

My experience was perhaps a bit odd (I haven’t read any of the other comments yet, so I suppose that this may not be true).

(1) The very first thing I noticed more or less instantly was the sin^2(x) +cos^2(x) term, and so I eliminated this by subtracting 1 from both sides of the equation, and I wrote this down on a piece of paper.

(2) I looked at the resulting 2sin(x)cos(x)=1, and did not at first spot that

the LHS of the new equation is equal to sin(2x).

(3) I rewrote the original equation as (sin(x) + cos(x))^2 = 2, and I didn’t

like this.

(4) I then remembered that 2sin(x)cos(x)= sin(2x), and followed the approach

2CD.

The whole process took about ten seconds or less.

May 6, 2010 at 6:19 am |

I forgot to mention that in between Steps 2 and 3 above, I rewrote the

equation 2sin(x)cox(x) = 1 as sin(x)cos(x) = 1/2, which of course did not help

matters.

May 6, 2010 at 4:06 pm |

These two small experiments inspired me to write some posts about how I solve problems. I have posted the a problem here, and in a couple of days I will explain how I solved it. I tried to leave a pingback here, but it didn’t work.

May 7, 2010 at 8:56 am

My second post is up now.

May 7, 2010 at 3:05 pm |

I did a variant of approach 2:

1. 2A: I spotted sin^2+cos^2=1 and got to sin(x)cos(x)=1/2.

2. I wrote sin*cos in terms of the exponential function and did the multiplication.

3. Removing the exponential terms, I rewrote the resulting equation as sin(2x)=1.

4. 2CD

May 8, 2010 at 8:00 am |

While the majority of the people seems to factorise the LHS, last time almost nobody did that, and I can´t help but wondering why. (Assuming people are truthful, of course). Especially since I myself did not see that the LHS was a square at all this time, while that was the first thing I noticed last time. Maybe it has something to do with the fact that I feel quite comfortable with ‘normal’ algebraic equations, I instantaneously get nervous when I see a sine or cosine somewhere. So with the first test I was like ‘Wiee, I’m good at this, let’s solve this baby’, and started eagerly at the LHS and so I was ‘doomed’ to factorise, because I didn’t even look at the RHS before noticing the LHS was a square. With this test, however, my initial thought was ‘Meh, sines and cosine, can’t do it’, which resulted in looking at the whole equation first, before even wanting to solve anything. And at that point I noticed one of the few things I actually do know about sines and cosines; sin^2 + cos^2 = 1.

May 8, 2010 at 9:18 pm |

I spotted the (A+B)^2 and sin(2x) patterns nearly simultaneously, but was already somewhat turned off from the first by the imminent appearance of the square root of 2. As a reflex, I thought of expressing everything in terms of trig functions without squares, e.g. writing cos^2(x) in terms of cos(2x). (I’ve been TAing some calculus so this kind of thing is on my mind right now.) Cosine is ideal for this kind of thing, so the first step was to write sine squared in terms of cosine, and then 2 revealed itself. Getting sin(2x) = 1 I knew the solution set would have period pi/2 and only one point in each pi/2-interval, but I didn’t bother to find the “offset”.

May 8, 2010 at 11:56 pm |

I was doing it all in my head and feeling pretty tired and hazy so was too lazy to finish it off, but I spotted the (sin x + cos x)^2 = 2 and that the solutions therefore occured at ‘critical points’ pi/4 + k.pi, I guess accidentally using Jensen’s inequality or something like that without realising it to make the problem feel ‘solved’ without actually needing to write anything down or do any ‘work’.

May 18, 2010 at 12:09 pm |

I looked at it and saw right away that sin^2(x) + cos^2(x) =1, and that 2sinxcosx = sin2x, and hence that sin2x=1, and hene that solutions occurred when x= kpi + pi/4

June 3, 2010 at 3:58 pm |

1. Wrote it down. While doing that, I noticed that sin(x)cos(x) simplifies to ‘something’, but I repressed my thought so I finish writing.

2. Now I looked at the whole thing and noticed that (sin(x))^2+(cos(x))^2 is 1 and I wrote down 2sin(x)cos(x)=1.

3. Now I tried to remember to what 2sin(x)cos(x) simplifies, so I started thinking about e^(2ix)=(cos(x)+isin(x))^2.

4. Then I took a break and got back to the initial equation and noticed that it is of the form (a+b)^2, so I wrote down sin(x)+cos(x)=sqrt(2).

5. Since nothing came to mind immediately, I turned back to e^(2ix), which I knew will mechanically give a solution.

6. I realized it’s the complex part I’m interested in so it’s sin(2x), rather than cos(2x). (Yes, I always have to think of e^(ix)e^(iy)… can’t remember otherwise.)

7. Thought of the trigonometric circle and write down pi/2+2k pi

8. Wrote down x=pi/4+k pi

I guess I write a lot for such simple problems.

June 9, 2010 at 10:41 pm |

2A1A2B2C2D

(Doing in my head, I decided that sinx + cosx is no fun and returned to method 2.)

June 11, 2010 at 5:31 pm |

Started with 1A and then checked whether the “standard” values of sin and cos solve the equation. They did, and I checked whether these are the only solutions by recollecting how the graphs of sin and cos look like.

June 28, 2010 at 4:32 am |

An alternative which I found just after the second approach is not actually too unobvious is to look at (sinx-cosx)^2 = 1-2sinxcosx = 0 on applying sin^2+cos^2=1 to the equation to be solved. Then trivially sinx=cosx and this obviously has the solution stated without having to know that 2sinxcosx=sin2x!

July 7, 2010 at 7:04 am |

I answered “none of the above”.

1. Factored LHS as (sin x + cos x)^2, concluded sin x + cos x = +/- sqrt2.

2. Let’s use intermediate value theorem to check whether there are solutions. There’s enough symmetry that I can just apply it to the interval [0, pi/2].

3. Whoops, sin x + cos x is 1 at both endpoints. But cos falls slower than sin rises, so it’ll be larger in the middle.

4. What’s the value at the midpoint? Great, it’s sqrt2! And it’s intuitively clear that sin x + cos x is maximized there…but how to prove it?

5. The level sets of x + y are lines of slope -1, and this is tangent to the unit circle at the upper-right-hand corner. The other solution thus happens at bottom left corner.

Some remarks:

A. I did not actually put the answer into numerical form (e.g. “pi/4″ or “45 degrees”) — I saw where it was on the circle, and putting it into numbers would have just invited mistakes (e.g. “is that point pi/4 or pi/2 or pi/8?”). In fact in step 2 I didn’t actually think about [0,pi/2], I just thought “the interval from the origin to where cos hits the x-axis”.

B. At step 4 I wasn’t thinking explicitly enough to rule out e.g. -pi/4 as a solution, but this became clear once I looked at the circle.

C. I would bet that if the question had been posed as:

“On the unit circle, solve x^2 + 2xy + y^2 = 2.”

almost everyone would have done it the way I did (factoring, then using tangent line to maximize x + y on the circle).

July 7, 2010 at 10:30 pm |

After looking at the equation, regrouped the left side visually to get (sin x + cos x)^2=2. I replaced my mental picture of the equation with sin x + cos x = +/- sqrt(2). Didn’t like the split into the cases. Looked at the equation again. Canceled sin^2 x and cos ^2 x with 1 on the right. Noticed Sin(2x) = 1. Pictured the unit circle, wrote 2x = pi/2 + 2k pi in my mental sketchpad. Divided by 2 to get x = pi/4 + k pi.

January 1, 2011 at 5:30 pm |

I first spotted the sin^2 + cos^2 = 1 identity and this lead me to sin(x)cos(x) = 1/2. I didn’t like this so I reformulated it to 2sin(x)cos(x) = 1 and then thought that I should have learned the addition theorems (is that the right notion?) for sinus and cosinus. Since I couldn’t remember them, I thought of approach 1 but the result looked too complicated so I returned to approach 2 and looked the additon theorems up (Yes, you might be thinking of this:

http://www.youtube.com/watch?v=ONvYPldXoZs#t=2m49s ).

July 6, 2011 at 12:36 pm |

I spotted both and quite simultaneously, without writing on paper. After that, the equation appeared in my mind. Then it was trivial to think that , so , for .

well, this immediate view of the problem is natural to me: I teach math at high school and, regarding trigonometry, I always tell my students to first think of the basic

trigonometricidenities that could be used in a problem. It seems interesting to me that many others (prof. Tao among them) started with approach 1. I tend to conclude that the moredifficultproblems one has solved, the more “global” or non-trivial ways one finds to deal with (even easy) problems…[a big THANKS for sharing your ideas, prof. Gowers, and for all the insight you give me to become a better teacher...]