I picked 1D in the poll, but I didn't follow through your "first approach." After noticing the square on the l.h.s. I realized that sin(x)+cos(x) actually has max/min equal to +/- sqrt(2), and these occur precisely at x = Pi/4 + multiples of Pi. (a quick side-thought to calculus just to confirm gave the derivative cos-sin which can only vanish at the mentioned spots, so there was no need for scratch paper).
So maybe I should have gone with "other," since I definitely didn't go with step 1B?
Brian - 14 years ago
Whenever I see sin^2 I always look for a cos^2 first to cancel. Thus, 1D was my approach (though I did notice it was a square somewhere in the back of my head, but never even proceeded in this direction). I did the problem in my head.
Gustav - 14 years ago
I voted as 1D even though I really only did 1A which told me that (sin x + cos x)^2 = 2. It therefore seemed natural to make a little sketch on a piece of paper in which I drew 4 right-angled triangles with sides 1,sin x,cos x so that they form a square with sides sin x + cos x with an empty square with sides 1 in the centre. Knowing that the area of the these four triangles around the outside should be 1, just like the square in the middle, then tells you that the angle x has to be pi/2. This is more obvious if you actually sketch this on a piece of paper. I guess the reason why my mind took the geometric route is that my mind is not able to remember trigonometric identities.
My thought process was very similar to Eric Astor's, so I too had to answer none of the above, and my sequence too was 2A1A2D.
Sarah - 14 years ago
I did 1A2A and then thought that both looked a bit fiddly. Then I thought - aha, if x=pi/4 then sin(x) and cos(x) are both (sqrt2)/2 - so that's one solution to sin(x) + cos(x) = +- root 2. And come to think of it, x=5pi/4 is also a solution. So in general x=pi(n+1/4) is a solution.
But have I missed some? I seem to recall that sin(x)+cos(x) is a wiggly line with period 2pi, so I could be missing half the solutions unless the maxima/minima fall at x=pi(n+1/4). Which (scribbles graphy things) they do :-)
Eric Astor - 14 years ago
Hm. I had to choose none of the above, because my brain got into a bit of a fight with itself...
Best description of mine is:
2A/2 (sin^2+cos^2, so 1+2sin(x)cos(x)=2... no, wait, maybe the perfect square would be more useful)
1A (Ah, so sin(x)+cos(x)=sqrt(2)... I'll correct for the odd root later... no, that seems harder. Back to the Pythagorean identity...)
2D (Okay, let's cancel the one... 2sin(x)cos(x)=1... Which double-angle identity is that again? Ah, sin(2x)=1... 2x=pi/2, x=pi/4, plus the repetitions... ah, x=pi/4+n*pi!)
I see something that looked quadratic, so I tried to complete the square. (Sheer force of habit. A fantastic lecturer I had said it's a good habit to pick up, so I did and he was right. Now it's pure instinct. It speeds me up more often than it slows me down.) Then... bonus! It was a perfect square ( sin(x) + cos(x) ). That turns out less useful to me: I have a pen on my desk in my room, but my scribbling paper is whole metres away. I ask myself: do I bite the bullet and transform the stuff inside the brackets? That feels like it would involve converting the cos(x) into sin(pi/2-x) and using an angle-sum formula. I remember that leads in circles, and don't bother. Step 1B didn't occur to me (I'm out of practice, was never all that good to start with, and still haven't bothered getting out of my chair to get the paper.)
So I have a decision to make: do I get up and grab the paper (high disutility, since I'm very lazy), or do I look for a lazier solution? I look at the equation again, this time switching of my "algebraic manipulations" brain and switching on my "trig manipulations" brain. Steps 2A–D drop out immediately. I congratulate myself for being so clever, and kick myself for being so stupid as not to spot this straight away.
I decide on whether or not to narrate the event, as I'm meeting a friend for coffee in just a few minutes. I decide on "yes", since you said explicitly that narrations would be interesting, and since I knew I'd forget all about this by the time I get back home from coffee.
Dan Grayson - 14 years ago
I got sin x + cos x = sqrt 2, as in 1A, ignored the negative square root, and then visualized a right triangle in the unit circle and wondered about the sum of the lengths of its sides. The first angle I tried was pi/4, and that worked. I thought it unlikely there were other solutions, so did not pursue them.
Hi Tim, love these polls :)
I picked 1D in the poll, but I didn't follow through your "first approach." After noticing the square on the l.h.s. I realized that sin(x)+cos(x) actually has max/min equal to +/- sqrt(2), and these occur precisely at x = Pi/4 + multiples of Pi. (a quick side-thought to calculus just to confirm gave the derivative cos-sin which can only vanish at the mentioned spots, so there was no need for scratch paper).
So maybe I should have gone with "other," since I definitely didn't go with step 1B?
Whenever I see sin^2 I always look for a cos^2 first to cancel. Thus, 1D was my approach (though I did notice it was a square somewhere in the back of my head, but never even proceeded in this direction). I did the problem in my head.
I voted as 1D even though I really only did 1A which told me that (sin x + cos x)^2 = 2. It therefore seemed natural to make a little sketch on a piece of paper in which I drew 4 right-angled triangles with sides 1,sin x,cos x so that they form a square with sides sin x + cos x with an empty square with sides 1 in the centre. Knowing that the area of the these four triangles around the outside should be 1, just like the square in the middle, then tells you that the angle x has to be pi/2. This is more obvious if you actually sketch this on a piece of paper. I guess the reason why my mind took the geometric route is that my mind is not able to remember trigonometric identities.
My thought process was very similar to Eric Astor's, so I too had to answer none of the above, and my sequence too was 2A1A2D.
I did 1A2A and then thought that both looked a bit fiddly. Then I thought - aha, if x=pi/4 then sin(x) and cos(x) are both (sqrt2)/2 - so that's one solution to sin(x) + cos(x) = +- root 2. And come to think of it, x=5pi/4 is also a solution. So in general x=pi(n+1/4) is a solution.
But have I missed some? I seem to recall that sin(x)+cos(x) is a wiggly line with period 2pi, so I could be missing half the solutions unless the maxima/minima fall at x=pi(n+1/4). Which (scribbles graphy things) they do :-)
Hm. I had to choose none of the above, because my brain got into a bit of a fight with itself...
Best description of mine is:
2A/2 (sin^2+cos^2, so 1+2sin(x)cos(x)=2... no, wait, maybe the perfect square would be more useful)
1A (Ah, so sin(x)+cos(x)=sqrt(2)... I'll correct for the odd root later... no, that seems harder. Back to the Pythagorean identity...)
2D (Okay, let's cancel the one... 2sin(x)cos(x)=1... Which double-angle identity is that again? Ah, sin(2x)=1... 2x=pi/2, x=pi/4, plus the repetitions... ah, x=pi/4+n*pi!)
1A2D.
I see something that looked quadratic, so I tried to complete the square. (Sheer force of habit. A fantastic lecturer I had said it's a good habit to pick up, so I did and he was right. Now it's pure instinct. It speeds me up more often than it slows me down.) Then... bonus! It was a perfect square ( sin(x) + cos(x) ). That turns out less useful to me: I have a pen on my desk in my room, but my scribbling paper is whole metres away. I ask myself: do I bite the bullet and transform the stuff inside the brackets? That feels like it would involve converting the cos(x) into sin(pi/2-x) and using an angle-sum formula. I remember that leads in circles, and don't bother. Step 1B didn't occur to me (I'm out of practice, was never all that good to start with, and still haven't bothered getting out of my chair to get the paper.)
So I have a decision to make: do I get up and grab the paper (high disutility, since I'm very lazy), or do I look for a lazier solution? I look at the equation again, this time switching of my "algebraic manipulations" brain and switching on my "trig manipulations" brain. Steps 2A–D drop out immediately. I congratulate myself for being so clever, and kick myself for being so stupid as not to spot this straight away.
I decide on whether or not to narrate the event, as I'm meeting a friend for coffee in just a few minutes. I decide on "yes", since you said explicitly that narrations would be interesting, and since I knew I'd forget all about this by the time I get back home from coffee.
I got sin x + cos x = sqrt 2, as in 1A, ignored the negative square root, and then visualized a right triangle in the unit circle and wondered about the sum of the lengths of its sides. The first angle I tried was pi/4, and that worked. I thought it unlikely there were other solutions, so did not pursue them.