Algebraic Manipulation Is Overrated

An intuition question.

Look at the function below. It may surprise you that it is a constant! For any value of x, the function g will have the same value. I’m wondering, now that you know this, if you can get a sense of why it would be a constant, without (a) using your graphing calculator, or (b) taking the derivative to show that it is 0 [that is what I did, and as a side note, I have to use this on a test or homework next year].


Can you find some geometric way to see that?

It took me somewhere between a half hour and an hour of playing around to get it. I can post my solution in a couple days, but right now I don’t have the energy to find a program to draw my solution [1]. But let me just tell you: it’s beautiful. You’ll be stunned when you first do it. Yeah, the calculus way tells you it is a constant, but seeing the “why” is still a mystery. The geometric way takes a bit, but whoa nellie, you won’t regret spending the time!

[1] Or maybe I should claim there is no room in the margin! (JK)

Update: I did finally write up my solution. I quickly did something I never have done before: do my work in powerpoint. It worked fine.

Update: Mr. K solved the problem in 3 minutes and found a way to show the geometric solution. Head over to his very excellent blog to see it in all it’s glory.

Update: Besides mine and Mr. Ks, a third and perhaps more elegant solution is up at 11011110.

Of the three, I think I like Mr. K’s visualization best, even though it might not be a proof in the formal sense.



  1. Mr. K’s solution is different from mine, in which (sin x + sin x+a)/2 and (cos x – cos x+a) are the height and base length (respectively) of a family of similar isosceles triangles with apex angle a:
    (in the diagram in that post, let AD be a horizontal line, and let B and C be points on the unit circle at angles x and x+a from horizontal).

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s