Today in my Pre-Calculus class we were using the Pythagorean Identities

I had an easy way of explaining why the first one was true, a method for explaining the second one with a little more effort, but no good way of getting at the third without an unintuitive tweak. Now I may have found a way at expressing the third identity, using a circle type I’ve never seen in a book before.

(EDIT: As Druin and Maria astutely point out in the comments, with algebra we can divide the first equation by and to get the second and third equations respectively. I’m meaning a direct demonstration using the Pythagorean Theorem.)

Gory details after the jump.

The way the unit circle is normally written in trigonometry is like so

so that, rather conveniently, and .

Also, since , substitution shows that .

Now, with some fiddling the traditional unit circle can be used to represent all the trigonometry functions, but there’s an alternate version of the unit circle that works better in demonstrating the tangent and secant.

Notice that the radius of the circle is still fixed at 1, but the triangle used to determine is drawn slightly differently, so that the line segment at 0 degrees is fixed at 1.

Now, and .

Using the Pythagorean theorem again, and by substitution .

What I used to do to explain was to modify my second unit circle slightly

and note that and . The Pythagorean Theorem gets again which now can be substituted with .

Unfortunately, that’s not consistent with how the other two unit circles work. What would be nice to have is for the original position to have a 1 in the denominator for the cotangent and cosecant.

The only way to do this is to drop having a unit circle concept, and just use circles of changing size. The site opposite of is fixed at 1, so when the angle grows (to 90 degrees), the circle shrinks:

and when the angle shrinks (from 90 to 0 degrees), the circle gets larger

but now without having to pick a different position for the angle.

This construction also makes it apparent why the cotangent and cosecant have asymptotes at 0 and 180 degrees. There’s no way to construct the triangle so the opposite side has a length of 1, it has to be 0.

I have no idea if this is new — it may be all over the place and I just haven’t noticed it — but I wanted to share.

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Filed under: Education, Mathematics

Druin, on March 27, 2008 at 9:18 pm said:I don’t know if this will help, but pre-calc kids should be able to realize that dividing through the equation by the same value produces a true equation.

If you start with sin^2 x + cos^2 x = 1 and divide through by sin^2 x, then you get 1 + cot^2 x = csc^2 x. If, instead, you divide through by cos^2 x, you get the other identity.

This was the way my trig teacher taught it *way* back when and I always liked it because I only had to know 1 of the equations and I could find the others. I hope it helps!

Maria H. Andersen, on March 28, 2008 at 9:12 am said:Actually, I was going to suggest the same as Druin. I find that only having to remember the first, and then having a vague recollection that the other two exist and knowledge of how to get them is great for my students.

Maria

Jason Dyer, on March 28, 2008 at 11:00 am said:I suppose I should have clarified with “demonstrate directly” be my meaning of “explain”, although the algebra is simple enough that it’d be fine to just show that.

jd2718, on April 22, 2008 at 5:40 am said:Jason, sorry I missed this one….

You have to use your new way….

Tan is … a tangent….

Sec is … an actual reallife secant.

This is hot stuff.

Jonathan

Jason Dyer, on April 23, 2008 at 12:24 pm said:I admit being curious about why “tangent” and “secant” had those names was what led me to the second unit circle in the first place. According to a schooled-in-the-old-ways teacher at our school, it used to be the usual way of drawing a unit circle.

Also, you get a beautiful (and much more intuitive) animation of the formation of the tangent graph.

I’m sort of a math etymology obsessive. I need to post up about “hypotenuse” sometime.

Thanks for the feedback!