You are here:
OakRoadSystems → Home → Articles → Math →
Trig without Tears
 → 6/Squared Identities

Trig without Tears Part 6:

The “Squared” Identities

revised 13 Apr 2014
Copyright © 1997–2014 Stan Brown, Oak Road Systems

Summary: This chapter begins exploring trigonometric identities. Three of them involve only squares of functions. These are called Pythagorean identities because they’re just the good old Theorem of Pythagoras in new clothes. Learn the really basic one, namely sin²A + cos²A = 1, and the others are easy to derive from it in a single step.

Students seem to get bogged down in the huge number of trigonometric identities. As I said earlier, I think the problem is that students are expected to memorize all of them. But really you don’t have to, because they’re all just forms of a very few basic identities. The next couple of chapters will explore that idea.

For example, let’s start with the really basic identity:


(38) sin²A + cos²A = 1


That one’s easy to remember: it involves only the basic sine and cosine, and you can’t get the order wrong unless you try.

right triangle, hypotenuse=1, sin A and cos A as the sides opposite and adjacent to angle A But you don’t have to remember even that one, since it’s really just another form of the Pythagorean theorem. (You do remember that, I hope?) Just think about a right triangle with a hypotenuse of one unit, as shown at right.

First convince yourself that the figure is right, that the lengths of the two legs are sin A and cos A. (Check back in the section on lengths of sides, if you need to.) Now write down the Pythagorean theorem for this triangle. Voilà! You’ve got equation 38.

What’s nice is that you can get the other “squared” or Pythagorean identities from this one, and you don’t have to memorize any of them. Just start with equation 38 and divide through by either sin²A or cos²A.

For example, what about the riddle we started with, the relation between tan²A and sec²A? It’s easy to answer by a quick derivation—easier than memorizing, in my opinion.

If you want an identity involving tan²A, remember equation 3: tan A is defined to be sin A/cos A. Therefore, to create an identity involving tan²A you need sin²A/cos²A. So take equation 38 and divide through by cos²A:

sin²A + cos²A = 1

sin²A/cos²A + cos²A/cos²A = 1/cos²A

(sin A/cos A)² + 1 = (1/cos A)²

which leads immediately to the final form:


(39) tan²A + 1 = sec²A


You should be able to work out the third identity (involving cot²A and csc²A) easily enough. You can either start with equation 39 above and use the cofunction rules equation 6 and equation 7, or start with equation 38 and divide by something appropriate. Either way, check to make sure that you get


(40) cot²A + 1 = csc²A


angle theta at center of unit circle, with line segments for values of the six trig functions

graphic courtesy of TheMathPage

It may be easier for you to visualize these identities geometrically.

In the diagram at right, remember that we showed that ED = tan A and AE = sec A. AD is a radius of the unit circle and therefore AD = 1. Since angle D is a right angle, you can use the Pythagorean theorem:

(ED)² + (AD)² = (AE)²

tan²A + 1 = sec²A

You can use triangle AFG in a similar way to prove equation 40.

next:  7/Sum and Difference


this page:  http://oakroadsystems.com/twt/pythag.htm

Donate via PayPal

Was this page useful? Visit my other Math Articles.
Want to show your appreciation? Please donate a few bucks to the author.