Trig without Tears Part 6:

# The “Squared” Identities

revised 17 Aug 2015

Copyright © 1997–2015 Stan Brown, BrownMath.com

Trig without Tears Part 6:

revised 17 Aug 2015

Copyright © 1997–2015 Stan Brown, BrownMath.com

**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
*sin*²A + *cos*²A = 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) *sin*²A + *cos*²A = 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.

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 *sin*²A or *cos*²A.

For example, what about the riddle we started
with, the relation between
*tan*²A and *sec*²A? It’s easy to answer by a quick
derivation—easier than memorizing, in my opinion.

If you want an identity involving *tan*²A, remember
equation 3: *tan* A is defined to be *sin* A/*cos* A.
Therefore, to create an identity involving *tan*²A you need
*sin*²A/*cos*²A. So take equation 38 and divide through
by *cos*²A:

*sin*²A + *cos*²A = 1

*sin*²A/*cos*²A + *cos*²A/*cos*²A =
1/*cos*²A

(*sin* A/*cos* A)² + 1 = (1/*cos* A)²

which leads immediately to the final form:

(39) *tan*²A + 1 = *sec*²A

You should be able to work out the third identity (involving
*cot*²A and *csc*²A) 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) *cot*²A + 1 = *csc*²A

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

*tan*²A + 1 = *sec*²A

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

next: 7/Sum and Difference

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