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Updated 20 Nov 2016 (What’s New?)

Trig without Tears Part 6:

The “Squared” Identities

Copyright © 1997–2024 by Stan Brown,

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 sin² A + cos² A = 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) sin² A + cos² A = 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 sin² A or cos² A.

For example, what about the riddle we started with, the relation between tan² A and sec² A? It’s easy to answer by a quick derivation—easier than memorizing, in my opinion.

If you want an identity involving tan² A, remember equation 3: tan A is defined to be sin A/cos A. Therefore, to create an identity involving tan² A you need sin² A/cos² A. So take equation 38 and divide through by cos² A:

sin² A + cos² A = 1

sin² A/cos² A + cos² A/cos² A = 1/cos² A

(sin A/cos A)² + 1 = (1/cos A

which leads immediately to the final form:

(39) tan² A + 1 = sec² A

You should be able to work out the third identity (involving cot² A and csc² A) 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) cot² A + 1 = csc² A

It may be easier for you to visualize these two identities geometrically. Start with the sin A, cos A, 1 right triangle above. Divide all three sides by cos A and you get the first triangle below; divide by sin A instead and you get the second one. You can then just read off the Pythagorean identities.

triangle with legs tan A and 1, and hypotenuse sec A triangle with legs cot A and 1, and hypotenuse csc A

From the first triangle, tan² A + 1 = sec² A; from the second triangle, cot² A + 1 = csc² A.

Practice Problems

To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.

Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.

You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.

1 If sin A = 3/4, find cos A.
2 tan B = −2√2. Find sec B.
3 tan C = √15. Find cos C.
4 tan D =√15, Find sin D.
5Prove: sin² x = tan² x / (tan² x + 1)
This assumes that x ≠ π/2 + kπ, for integer k—or 90° + 180k°, if you prefer—because the tangent is undefined for those angles.

What’s New

next:  7/Sum and Difference

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