Trigonometry, Radicals, and a Very Common Error

(A new question of the week)

While I was looking through recent questions to choose one to post, I ran across one that deals with an error we see very commonly – in fact, a student I had worked with that very afternoon in face-to-face tutoring had done the same sort of thing. The context here deals with trigonometric identities, but it could just as well occur in working with the Pythagorean Theorem in geometry, or solving an equation in algebra, or even in calculus. We’ll also see a number of other pitfalls for beginning algebra students.

Why is the sine squared?

This is the question, sent to us by Simran in mid-August:

Hello math doctors,

When we write cos A as sin A, that is,

cos A = 1/√1 – sin²A

why is it sin² A and not sin A? Shouldn’t it be sin A in the root as we have taken the square root?

Sorry, if the doubt is too dumb.

Thanking you,

Regards,

Simran

Our first task, besides reassuring Simran, is to determine which of several issues is the central doubt.

Doctor Rick answered with a variation on our common statement that “the only stupid question is the one you don’t ask”:

A doubt unexpressed is dumb (literally – “dumb” means silent). It is smart to express your doubts so that they can be cleared up.

Simran is doing just what needs to be done. I often discuss with tutees the fact that asking questions is the only way to truly learn. If you feel you can’t ask a question in class, ask a tutor, or write to a site like ours.

The word “dumb” originally meant “mute” (unable to speak), but came increasingly to mean “stupid” (having nothing to say), and its use to refer to mute people became offensive. But either way, if you ask questions, you are not dumb!

Clarifying the expression

What you wrote has several problems in terms of order of operations, and some other problems as well, so I can’t be exactly sure what you are thinking. You wrote:

When we write cos A as sin A, that is,

cos A = 1/√1 – sin²A

why is it sin² A  and not sin A? Shouldn’t it be sin A in the root as we have taken the square root?

I assume that you meant “when we write cos A in terms of sin A”.

This is a phrase many students need to be introduced to! In fact, again, just today I helped a student understand what she was being asked to do, when the question said to express one function in terms of another. It means to use the latter to state the former.

Now, what does “cos A = 1/√1 – sin²A” mean?

What you wrote on the second line is not correct. Here is how the correct equation is derived from the Pythagorean identity:

sin2 A + cos2 A = 1

cos2 A = 1 – sin2 A

cos A = ± √(1 – sin2 A)

The square root is not in the denominator as you put it. Also, notice the plus-or-minus sign. That is needed to make the equation true for all angles — for instance, if A = 120°, then sin A = √3/2 and cos A = 1/2.

The parentheses are also important! This is a common problem in typing radicals, because we can’t draw the bar (called a vinculum) over the radicand, to show what we are taking a root of. In this setting, we need to use parentheses in place of the untypable vinculum. As Simran had typed it, “1/√1 – sin2 A” would just mean \(\frac{1}{\sqrt 1}-\sin^2(A)\) rather than what is presumably intended, \(\frac{1}{\sqrt{1-\sin^2(A)}}\), or what it should be, \(\sqrt{1-\sin^2(A)}\). But perhaps he did mean what he wrote, as we’ll be seeing.

Focusing on the algebra

But these things aren’t what you’re asking about; you are focused only on whether sin A should be squared. Could you say more about why you think sin A should not be squared? What if we put trig aside, and just solve the equation

x2 + y2 = 1

for y in terms of x?

y2 = 1 – x2

y = ± √(1 – x2)

Perhaps you’re thinking that, to take the square root of 1 – x2, we can just take the square root of 1 and the square root of x2, getting

√(1 – x2) = √1 – √(x2) = 1 – x

If that’s what you’re thinking, we can talk it through thoroughly — but if it’s something else, I want to know, so I don’t waste your time. So let me know what you’re thinking, and we’ll talk about it.

The suggestion here is that Simran might be thinking that the square root “distributes over the addition”, so that \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\). As we’ll see, this is a common mistake.

Simran replied,

Hi Doctor Rick,

Actually I thought that when we will square it the square will go away. I don’t understand the square still remaining in the expression. Since cos A is written without square so shouldn’t 1 – sin A be written without square too? I don’t understand how after writing square root of a number we can write the square along with it. If possible can you explain me in simple terms.

I hope it’s understandable because I am myself confused how to express the doubt.

And thankyou I often think my doubts are too dumb to be answered.

Regards,

Simran

It appears that he did mean \(\cos(A) = 1 – \sin(A)\), and the distribution idea is probably part of the problem. We’re getting closer.

And he is saying “square it” probably because of the awkwardness of expressing a square root in English; I find many students say “square root it”, which is not a standard verb, rather than the proper but wordy “extract the square root” or “take the square root”. It’s also easy to omit the word “root”, as he has done here.

Doctor Rick wrote back, pointing out the wrong word usage (which, again, I find to be quite common even in native speakers of English):

I cannot understand what you are saying. “When we square it …” What did we square? Did you mean “when we take the square root“? If we’re going to straighten out your confusion, we have to write carefully. I know that’s hard when you’re confused!

Perhaps you are thinking that, for example, √(x2) = x. You may know that this isn’t quite correct – if x can be negative, then we should write √(x2) = |x|. But I don’t want to cause more confusion, so I’ll keep it simple: if x is positive, then squaring it and then taking the square root of the result gets you back to x.

Note, though, that when we do this, both the square-root sign and the exponent 2 go away. If we still need the radical (square-root sign) then we also still need the square.

That is, we can’t say that \(\sqrt{x^2}=\sqrt{x}\). But more important, in the expression we’re discussing, we aren’t taking the square root of the square itself at all:

And if we do something else to the square before we take the square root, then we can’t just “cancel” the two symbols. We have to first write the expression just as it is, and look to see if there is any valid property or fact that we can use to simplify it. If I have

y = √(1 – x2)

there is no property of exponents or radicals that I can use to simplify this!

I very often find that students need to consciously ask themselves whether there is more that can be done, forcing themselves to stop and think, rather than let their “simplifying momentum” carry them beyond what is legal.

A numerical example

When you don’t know whether a step is valid, one way to check is to try it with specific numbers.

Let’s plug in some numbers. If x is greater than 1, then 1 – x2 will be less than zero, and we can’t take the square root of a negative number. (Well, we can, but we’d need to talk about imaginary numbers then, and that’s irrelevant to the trig context.) So I’ll choose a number between 0 and 1 – let’s say x = 0.75. Then

y = √(1 – (0.75)2)

   = √(1 – 0.5625)

   = √0.4375

   = 0.6614

I think (though you did not confirm this) that you are thinking √(1 – x2) is the same as √1 – √(x2), which is the same as 1 – x. But if x = 0.75, then 1 – x = 0.25, not 0.6614.

So, $$\sqrt{1-0.75^2} = 0.6614$$ but $$1-0.75 = 0.25$$

They are not equal. So when you come to an expression like the former (but with a variable so you can’t just evaluate it), you must put on your brakes and stop. Don’t keep simplifying when the next step, though it feels natural, is illegal!

If the values had been equal, it would not prove the general statement to be true; but if it fails for one number, then it can’t be generally true!

Again, I have to ask: Is this what you’re thinking? If the problem lies here, then we can solve your confusion without bringing trigonometry into it. If not, then keep trying to explain your thinking as clearly as you can. For instance, don’t use words like “it” without saying exactly what “it” is.

Success, and a deeper look

Simran replied, confirming the interpretation of his issue:

Hi Doctor Rick,

Thank you!! Yes I was assuming we took the root of x² that’s why I was confused and by square I meant that 1 – x² is present in square form so when we take the root we should be writing as ✓1 – x  but now I get it.

Thanks a lot!!

Regards,

Simran

Doctor Rick could say more now:

Good! Since now I know what your misconception was, let me say a bit more about it.

It is not unusual for students to think that √(a2 + b2) = a + b. If they think about it, though, this would turn the Pythagorean theorem, c2 = a2 + b2 where a, b, and c are sides of a right triangle, into c = a + b. That’s not right! So it’s mostly when a student isn’t really thinking about it that they fall into this trap.

For example, taking the familiar 3-4-5 right triangle, the hypotenuse can be found as $$\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$$ but if we could distribute the root, we’d have $$\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{9}+\sqrt{16}=3+4=7$$ which is wrong.

If we compare problem solving to a race, it is easy to coast at the end, when the goal is just ahead; but we must keep thinking carefully to the end. Never stop thinking!

In the same way, if a student is going too quickly though a problem, he might write (a + b)2 as a2 + b2. It’s really the same error. If we expand the square properly, we get (a + b)2 = a2 + 2ab + b2. And if we take the square root of both sides of this equation, we find that a + b = √(a2 + 2ab + b2), not √(a2 + b2).

If it were true that \(\sqrt{a^2+b^2}=a+b\), then it would be true that \(a^2+b^2=(a+b)^2\); but that is missing the middle term.

Simran closed:

Hello doctor Rick,

Thank you, the example is really helpful and as I always solve questions in hurry I would be cautious of this.

Regards,

Simran

There’s another lesson learned! And this is what we’re here for.

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