Question

What conditions must be imposed on the real number, x, to make the following
equation true? sqrt (1−sin^2 x) = cos x
If a, b, and x are all real, it is incorrect to say that the equation ax = b has exactly
one solution. Explain.

Answer 1

Do you know what the restrictions on roots are?

Answer 2

Like the restrictions on the possible values? @e.mccormick

Answer 3

I'm not sure if I know what you mean @e.mccormick

Answer 4

Well, potential input to get real solutions. Does it apply here?

Answer 5

I have no idea. Honestly, this is on my study guide for my history of mathematics course and I don't know what its asking or how to solve it. @e.mccormick

Answer 6

Ah, OK. Well, I was trying to point out something this question is not talking about. For real solutions, you can't have a negative under the radical. So the "conditions" they are talking about are not the obvious ones of the fact that it would be a root.

Answer 7

They talk about "What conditions must be imposed" but \(\sin^2x\) can only output a maximum value of 1, so there is no complex part to deal with. So if it is not because of the root, what else might cause there to be some sort of restriction or condition?

Answer 8

Okay. I think I understand what you've said so far. So, we still don't know what the conditions may be? @e.mccormick

Answer 9

Sorry. You clearly understand this a lot more than I do. @e.mccormick

Answer 10

Well, lets look at where that came from, the Pythagorean Identity.\[cos^2+sin^2=1\]Right?

Answer 11

Yes @e.mccormick

Answer 12

Think about what the squares mean. What they do to what is input into the equation. Well, then we use algebra to move this around:
\[\cos^2 x+\sin^2 x=1\implies \\
\cos^2 x=1-\sin^2 x\implies \\
\\sqrt{cos^2 x}=\sqrt{1-\sin^2 x}\implies \\
\cos x=\sqrt{1-\sin^2 x}
\]We took the root of cos squared! What would have happened if \(\cos x\) was originally a negative value?

Answer 13

Okay. I understand. If cos x was originally negative, well that wouldn't be able to be computed since we can't do the square root of a negative value. @e.mccormick

Answer 14

Well, it had been squared! So the - would become positive. That means your negative values get masked. It is like absolutle value. \(|a|=\sqrt{a^2}\).

Answer 15

Oh. You mean what if we had: - cos^2 x and then we square rooted it? @e.mccormick

Answer 16

So lets say \(\cos x=-\frac{1}{2}\). That is a very possible answer. It happens at \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\). Well, when it was squared and rooted, it became \(\frac{1}{2}\). Well, those answers would be \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\)! Completely different points on the unit circle!

Answer 17

okay. makes sense. so no matter the value of cos x, positive or negative, once its squared and rooted, it will be the absolute value of its original value. @e.mccormick

Answer 18

I know I kind of took the long way around here, but I am trying to get you to think about what roots and squares do inside the equations. If it were only roots, you just need to worry about \(\sqrt{a}\, |\,a<0\). However, this is the combination of squares and roots, so it is a different issue even though they both involve a root.

Now you get that point, what does it mean about the domain? If the range is always a + value from the cosine, what domain is the only place that can give a valid answer? because that is a condition that would be required to make this true.

Answer 19

x = 0? @e.mccormick

Answer 20

Not, quite. Have you gone over the restrictions on arcsin, arccos, and so on?

Answer 21

No. Are you asking me what values of x (the domain) make \[\sqrt{1-\sin ^{2}x} = cox(x)\] true? @e.mccormick

Answer 22

yes. Because that is your question. "What conditions must be imposed on the real number, x, to make the following equation true? sqrt (1−sin^2 x) = cos x"

Everything up to this point has been the background related to that question.

Answer 23

Okay. So when I plugged in 0 for x, I got:

\[\sqrt{1-\sin ^{2}} (0) = \cos (0)\]
\[\sqrt{1-(\sin(0) \sin(0))} = 1\]
\[\sqrt{1-(0)(0)} = 1\]
1=1

@e.mccormick

Answer 24

Thats wrong? @e.mccormick

Answer 25

It is right. It is incomplete. It is not the only possibility.