Expand and Simplify

[sweiss]

Question is:

Expand and simplify, if possible. [tan^2(x) +1] * [1-cos^2(x)]

I seem to be arriving at two related but separate answers and am unsure as to which one is correct...

Here's what I have done so far with the problem:

[tan^2(x) + 1] = [sec^2(x)]
[1 - cos^2(x)] = [sin^2(x)]
This leads to an answer of: [sec^2(x)] * [sin^2(x)]

Then I figured I could simplify further:
[1/cos^2(x)] * [sin^2(x)] = [sin^2(x) * cos^2(x)] / [cos^2(x)] (placed over a common denominator)
Reduces = sin^2(x)

Just not sure which of these solutions (if either) are correct...

Thank you for any input,

S. Weiss

 

[tkhunny]

You're gold to here: [1/cos^2(x)] * [sin^2(x)]

After that, you seem to have missed that your two expressions are joined by multiplication.
You have treated them as if they were joined by addition.
Try that again.

 

[sweiss]

You're gold to here: [1/cos^2(x)] * [sin^2(x)]

After that, you seem to have missed that your two expressions are joined by multiplication.
You have treated them as if they were joined by addition.
Try that again.

Thanks for responding!
Once I have [1/cos^2(x)] * [sin^2(x)], I am placing over a common denominator, which gives me [sin^2(x) * cos^2(x)] / cos^2(x) + 1 / cos^2(x)
from that point I am kinda lost on how to reduce and complete...

 

[Dr.Peterson]

Thanks for responding!
Once I have [1/cos^2(x)] * [sin^2(x)], I am placing over a common denominator, which gives me [sin^2(x) * cos^2(x)] / cos^2(x) + 1 / cos^2(x)
from that point I am kinda lost on how to reduce and complete...

There is no addition here! You don't need a common denominator!

What you have is [MATH]\frac{1}{\cos^2(x)}\cdot \sin^2(x) = \frac{1}{\cos^2(x)}\cdot \frac{\sin^2(x)}{1}[/MATH]. How do you multiply fractions?

 

[JeffM]

Thanks for responding!
Once I have [1/cos^2(x)] * [sin^2(x)], I am placing over a common denominator, which gives me [sin^2(x) * cos^2(x)] / cos^2(x) + 1 / cos^2(x)
from that point I am kinda lost on how to reduce and complete...

The first problem that you are having is that, as tkhunny mentioned, you have no need to create a common denominator when multiplying.

[MATH]y = \dfrac{1}{cos^2(x)} * sin^2(x) = \dfrac{1}{cos^2(x)} * \dfrac{sin^2(x)}{1} = \\ \\ \dfrac{1}{cos^2(x)} * \dfrac{sin^2(x) * cos^2(x)}{cos^2(x)} = \dfrac{1 * sin^2(x) * \cancel {cos^2(x)}}{cos^2(x) * \cancel {cos^2(x)}} = \dfrac{sin^2(x)}{cos^2(x)}.[/MATH]
That common denominator process is not technically wrong; it simply adds unnecessary work. The answer that you got is exactly what you would have got had you simply multiplied.

In your original post, you somehow equated [MATH]\dfrac{sin^2(x)}{cos^2(x)} = sin^2(x)[/MATH]
which is simply a mistake.

So let's agree that [MATH]\dfrac{sin^2(x)}{cos^2(x)}[/MATH] is indeed a correct answer BUT

[MATH]\dfrac{sin^2(x)}{cos^2(x)} = sin^2(x) * \left ( \dfrac{1}{cos(x)} \right )^2 = sin^2(x) * sec^2(x).[/MATH]
And that second answer was what you got using your other method. So your second problem was failing to recognize when two trigonometric expressions ARE EQUIVALENT. You need to be alert to that possibility, or you will drive yourself nuts. That is in fact one reason why you learn trig identities: to help you figure out when two different expressions have the exact same meaning.

 

[sweiss]

There is no addition here! You don't need a common denominator!

What you have is [MATH]\frac{1}{\cos^2(x)}\cdot \sin^2(x) = \frac{1}{\cos^2(x)}\cdot \frac{\sin^2(x)}{1}[/MATH]. How do you multiply fractions?

Wow! Definitely was messing that one up...
When I multiply them together I get [sin^2(x)] / [cos^2(x)]

 

[sweiss]

The first problem that you are having is that, as tkhunny mentioned, you have no need to create a common denominator when multiplying.

[MATH]y = \dfrac{1}{cos^2(x)} * sin^2(x) = \dfrac{1}{cos^2(x)} * \dfrac{sin^2(x)}{1} = \\ \\ \dfrac{1}{cos^2(x)} * \dfrac{sin^2(x) * cos^2(x)}{cos^2(x)} = \dfrac{1 * sin^2(x) * \cancel {cos^2(x)}}{cos^2(x) * \cancel {cos^2(x)}} = \dfrac{sin^2(x)}{cos^2(x)}.[/MATH]
That common denominator process is not technically wrong; it simply adds unnecessary work. The answer that you got is exactly what you would have got had you simply multiplied.

In your original post, you somehow equated [MATH]\dfrac{sin^2(x)}{cos^2(x)} = sin^2(x)[/MATH]
which is simply a mistake.

So let's agree that [MATH]\dfrac{sin^2(x)}{cos^2(x)}[/MATH] is indeed a correct answer BUT

[MATH]\dfrac{sin^2(x)}{cos^2(x)} = sin^2(x) * \left ( \dfrac{1}{cos(x)} \right )^2 = sin^2(x) * sec^2(x).[/MATH]
And that second answer was what you got using your other method. So your second problem was failing to recognize when two trigonometric expressions ARE EQUIVALENT. You need to be alert to that possibility, or you will drive yourself nuts. That is in fact one reason why you learn trig identities: to help you figure out when two different expressions have the exact same meaning.

Thanks for all the helpful information! I don't know what I was thinking within that one reduction I made. So sin^2x / cos^2x is technically the answer I would be looking for in a problem like that?

 

[JeffM]

Thanks for all the helpful information! I don't know what I was thinking within that one reduction I made. So sin^2x / cos^2x is technically the answer I would be looking for in a problem like that?

You are 98% of the way there, but may be missing two subtle points.

First, both [MATH]\dfrac{sin^2(x)}{cos^2(x)}[/MATH] and [MATH]sin^2(x) * sec^2(x)[/MATH]
mean the same thing and are equally correct for all mathematical purposes. I dislike working with fractions so I personally prefer the second form that uses the secant function. Others dislike using any trigonometric functions but the sine and cosine (or sine, cosine, and tangent). For those people, they would prefer the fractional form. In terms of your mathematical understanding, you should recognize that both expressions are equally valid. In terms of getting a decent grade, you need to find out whether whoever is grading your test wants answers in terms of two trig functions, three trig functions, or six. So ASK.

Furthermore, you need to be aware that on a multiple choice test, you may come up with a correct answer that is not listed, but an equivalent answer is listed. You do not want to pick "None of the above" in that case. This problem is a great way to learn about the danger of equivalent answers on tests.

Second, in terms of simplification, what seems simplest to me is

[MATH]\dfrac{sin^2(x)}{cos^2(x)} = \left ( \dfrac{sin(x)}{cos(x)} \right )^2 = tan^2(x).[/MATH]
Unless whoever is grading is one of those who wants answers only in terms of sine and cosine, I think the expression using the tangent function is simplest. And that is a third correct solution to the original problem.

 

[sweiss]

You are 98% of the way there, but may be missing a subtle point.

Both [MATH]\dfrac{sin^2(x)}{cos^2(x)}[/MATH] and [MATH]sin^2(x) * sec^2(x)[/MATH]
mean the same thing and are equally correct for all mathematical purposes. I dislike working with fractions so I personally prefer the second form that uses the secant function. Others dislike using any trigonometric functions but the sine and cosine (or sine, cosine, and tangent). For those people, they would prefer the fractional form. In terms of your mathematical understanding, you should recognize that both expressions are equally valid. In terms of getting a decent grade, you need to find out whether whoever is grading your test wants answers in terms of two trig functions, three trig functions, or six. So ASK.

Furthermore, you need to be aware that on a multiple choice test, you may come up with a correct answer that is not listed, but an equivalent answer is listed. You do not want to pick "None of the above" in that case. This problem is a great way to learn about the danger of equivalent answers on tests.

That certainly makes sense. Definitely valuable to be able to recognize the equivalent trig functions. I think my course prefers answers that utilize sine and cosine, but I will look into that. Thanks so much for taking the time to assist!

 

[JeffM]

That certainly makes sense. Definitely valuable to be able to recognize the equivalent trig functions. I think my course prefers answers that utilize sine and cosine, but I will look into that. Thanks so much for taking the time to assist!

I edited the post that you quoted. I overlooked something important in my first go round.