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- Thread starter pbonnie
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- #2

Mark44

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## Homework Statement

(question attached)

## Homework Equations

## The Attempt at a Solution

Checking solution.. pretty sure I did this wrong.

(solution attached)

Yes, you did. I see several mistakes.

On the left side, tanθ - 1 = sinθ/cosθ - 1, but that's not equal to (sinθ - 1)/cosθ, as you show.

On the right side there are two mistakes that I see: you factored the denominator wrong in the next to the last line, and I have no idea how you ended with what you did in the last line.

- #3

Mark44

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- #5

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- #6

Mark44

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Oh wow okay, I think I got it. Thank you very much

No, you're not getting it at all. On the right side, you have this:

$$ \frac{sin^2(θ) - cos^2(θ)}{sin(θ)cos(θ) + cos^2(θ)} = \frac{sin^2(θ) - 1}{sin(θ)cos(θ)}$$

This is wrong. It appears that you are cancelling the cos

The only time you can cancel is when the same

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No matter what I do I can't get the answer. I don't understand.

- #8

Mark44

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Can you factor ab + b

You need to know how to make these factorizations so that you can simplify this:

$$ \frac{sin^2(θ) - cos^2(θ)}{sin(θ)cos(θ) + cos^2(θ)} $$

The way you did it is not valid.

- #9

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So the numerator would be (sinƟ+cosƟ)(sinƟ+cosƟ)

But I don't know how to factor ab + b^2

But I don't know how to factor ab + b^2

- #10

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Oh! ab + b^2 would equal b(a+b)?

So the denominator would be cosƟ(sinƟ + cosƟ)?

So the denominator would be cosƟ(sinƟ + cosƟ)?

- #11

Mark44

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It's because you do not understand how to simplify fractions and rational expressions.No matter what I do I can't get the answer.

In post #1 you did this:

$$ \frac{sin^2(θ) - (1 - sin^2(θ))}{sin(θ) cos(θ) (1 - sin^2(θ))} ~~\text{(1)}$$

$$ \frac{sin^2(θ) - 1}{sin(θ) cos(θ)} ~~\text{(2)}$$

$$ \frac{sin(θ) - 1 }{cos(θ)} ~~\text{(3)}$$

In (1) you have the wrong sign in the denominator.

In (2), you apparently cancelled (1 - sin

In (3), you apparently cancelled sin(θ), which is also invalid since sin(θ) is not a factor of the numerator.

In post #4 you did this:

$$ \frac{sin^2(θ) - cos^2(θ)}{sin(θ) cos(θ) + cos^2(θ)} ~~\text{(4)}$$

$$ \frac{sin^2(θ) - 1)}{sin(θ) cos(θ) } ~~\text{(5)}$$

$$ \frac{sin(θ) - 1}{cos(θ) } ~~\text{(6)}$$

In (4), you have the sign in the denominator correct this time.

In (5), you cancelled cos

In (6), you sin(θ), which is also not valid. sin(θ) is not a factor in the numerator.

- #12

Mark44

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If you multiply this back out do you get sinSo the numerator would be (sinƟ+cosƟ)(sinƟ+cosƟ)

Yes!Oh! ab + b^2 would equal b(a+b)?

So the denominator would be cosƟ(sinƟ + cosƟ)?

- #13

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So now I can actually cancel out (sinƟ + cosƟ) once on the top and once on the bottom, to get

sinƟ + cosƟ/cosƟ

From here I'm stuck again. I know the sinƟ/cosƟ would equal tan, but I'm not sure where the -1 comes from

- #14

Mark44

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Does (sinƟ+cosƟ)(sinƟ+cosƟ) multiply out to sinOkay so now I have (sinƟ+cosƟ)(sinƟ+cosƟ)/cosƟ(sinƟ + cosƟ)

So now I can actually cancel out (sinƟ + cosƟ) once on the top and once on the bottom, to get

sinƟ + cosƟ/cosƟ

From here I'm stuck again. I know the sinƟ/cosƟ would equal tan, but I'm not sure where the -1 comes from

- #15

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(sinƟ+cosƟ)(sinƟ-cosƟ)/cosƟ(sinƟ + cosƟ)

Then sinƟ-cosƟ/cosƟ

Now I'm stuck again. I'm doing different variations of substituting the quotient identity into the equation but I can't get the right order

- #16

Mark44

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So far, so good, but you need some more parentheses around the entire denominator, like so:Ah sorry I had the right answer in my head but I wrote it out wrong. Didn't even notice.

(sinƟ+cosƟ)(sinƟ-cosƟ)/cosƟ(sinƟ + cosƟ)

(sinƟ+cosƟ)(sinƟ-cosƟ)/

Here you need some parentheses around the entire numerator, like so:Then sinƟ-cosƟ/cosƟ

(A + B)/C is the same as A/C + B/C, and (A - B)/C is the same as A/C - B/C.Now I'm stuck again.

I'm doing different variations of substituting the quotient identity into the equation but I can't get the right order

You're having a lot of difficulties working with fraction-type expressions. It would be a good idea for you to take some time reviewing how to simplify fractions and rational expressions. If you don't still have the textbook you studied from, there are a number of online sites that are helpful, such as http://khanacademy.org.

- #17

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= tanƟ-1

Ahh thank you so much. I definitely need a refresher it's been a long time since I've taken a math course. Thank you for the link and for you patience and help!!!!

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