What am I missing in this Trigonometric Identity Simplification?

[jddoxtator]

I am still trying to learn how to fully navigate Trigonometric Expression Simplifications.
It seems I always have a hard time when I get down to a combination of a trigonometric function and a negative 1 or positive 1 that is not a direct identity.
I have attached one that I am stuck on, I am sure there is a very specific rule that I am either unaware of or not implementing correctly.

 

[khansaheb]

I am still trying to learn how to fully navigate Trigonometric Expression Simplifications.
It seems I always have a hard time when I get down to a combination of a trigonometric function and a negative 1 or positive 1 that is not a direct identity.
I have attached one that I am stuck on, I am sure there is a very specific rule that I am either unaware of or not implementing correctly.

You have not included what were you supposed to do....

 

[mario99]

I am still trying to learn how to fully navigate Trigonometric Expression Simplifications.
It seems I always have a hard time when I get down to a combination of a trigonometric function and a negative 1 or positive 1 that is not a direct identity.
I have attached one that I am stuck on, I am sure there is a very specific rule that I am either unaware of or not implementing correctly.

I can see that you can simplify the last step further!

 

[jddoxtator]

You have not included what were you supposed to do....

The question only states to simplify the expression, no term were given, just the original equation.

 

[jddoxtator]

I can see that you can simplify the last step further!

And that is where I am stuck.
If the base is a binomial, I cannot work on either side of it independently, and sin^2 (theta) - 1 has no clear path to any identity I know of.

 

[mario99]

And that is where I am stuck.
If the base is a binomial, I cannot work on either side of it independently, and sin^2 (theta) - 1 has no clear path to any identity I know of.

You already know that:

[imath]\cos^2 \theta = 1 - \sin^2\theta[/imath]

What happen if you multiply both side by a negative one?

 

[Otis]

a trigonometric function and a negative 1 or positive 1 that is not a direct identity

Hi jd. Do you consider variations of identities that you've seen listed. For example, we could vary the double angle formula by adding 1 to each side, if we were in need of an identity for cos(2x)+1.

cos(2x) = 2 cos²(x) – 1

cos(2x) + 1 = 2 cos²(x)

You could do something similar, using a variation of another identity, to write your answer in terms of sine and cosine.

The question only states to simplify the expression

In that case, it's a matter of personal taste. You have simplified in terms of the sine function.

 

[jddoxtator]

Hi jd. Do you consider variations of identities that you've seen listed. For example, we could vary the double angle formula by adding 1 to each side, if we were in need of an identity for cos(2x)+1.

cos(2x) = 2 cos²(x) – 1

cos(2x) + 1 = 2 cos²(x)

You could do something similar, using a variation of another identity, to write your answer in terms of sine and cosine.


In that case, it's a matter of personal taste. You have simplified in terms of the sine function.

Alright, that is what I was thinking.
The only reason why I was avoiding splitting the terms over the equals sign was because all other examples factored completely to one term without doing so.

Earlier in this textbook, it did mention that an expression is only simplified when it is in terms of a single trigonometric function or written as a numerical value.

 

[Otis]

Another method is to rewrite everything in terms of sine and cosine at the beginning.

\(\displaystyle \frac{\frac{\cancel{\cos(\theta)}}{\sin(\theta)}\cdot\frac{1}{\cancel{\cos(\theta)}}}{\;\;\frac{\sin^2(\theta)}{\sin^2(\theta)}-\frac{1}{\sin^2(\theta)}\;\;}\)

\(\displaystyle \frac{\frac{1}{\sin(\theta)}}{\;\;\frac{\sin^2(\theta)-1}{\sin^2(\theta)}\;\;}\)

\(\displaystyle \frac{1}{\cancel{\sin(\theta)}}\cdot\frac{\cancel{\sin^2(\theta)}^{\;\;\sin(\theta)}}{\;\;\;sin^2(\theta)-1\;\;\;}\)

\(\displaystyle \frac{\sin(\theta)}{\;\;\sin^2(\theta)-1\;\;}\)
[imath]\;[/imath]

 

[Dr.Peterson]

Earlier in this textbook, it did mention that an expression is only simplified when it is in terms of a single trigonometric function or written as a numerical value.

In general, this is not true; something is simplified when it is simple enough for your purposes. (I say "simplicity is in the eye of the beholder".)

But if they have told you that, for these problems, you can expect to get down to a single trig function, then you do need to keep going.

Do you see how your denominator can be simplified using a Pythagorean identity, after a small rearrangement?

Then, you'll be able to combine numerator and denominator into a single function, and you'll be done.

Simplification using identities is an art you have to get used to through lots of experience.

 

[Otis]

in this textbook, it did mention that an expression is only simplified when it is in terms of a single trigonometric function

That's one opinion. In a different exercise, if I were simplifying to hopefully cancel some factor in a difference of squares, then I would be pleased to stop where you did.
[imath]\;[/imath]

 

[jddoxtator]

In general, this is not true; something is simplified when it is simple enough for your purposes. (I say "simplicity is in the eye of the beholder".)

But if they have told you that, for these problems, you can expect to get down to a single trig function, then you do need to keep going.

Do you see how your denominator can be simplified using a Pythagorean identity, after a small rearrangement?

Then, you'll be able to combine numerator and denominator into a single function, and you'll be done.

Simplification using identities is an art you have to get used to through lots of experience.

I think I have it.
I was unsure if the identity cos^2 (theta) + sin^2 (theta) = 1, could be used with a mixed polarity of theta, so I used the graphing calculator to verify.
Apparently it can, according to the calculator.
This should satisfy the condition of one term.

 

[Dr.Peterson]

I think I have it.
I was unsure if the identity cos^2 (theta) + sin^2 (theta) = 1, could be used with a mixed polarity of theta, so I used the graphing calculator to verify.
Apparently it can, according to the calculator.
This should satisfy the condition of one term.

No, go back to what you had originally, and continue from the correct result of \(\displaystyle \frac{\sin(\theta)}{\;\;\sin^2(\theta)-1\;\;}\).

You made a careless error here:

​

In addition to somehow having changed [imath]\theta[/imath] to [imath]-\theta[/imath] (or is it -e?), and changed the subtraction to an addition, you omitted parentheses when you replaced [imath]\csc^2(\theta)[/imath] with its equivalent.

There's a related error here:

​

Both involve not paying close attention to the order of operations.

Can you explain what you mean by "mixed polarity of theta"? There may be a misunderstanding there we can correct, so you won't need to wonder in the future. I think it's just that negative sign you have floating around that seems to come and go at random.

 

[jddoxtator]

No, go back to what you had originally, and continue from the correct result of \(\displaystyle \frac{\sin(\theta)}{\;\;\sin^2(\theta)-1\;\;}\).

You made a careless error here:

View attachment 38285​

In addition to somehow having changed [imath]\theta[/imath] to [imath]-\theta[/imath] (or is it -e?), and changed the subtraction to an addition, you omitted parentheses when you replaced [imath]\csc^2(\theta)[/imath] with its equivalent.

There's a related error here:

View attachment 38286​

Both involve not paying close attention to the order of operations.

Can you explain what you mean by "mixed polarity of theta"? There may be a misunderstanding there we can correct, so you won't need to wonder in the future. I think it's just that negative sign you have floating around that seems to come and go at random.

The polarity of the degree itself as in (theta) and (-theta).

The expression before in the denominator was 1 - csc^2 (theta).
I was figuring I could make it 1 + csc^2 (-theta), but I checked and it is indeed not the case.
I could still turn - csc^2(theta) into -(cot^2 (theta) + 1), but I will try again from the previous spot on the previous attempt.
But I am not 100% sure how to get an identity out of the denominator sin^2(theta) -1.
Wouldn't there have to be a cos^2 to complete the identity?
I tried, but just made a whole page of circular logic.

 

[Steven G]

Let o be theta.
csc^2(o) = 1/sin^2(o) = 1/[sin(o)sin(o)] = 1/[sin(-o)sin(-o)] = 1/sin^2(-o) = csc^2(-o). So yes, 1 - csc^2 (o) is NOT 1+csc^2(-o).

 

[Dr.Peterson]

The polarity of the degree itself as in (theta) and (-theta).

That would get in the way of applying the identity. But there never should have been a negative there.

The expression before in the denominator was 1 - csc^2 (theta).
I was figuring I could make it 1 + csc^2 (-theta) + 1, but I checked and it is indeed not the case.

Yes, [imath]1+\csc^2(-\theta)[/imath] is equal to [imath]1+\csc^2(\theta)[/imath], because squaring eliminates the negative, so this doesn't get rid of the subtraction. But you didn't need that; one of the identities can be written as [imath]\csc^2(\theta)-1=\cot^2(\theta)[/imath], so subtraction is exactly what you need. It looks like you need some practice with signs. (This is often a place where students with somewhat weak algebra skills are forced to improve them!)

I could still turn - csc^2(theta) into -(cot^2 (theta) + 1), but I will try again from the previous spot on the previous attempt.

Yes, that's a valid way to go.

But I am not 100% sure how to get an identity out of the denominator sin^2(theta) -1.
Wouldn't there have to be a cos^2 to complete the identity?
I tried, but just made a whole page of circular logic.

Again, work on those signs. You should be able to see that [imath]\sin^2(\theta) -1=-(1-\sin^2(\theta)) = -\cos^2(\theta)[/imath].

 

[jddoxtator]

That would get in the way of applying the identity. But there never should have been a negative there.

Yes, [imath]1+\csc^2(-\theta)[/imath] is equal to [imath]1+\csc^2(\theta)[/imath], because squaring eliminates the negative, so this doesn't get rid of the subtraction. But you didn't need that; one of the identities can be written as [imath]\csc^2(\theta)-1=\cot^2(\theta)[/imath], so subtraction is exactly what you need. It looks like you need some practice with signs. (This is often a place where students with somewhat weak algebra skills are forced to improve them!)

Yes, that's a valid way to go.

Again, work on those signs. You should be able to see that [imath]\sin^2(\theta) -1=-(1-\sin^2(\theta)) = -\cos^2(\theta)[/imath].

Oh! You can rearrange the base Pythagorean Identities?
I didn't know that, that explains a LOT of the issues I have been having.

 

[Dr.Peterson]

Oh! You can rearrange the base Pythagorean Identities?
I didn't know that, that explains a LOT of the issues I have been having.

It does. You get to use all your algebra skills (and maybe some new ones) when you apply identities.

 

[jddoxtator]

It does. You get to use all your algebra skills (and maybe some new ones) when you apply identities.

That changes everything. I thought they were static rules that you have to make the equation conform to.
Textbook never showed an example that did that, so I didn't even think of doing that.

 

[jddoxtator]

Doing these trigonometric simplifications, I have also discovered a weakness in my Algebra.
Turns out I have a hard time seeing AB + AC = A ( B + C ), especially when it is in trigonometric terms.
I can see A ( B + C ) = AB + AC, just fine, but the reverse order gives me trouble for some reason.
I speculate it is because the reverse order is not seen as frequently in Algebra until this point.
At any rate, more practice is needed. Improvements are clearly being made if I am seeing my blind spots.