Changing Trigonometry Cofunction Identities Over the Equals Sign?

[jddoxtator]

Simple question,
If we have to change the cofunction Identity of a trigonometric function in an equation, do we have to do anything to the other side of the equation to keep it balanced or is it a direct swap of equal identities?

Example: Changing y*4 = sin (theta) to y*4 = cos(theta)

 

[Dr.Peterson]

Simple question,
If we have to change the cofunction Identity of a trigonometric function in an equation, do we have to do anything to the other side of the equation to keep it balanced or is it a direct swap of equal identities?

Example: Changing y*4 = sin (theta) to y*4 = cos(theta)

I see no identity at all here. I'm not sure you know the meaning of "identity"; perhaps you just mean "expression".

The cofunction identity is the fact that [imath]\sin(\theta)=\cos(\frac{\pi}{2}-\theta)[/imath] for any [imath]\theta[/imath].

What you have done is to replace [imath]\sin(\theta)[/imath] with an entirely different expression, [imath]\cos(\theta)[/imath]. You just can't do that, and there is nothing you could do to the other side to make the new equation equivalent to the other.

What you could do is to change [imath]y*4 = \sin (\theta)[/imath] to [imath]y*4 = \cos(\frac{\pi}{2}-\theta)[/imath].

But also, can you clarify what you mean by "y*4"? We'd normally mean multiplication, but then we'd write it as "4y".

 

[khansaheb]

equal identities

cos(Θ) and sin(Θ) are NOT equal - you cannot legally swap those except for special values of Θ (e.g. Θ = 45º, etc.).

You could however write:

sin(Θ) = cos(π/2 - Θ) ...... (Dr.P beat me to it)

 

[jddoxtator]

I see no identity at all here. I'm not sure you know the meaning of "identity"; perhaps you just mean "expression".

The cofunction identity is the fact that [imath]\sin(\theta)=\cos(\frac{\pi}{2}-\theta)[/imath] for any [imath]\theta[/imath].

What you have done is to replace [imath]\sin(\theta)[/imath] with an entirely different expression, [imath]\cos(\theta)[/imath]. You just can't do that, and there is nothing you could do to the other side to make the new equation equivalent to the other.

What you could do is to change [imath]y*4 = \sin (\theta)[/imath] to [imath]y*4 = \cos(\frac{\pi}{2}-\theta)[/imath].

But also, can you clarify what you mean by "y*4"? We'd normally mean multiplication, but then we'd write it as "4y".

I get that it is supposed to be an identity, my issue was the direction of change implied by the identity.

In my textbook it is written as sin[(Pi/2) - theta] = cos theta.

This lead me to believe that sin first had to have 90 degrees added to negative theta it in order to become cos, which in my mind meant we were changing the value of one side of the equation if we were keeping the sin identity.

I see now that, in reality, it is now cos. So when you do add the (90 - theta) to sin, it must be cos as it is the equal identity.

That does clear it up a bit. I was doing unnecessary mental gymnastics in my head.

Edit: Also here is a problem with this I have been working on in the attached picture. Did I do it correctly?

 

[blamocur]

Edit: Also here is a problem with this I have been working on in the attached picture. Did I do it correctly?

I didn't find any errors, but the expression can be made simpler than your result.

 

[jddoxtator]

I didn't find any errors, but the expression can be made simpler than your result.

Is it taking the root of both sides?
I just noticed that.

 

[Dr.Peterson]

I get that it is supposed to be an identity, my issue was the direction of change implied by the identity.

In my textbook it is written as sin[(Pi/2) - theta] = cos theta.

It can be written several different ways; the cosine is the sine of the complement, and the sine is the cosine of the complement.

Your issue was that you didn't actually apply any identity at all.

I see now that, in reality, it is now cos. So when you do add the (90 - theta) to sin, it must be cos as it is the equal identity.

What is "now cos"? It isn't clear what you are thinking. And you don't add to sine; you rewrite the entire expression. But let's see what you did in the actual problem, which will probably clarify what you are saying:


I would not use cofunctions at all! You have not "simplified in terms of cos(θ)", since the LHS is no longer I, and the argument of the cosine is not θ.

What they probably want you to do is to take your expression [imath]I_0-I_0\sin^2(\theta)[/imath] and factor out [imath]I_0[/imath], then apply a Pythagorean identity, which will involve [imath]\cos(\theta)[/imath].

 

[jddoxtator]

It can be written several different ways; the cosine is the sine of the complement, and the sine is the cosine of the complement.

Your issue was that you didn't actually apply any identity at all.


What is "now cos"? It isn't clear what you are thinking. And you don't add to sine; you rewrite the entire expression. But let's see what you did in the actual problem, which will probably clarify what you are saying:

View attachment 38263
I would not use cofunctions at all! You have not "simplified in terms of cos(θ)", since the LHS is no longer I, and the argument of the cosine is not θ.

What they probably want you to do is to take your expression [imath]I_0-I_0\sin^2(\theta)[/imath] and factor out [imath]I_0[/imath], then apply a Pythagorean identity, which will involve [imath]\cos(\theta)[/imath].

That would make sense for the RHS of the equation, yes.
But how would you then use the Pythagorean Identity when the LHS of the equation still in terms of " I " ?

I'm going to attach a photo of the whole question because I think everyone is missing some information here.
It is a lengthy one that probably needs context.

My answer for B was I = 1/4Io which mental math and what I remember of physics seems to make sense.

 

[Dr.Peterson]

But how would you then use the Pythagorean Identity when the LHS of the equation still in terms of " I " ?

An identity is a tool to rewrite an expression. It is applied only to one side of an equation (at a time), replacing the existing expression with a new, equivalent, one. You don't need to change the other side of an equation, because the identity doesn't change the value of the side it is applied to.

That will be clearer when you show what you do to apply this identity.

I'm going to attach a photo of the whole question because I think everyone is missing some information here.

I don't think it adds anything to the discussion so far. When we have finished part (a), we can see how you apply it to part (b) (though I don't think you need to use (a) to answer the question in (b)).

 

[jddoxtator]

An identity is a tool to rewrite an expression. It is applied only to one side of an equation (at a time), replacing the existing expression with a new, equivalent, one. You don't need to change the other side of an equation, because the identity doesn't change the value of the side it is applied to.

That will be clearer when you show what you do to apply this identity.


I don't think it adds anything to the discussion so far. When we have finished part (a), we can see how you apply it to part (b) (though I don't think you need to use (a) to answer the question in (b)).

The example of Pythagorean Identities this textbook gives are in the form of an equation.
Does that mean we have to solve the Pythagorean Identity equation outside of the original equation to substitute an equal identity?
A nested equation so to speak?
Kind of like a function of a function, f[f(x)], but only affecting one element of one side?

 

[Dr.Peterson]

The example of Pythagorean Identities this textbook gives are in the form of an equation.
Does that mean we have to solve the Pythagorean Identity equation outside of the original equation to substitute an equal identity?
A nested equation so to speak?
Kind of like a function of a function, f[f(x)], but only affecting one element of one side?

The identity says that [imath]\cos^2(\theta)+\sin^2(\theta)=1[/imath] for any value of [imath]\theta[/imath].

Surely your book has demonstrated how to use such an identity; look for a section with examples of simplifying trig expressions.

One way is to solve for, say, [imath]\sin^2(\theta)=1-\cos^2(\theta)[/imath], and then replace [imath]\sin^2(\theta)[/imath] in your expression with [imath]1-\cos^2(\theta)[/imath]. Try that.

Another way would be if you rearranged your equation so that it contained [imath]1-\sin^2(\theta)[/imath], and replace that with [imath]\cos^2(\theta)[/imath], knowing that [imath]1-\sin^2(\theta)=\cos^2(\theta)[/imath].

(You are still using terminology in incorrect ways. You don't "substitute an equal identity"; you substitute an expression with an equivalent expression. The identity is the equation that says these expressions are (always) equal.)

 

[jddoxtator]

I am just trying to mentally compartmentalize the rules with language in my head.
If I can describe it, I can know it.
If an expression is an equivalent expression, can we not say for the specific expression of identities, that we can substitute an identity with an equivalent identity? I suppose you can say identity itself means two equivalent expressions, but the word identity also helps compartmentalize the data to trigonometric functions in my mind.

I have made another attempt at the question and marked what I am unsure of on the paper.
The result is very different than what I had before.

 

[Dr.Peterson]

If an expression is an equivalent expression, can we not say for the specific expression of identities, that we can substitute an identity with an equivalent identity? I suppose you can say identity itself means two equivalent expressions, but the word identity also helps compartmentalize the data to trigonometric functions in my mind.

You should learn standard terminology in order to communicate with others.

I have made another attempt at the question and marked what I am unsure of on the paper.
The result is very different than what I had before.

​

You can go a little further in (a): distribute and combine like terms. You'll be amazed.

Then there will be less work in (b) (but the same result).

 

[jddoxtator]

You should learn standard terminology in order to communicate with others.

View attachment 38270​

You can go a little further in (a): distribute and combine like terms. You'll be amazed.

Then there will be less work in (b) (but the same result).

So, I was correct to root both sides of of the Pythagorean identity before substituting it back into the equation? That point I was really unsure of as we have taken a square and substituted it with it's root without changing the original equation.

I am really interested in knowing this for sure because leaving it as a square leads to the answer (3/4)Io, which is the same answer I got with the Cofunction Identity in the very first example.

I see the further simplification as well, it is I = cos(theta)Io

 

[Dr.Peterson]

So, I was correct to root both sides of of the Pythagorean identity before substituting it back into the equation? That point I was really unsure of as we have taken a square and substituted it with it's root without changing the original equation.

Oops. I didn't read the fine print closely enough, probably not thinking you could make such a major mistake.
You wrote

​

That's wrong; the square root of a difference is not the difference of the roots. The square root was unnecessary, and in doing it, you did it wrong.

Then, you replaced [imath]\sin^2(\theta)[/imath] with the expression you'd just found (wrongly) for [imath]\sin(\theta)[/imath]:

​

​

​

I just saw what I expected to see, which would have been this:

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

[imath]I=I_0-I_0(1-\cos^2(\theta))[/imath]​

I missed that it wasn't squared, so I didn't comment on it.

I guess I should doubt you more ...

 

[jddoxtator]

Oops. I didn't read the fine print closely enough, probably not thinking you could make such a major mistake.
You wrote

View attachment 38271​

That's wrong; the square root of a difference is not the difference of the roots. The square root was unnecessary, and in doing it, you did it wrong.

Then, you replaced [imath]\sin^2(\theta)[/imath] with the expression you'd just found (wrongly) for [imath]\sin(\theta)[/imath]:

​

View attachment 38272​

View attachment 38273​

I just saw what I expected to see, which would have been this:

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

[imath]I=I_0-I_0(1-\cos^2(\theta))[/imath]​

I missed that it wasn't squared, so I didn't comment on it.

I guess I should doubt you more ...

So, the square root should have had an irrational component then, as in, the root symbol encompassing the whole RHS of the equation and on top of that, it was not necessary as we needed a square to substitute back in. I clearly messed up my terms and binomials in the root, as a root must affect a whole term and a binomial is a whole term, but I didn't even need it anyways. Chalk it up to overthinking and being stressed, I just finished an intense workout before that attempt.

That is clear, thank you.

I realize you are busy, have many demands of your time, and do not have to do this.
So please, do not interpret my questions as arrogance. I will never presume to know better than someone who has been doing this their whole life.
I only ask to verify my own conclusions. Sometimes, a review of your own thoughts from an outside source are required to make progress.

I have redone the question with these corrections and it now gives me a result that mirrors the first attempt, in part B.
My only question is, why was it wrong to use the Cofunction identity in the first place? Is it just because it does not leave the equation in terms of "I"? The answer in part B was the same, the only problem seems to be the structure of the equation.

I am now going to do all the practice questions in this chapter to cement this knowledge and hopefully gain more insight.

 

[Cubist]

Oops. I didn't read the fine print closely enough, probably not thinking you could make such a major mistake.
...
I guess I should doubt you more ...

Are we trying to help people on the forum or trying to make them feel bad about a mistake?
But it seems from post#16 that OP doesn't feel bad

@jddoxtator when we take roots of both sides of an equation we must take the root of the whole LHS and whole RHS like this...
[math]\sin^2(\theta) = 1 - \cos^2(\theta)[/math][math]\sqrt{ \sin^2(\theta)} = \sqrt{1 - \cos^2(\theta)}[/math]We can't take roots like this:-
[math]\sqrt{ \sin^2(\theta)} \ne \sqrt{1} - \sqrt{\cos^2(\theta)}[/math]Think about it, sqrt(1) doesn't equal sqrt(2) - sqrt(1)

 

[Dr.Peterson]

I have redone the question with these corrections and it now gives me a result that mirrors the first attempt, in part B.

​

Good. You got to the simple form, and applied it correctly.

My only question is, why was it wrong to use the Cofunction identity in the first place? Is it just because it does not leave the equation in terms of "I"? The answer in part B was the same, the only problem seems to be the structure of the equation.

The main problem is that you didn't do what you were told to do:

• simplify the expression for I (which, by the say, is not called "in terms of I", but "solved for I", or, in some countries, "with I as the subject"),
• and, specifically, to express it in terms of [imath]\cos(\theta)[/imath].

The latter is why the cofunction identity was inappropriate.

Keep in mind that if you were doing work on your own, not following instructions, there are all sorts of other things you might do, possibly including the cofunction identity. But you need a reason for whatever you do; for example, the only reason I can see for using the cofunction identity would be if the complement of the given angle were more useful than the theta they defined for you.

And, as I mentioned, it wasn't necessary to simplify in any form to get an answer for (b). There, too, the only reason you had to use the answer from (a) is that they told you to! (Which they presumably did because they knew that was a nice form to use.)

 

[jddoxtator]

View attachment 38275​

Good. You got to the simple form, and applied it correctly.


The main problem is that you didn't do what you were told to do:

• simplify the expression for I (which, by the say, is not called "in terms of I", but "solved for I", or, in some countries, "with I as the subject"),
• and, specifically, to express it in terms of [imath]\cos(\theta)[/imath].

The latter is why the cofunction identity was inappropriate.

Keep in mind that if you were doing work on your own, not following instructions, there are all sorts of other things you might do, possibly including the cofunction identity. But you need a reason for whatever you do; for example, the only reason I can see for using the cofunction identity would be if the complement of the given angle were more useful than the theta they defined for you.

And, as I mentioned, it wasn't necessary to simplify in any form to get an answer for (b). There, too, the only reason you had to use the answer from (a) is that they told you to! (Which they presumably did because they knew that was a nice form to use.)

AH! You see my weakness! I am horrendous at following instructions and tend to find my own way.
It is a nasty habit that has got me in lots of trouble over the years.
But has also lead to discoveries that would otherwise be left unknown.
A blessing and a curse you might say.

It is helpful to know that the cofunction identity is really only useful for finding compliments. It took me a couple of trials to figure out the amount in the bracket was in radians and not an exact decimal value. I will use it only when absolutely needed then.

 

[mario99]

Because you are a hard working student, I will give you this as a bonus. You can memorize them for now, but later when you get better in trigonometry, you will be able to derive them quickly from the unit circle.

[imath]\displaystyle \sin(\theta) = \cos\left(\theta - \frac{\pi}{2}\right) = -\sin(\theta - \pi) = -\cos\left(\theta - \frac{3\pi}{2}\right) = \sin\left(\theta - 2\pi\right) = -\cos\left(\theta + \frac{\pi}{2}\right) = -\sin\left(\theta + \pi\right) = \cos\left(\theta + \frac{3\pi}{2}\right) = \sin\left(\theta + 2\pi\right)[/imath]

-------------

[imath]\displaystyle \cos(\theta) = -\sin\left(\theta - \frac{\pi}{2}\right) = -\cos\left(\theta - \pi\right) = \sin\left(\theta - \frac{3\pi}{2}\right) = \cos\left(\theta - 2\pi\right) = \sin\left(\theta + \frac{\pi}{2}\right) = -\cos\left(\theta + \pi\right) = -\sin\left(\theta + \frac{3\pi}{2}\right) = \cos\left(\theta + 2\pi\right)[/imath]

------------

Because the cosine function is even, we can do this: [imath]\displaystyle \cos(-\theta) = \cos(\theta)[/imath]. This means [imath]\displaystyle \cos\left(\theta - \frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2} - \theta\right)[/imath].


Because the sine function is odd, we can do this: [imath]\displaystyle \sin(-\theta) = -\sin(\theta)[/imath]. This means [imath]\displaystyle \sin\left(\theta - \frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2} - \theta\right)[/imath].