Verifying Trig Identites: (cos(a)+cos(b))^2 + (sin(a)-sin(b))^2 = 2+2cos(a+b) and...

[rojareina]

Hello everyone, I recently posted something very similar to this. The trig problems from the previous post turned out to be a mistake on my teacher's part and so she gave us a new set of problems. As anyone can see, I'm still bad at trig. The instructions are as follows: verify the following trigonometric identities.

1) (cos(a)+cos(b))^2 + (sin(a)-sin(b))^2 = 2+2cos(a+b)

2) (tan(b))/(1-cos(b)) = (csc(b))(1+sec(b))

For both of these problems, I'm just confused on to how to start proving them. I don't know which side, whether the right or left, to manipulate (like which side would be easier to manipulate). Any help and explanation as to why you would go about the problems in that why will be very appreciated!

 

[Dr.Peterson]

Hello everyone, I recently posted something very similar to this. The trig problems from the previous post turned out to be a mistake on my teacher's part and so she gave us a new set of problems. As anyone can see, I'm still bad at trig. The instructions are as follows: verify the following trigonometric identities.

1) (cos(a)+cos(b))^2 + (sin(a)-sin(b))^2 = 2+2cos(a+b)

2) (tan(b))/(1-cos(b)) = (csc(b))(1+sec(b))

For both of these problems, I'm just confused on to how to start proving them. I don't know which side, whether the right or left, to manipulate (like which side would be easier to manipulate). Any help and explanation as to why you would go about the problems in that why will be very appreciated!

First, let's talk about how to verify an identity in general. I saw your previous post and the comments on it, and though I think the style you used is misleading and potentially dangerous, it can be valid if you think and explain yourself carefully. As the following pages explain, the usual technique we recommend is to start with one side and transform it step by step to the other, WITHOUT repeatedly copying the other side, or saying "= RHS" or whatever. What you did (and your teacher evidently does) is to transform the entire equation step by step until it becomes obviously true. The danger in this is that it looks as if you were solving the equation; but it is not proper to start with an equation that you don't know to be true. What you are really doing is showing that the original is equivalent to something known to be true. Also, some people read what you wrote as if you were making a series of statements about the OTHER side, which is not what you were doing. Here are a few explanations of different styles of proof, and how they are related:

http://mathforum.org/library/drmath/view/54112.html

http://mathforum.org/library/drmath/view/60762.html

http://mathforum.org/library/drmath/view/69423.html

Now, the general idea is to start on the more complicated side of the identity, and just do whatever you can do, to see whether it helps! With experience, I can often see what will work ahead of time, but at first you need to gain that experience by just trying lots of things. Even with experience, it will never be a rote process; you might even discover a better method than mine.

Typically, you want to make one side look more like the other. If you have a sum of fractions, and the other side is a single fraction, use a common denominator to combine them. If you have tangents and secants and the other side has something else, rewrite everything in terms of sine and cosine.

In you first example, you have squares of binomials on the left, so I would expand them ("FOIL"). Then I'd hope that some familiar identities can be used. (They can!) At the same time, the right side involves a function of a sum of variables; you might want to use an identity to rewrite that in terms of a and b only, so that you will recognize when you are almost finished on the left side. (This technique is discussed in one of the pages I gave above: we are starting at both ends and meeting in the middle.)

Your second problem is one where I would start by writing everything in terms of sine and cosine, and just simplify each side. Neither side is more complicated, so it makes sense to start on both sides.

So, try something, and try something else, until you find something that seems to be going in the right direction. Then write back and show what you've done, so we can either correct some detail, or suggest another way to go, or just help you on to the next step.

 

[rojareina]

First, let's talk about how to verify an identity in general. I saw your previous post and the comments on it, and though I think the style you used is misleading and potentially dangerous, it can be valid if you think and explain yourself carefully. As the following pages explain, the usual technique we recommend is to start with one side and transform it step by step to the other, WITHOUT repeatedly copying the other side, or saying "= RHS" or whatever. What you did (and your teacher evidently does) is to transform the entire equation step by step until it becomes obviously true. The danger in this is that it looks as if you were solving the equation; but it is not proper to start with an equation that you don't know to be true. What you are really doing is showing that the original is equivalent to something known to be true. Also, some people read what you wrote as if you were making a series of statements about the OTHER side, which is not what you were doing. Here are a few explanations of different styles of proof, and how they are related:

http://mathforum.org/library/drmath/view/54112.html

http://mathforum.org/library/drmath/view/60762.html

http://mathforum.org/library/drmath/view/69423.html

Now, the general idea is to start on the more complicated side of the identity, and just do whatever you can do, to see whether it helps! With experience, I can often see what will work ahead of time, but at first you need to gain that experience by just trying lots of things. Even with experience, it will never be a rote process; you might even discover a better method than mine.

Typically, you want to make one side look more like the other. If you have a sum of fractions, and the other side is a single fraction, use a common denominator to combine them. If you have tangents and secants and the other side has something else, rewrite everything in terms of sine and cosine.

In you first example, you have squares of binomials on the left, so I would expand them ("FOIL"). Then I'd hope that some familiar identities can be used. (They can!) At the same time, the right side involves a function of a sum of variables; you might want to use an identity to rewrite that in terms of a and b only, so that you will recognize when you are almost finished on the left side. (This technique is discussed in one of the pages I gave above: we are starting at both ends and meeting in the middle.)

Your second problem is one where I would start by writing everything in terms of sine and cosine, and just simplify each side. Neither side is more complicated, so it makes sense to start on both sides.

So, try something, and try something else, until you find something that seems to be going in the right direction. Then write back and show what you've done, so we can either correct some detail, or suggest another way to go, or just help you on to the next step.

Thank you so much for your previous advice, Dr. Peterson. I think managed to solve the first question, but I would like to be sure. I focused on working on the left side in order to make it equivalent to the right side.

1) (cos(a)+cos(b))^2 + (sin(a)-sin(b))^2 = 2+2cos(a+b)
(cos(a)+cos(b))(cos(a)+cos(b)) + (sin(a)-sin(b))(sin(a)-sin(b))
cos^2(a)+2cos(a)cos(b)+cos^2(b) + sin^2(a)-2sin(a)sin(b)+sin^2(b)
By rearranging the order as follows:
cos^2(a)+sin^2(a)+cos^2(b)+sin^2(b)+2cos(a)cos(b)-2sin(a)sin(b)
You can simplify to:
1+1+2cos(a)cos(b)-2sin(a)sin(b)
2+2(cos(a)cos(b)-sin(a)sin(b))
2+2cos(A+B)
You can do this because of the formula (cos(a)cos(b)-sin(a)sin(b)) which equals cos(a+b).
Thus, you will end up proving the trig identity true.
2+2cos(a+b) = 2+2cos(a+b)

As for the #2, I got a bit stuck. I read through the other pages you provided for a better grasp of trig identities and the method of meeting in the middle confused me even more. I think I would like to focus on changing just one side, that seems to work best for me. But anyways, for the problem I focused on manipulating the left side. This is how far I got:

2) (tan(b))/(1-cos(b)) = (csc(b))(1+sec(b))
And I know you said I should have reduced the tan to sin and cos, but after I tried that I was just lost on what to do next, so I instead multiples by the conjugate of the denominator.
(tan(b))/(1-cos(b))*(1+cos(b))/(1+cos(b))
(tan(b)(1+cos(b)))/(1-cos^2(b))
(tan(b)(1+cos(b)))/sin^2(b)
And that's how fat I got. I know I could change tan(b) to (sin(b))/(cos(b)), but I feel like that would complicate the problem even more.

 

[stapel]

...the method of meeting in the middle confused me even more.

All it means is that you start on one side, work your way down for a while, give up, and then start on the other side. You work your way down for a while, and then notice, hey!, the ends of the two sides match at some point. So then you pull out a clean sheet of paper, and start over. You pick one of the two sides, copy down your steps from that side until you get to that line that matched. Then you hop over to the other side of your scratch-work, but now you're copying up those steps, until you get to the original form of the other side of the equation. They've got an example here.

 

[Dr.Peterson]

1) (cos(a)+cos(b))^2 + (sin(a)-sin(b))^2 = 2+2cos(a+b)
(cos(a)+cos(b))(cos(a)+cos(b)) + (sin(a)-sin(b))(sin(a)-sin(b))
cos^2(a)+2cos(a)cos(b)+cos^2(b) + sin^2(a)-2sin(a)sin(b)+sin^2(b)
By rearranging the order as follows:
cos^2(a)+sin^2(a)+cos^2(b)+sin^2(b)+2cos(a)cos(b)-2sin(a)sin(b)
You can simplify to:
1+1+2cos(a)cos(b)-2sin(a)sin(b)
2+2(cos(a)cos(b)-sin(a)sin(b))
2+2cos(A+B)
You can do this because of the formula (cos(a)cos(b)-sin(a)sin(b)) which equals cos(a+b).
Thus, you will end up proving the trig identity true.
2+2cos(a+b) = 2+2cos(a+b)

Excellent. In this case, the "meeting in the middle" means that, if you didn't recognize cos(a)cos(b)-sin(a)sin(b) in your work, you could have moved over to the right side and rewritten 2+2cos(a+b) as 2+2(cos(a)cos(b)-sin(a)sin(b)), so that you would recognize it when you saw it.

As for the #2, I got a bit stuck. I read through the other pages you provided for a better grasp of trig identities and the method of meeting in the middle confused me even more. I think I would like to focus on changing just one side, that seems to work best for me. But anyways, for the problem I focused on manipulating the left side. This is how far I got:

2) (tan(b))/(1-cos(b)) = (csc(b))(1+sec(b))
And I know you said I should have reduced the tan to sin and cos, but after I tried that I was just lost on what to do next, so I instead multiples by the conjugate of the denominator.
(tan(b))/(1-cos(b))*(1+cos(b))/(1+cos(b))
(tan(b)(1+cos(b)))/(1-cos^2(b))
(tan(b)(1+cos(b)))/sin^2(b)
And that's how far I got. I know I could change tan(b) to (sin(b))/(cos(b)), but I feel like that would complicate the problem even more.

I didn't say "you should", I said "I would"! There are many ways to do these, none really better than the rest. Your approach is another I would be very likely to try.

But since you're stuck, it's time to try other things! I like to compare the process to finding your way through unfamiliar territory. You try going one direction that seems likely, but if it becomes impassable, you don't sit down and stop; you try another way, perhaps first climbing a tree to get more information.

The reason I suggested using sin and cos is that you have many different trig functions here, and that makes it very hard to choose the right direction. From where you stopped, you have to get rid of tan and try to make csc and sec.

Also, because each side is rather complex, I would, as suggested, try just simplifying each side, hoping that the simple forms I get will be the same. If necessary, I would look from one side to the other to get ideas; for example, if one side is a single fraction and the other a sum, I would take the hint and make both sides into single fractions so they are more alike.

So, try something new and show me what you do. I strongly recommend you take my suggestion. It will work!

 

[JeffM]

As Dr Peterson said, you ought not write down an UNPROVEN identity in the form of an equation because, UNTIL YOU HAVE PROVED IT, you do not have any basis whatsoever to claim an equality. In terms of presentation, pick the expression on one side or the expression on the other side, and transform that expression into the other step by step.

In terms of figuring out whether the identity is true rather than presenting the proof that it is true, you can try transforming both expressions until you find the same expression.

So here is how I would PRESENT the PROOF of your first identity:

\(\displaystyle ((cos(a) + cos(b))^2 + (sin(a) - sin(b))^2 =\)

\(\displaystyle cos^2(a) + 2cos(a)cos(b) + cos^2(b) + sin^2(a) - 2sin(a)sin(b) + sin^2(b) =\)

\(\displaystyle (cos^2(a) + sin^2(a)) + (cos^2(b) + sin^2(b)) + 2(cos(a)cos(b) - sin(a)sin(b)) =\)

\(\displaystyle 1 + 1 + 2(cos(a)cos(b) - sin(a)sin(b)) =\)

\(\displaystyle 2+ 2cos(a + b).\)

\(\displaystyle \therefore (cos(a) + cos(b))^2 + (sin(a) - sin(b))^2 \equiv 2 + 2cos(a + b).\)

EDIT: At the site shown below, the referenced pdf is an excellent summary of what you need to memorize in trigonometry.

http://mathhelpforum.com/trigonometry/185935-trigonometry-memorize-trigonometry-derive.html

 

[rojareina]

Excellent. In this case, the "meeting in the middle" means that, if you didn't recognize cos(a)cos(b)-sin(a)sin(b) in your work, you could have moved over to the right side and rewritten 2+2cos(a+b) as 2+2(cos(a)cos(b)-sin(a)sin(b)), so that you would recognize it when you saw it.



I didn't say "you should", I said "I would"! There are many ways to do these, none really better than the rest. Your approach is another I would be very likely to try.

But since you're stuck, it's time to try other things! I like to compare the process to finding your way through unfamiliar territory. You try going one direction that seems likely, but if it becomes impassable, you don't sit down and stop; you try another way, perhaps first climbing a tree to get more information.

The reason I suggested using sin and cos is that you have many different trig functions here, and that makes it very hard to choose the right direction. From where you stopped, you have to get rid of tan and try to make csc and sec.

Also, because each side is rather complex, I would, as suggested, try just simplifying each side, hoping that the simple forms I get will be the same. If necessary, I would look from one side to the other to get ideas; for example, if one side is a single fraction and the other a sum, I would take the hint and make both sides into single fractions so they are more alike.

So, try something new and show me what you do. I strongly recommend you take my suggestion. It will work!

Thank you for responding again Dr. Peterson, I appreciateyour advice very much! And thank you to everyone else who has offered their ownexplanations as well.
Anyways, back to #2. I decided to start the problem over(still manipulating the left side) and I think I managed to prove the trig identitiestrue this time around. Here is what I did:
(tanB)/(1-cosB) = cscB(1+secB)
Using the conjugate, I did the following:
(tanB)/(1-cosB)*(1+cosB)/(1+cosB)
((tanB)(1+cosB))/((1-cosB)(1+cosB)
((tanB)(1+cosB))/(1-cos^2B)
((sinB)(1+cosB))/((cosB)(1-cos^2))
((sinB)(1+cosB))/((cosB)(sin^2B))
((sinB)/(sin^2B))*((1+cosB)/(cosB))
(1/sinB)*(1/cosB)+(cosB/cosB)
cscB(secB+1)
Which can be rearranged as:
cscB(1+secB)
So the trig identity is proved true:

cscB(1+secB) = cscB(1+secB)

 

[Dr.Peterson]

Thank you for responding again Dr. Peterson, I appreciate your advice very much! And thank you to everyone else who has offered their own explanations as well.
Anyways, back to #2. I decided to start the problem over (still manipulating the left side) and I think I managed to prove the trig identities true this time around. Here is what I did:
(tanB)/(1-cosB) = cscB(1+secB)
Using the conjugate, I did the following:
(tanB)/(1-cosB)*(1+cosB)/(1+cosB)
((tanB)(1+cosB))/((1-cosB)(1+cosB)
((tanB)(1+cosB))/(1-cos^2B)
((sinB)(1+cosB))/((cosB)(1-cos^2(B))) <-- I added the argument so this makes sense.
((sinB)(1+cosB))/((cosB)(sin^2B))
((sinB)/(sin^2B))*((1+cosB)/(cosB))
(1/sinB)*[(1/cosB)+(cosB/cosB)] <-- I added the brackets, which are needed.
cscB(secB+1)
Which can be rearranged as:
cscB(1+secB)
So the trig identity is proved true:

cscB(1+secB) = cscB(1+secB)

Good work!

Notice that although you held off, you did eventually write the tan in terms of sin and cos as I recommended! But you didn't have to start with the right side as I suggested you might; you just saw where you could pick up the csc, and then discovered a sec waiting for you. Knowing that they were needed, you took the opportunity. And that's just how this sort of thing works.

My own work was essentially the same as yours, just in a somewhat different order.