Half variable identities

[HATLEY1997]

I have managed part i but struggling with ii). I have attached some of my initial thinking. The square is throwing me off I think compared to previous examples I have done without this

 

[Dr.Peterson]

I have managed part i but struggling with ii). I have attached some of my initial thinking. The square is throwing me off I think compared to previous examples I have done without this

Please state the half-variable identities, as you learned them.

I'm not sure what they intend you to do, but clearly they want you to use the result of part (i) together with this. I suspect that you can use the half-variable identity (in a particular form) to rewrite cosh^2(x), and then use part (i). It could also be that you can use part (i) first to rewrite part of the integrand first; but that just doesn't look to me like it goes in the right direction.

 

[blamocur]

I have managed part i but struggling with ii). I have attached some of my initial thinking. The square is throwing me off I think compared to previous examples I have done without this

I second @Dr.Peterson's post, but your formula for [imath]\cosh^2(x)[/imath] in the last line is wrong.

 

[HATLEY1997]

View attachment 38229
Please state the half-variable identities, as you learned them.

I'm not sure what they intend you to do, but clearly they want you to use the result of part (i) together with this. I suspect that you can use the half-variable identity (in a particular form) to rewrite cosh^2(x), and then use part (i). It could also be that you can use part (i) first to rewrite part of the integrand first; but that just doesn't look to me like it goes in the right direction.

Here is part i

 

[HATLEY1997]

I second @Dr.Peterson's post, but your formula for [imath]\cosh^2(x)[/imath] in the last line is wrong.

View attachment 38229
Please state the half-variable identities, as you learned them.

I'm not sure what they intend you to do, but clearly they want you to use the result of part (i) together with this. I suspect that you can use the half-variable identity (in a particular form) to rewrite cosh^2(x), and then use part (i). It could also be that you can use part (i) first to rewrite part of the integrand first; but that just doesn't look to me like it goes in the right direction.

Here are the identities we have been given

 

[Dr.Peterson]

I second @Dr.Peterson's post, but your formula for [imath]\cosh^2(x)[/imath] in the last line is wrong.

I hadn't looked closely at the work, but it did cross my mind that the half-angle formula I'd found looked similar to something I'd seen there. Just a typo in writing out the intended version of the formula?

View attachment 38230
Here is part i

Yes, I'd seen (that is, trusted!) that you did this in a previous thread; here we just need the result.

View attachment 38231
Here are the identities we have been given

Thanks. That's what I was hoping for (and includes the correct version of what you miswrote):



When I think of "half variable identity" I expect one that starts "cosh(x/2) = ..."; what this shows is the form you need to use.

 

[HATLEY1997]

I hadn't looked closely at the work, but it did cross my mind that the half-angle formula I'd found looked similar to something I'd seen there. Just a typo in writing out the intended version of the formula?

Yes, I'd seen (that is, trusted!) that you did this in a previous thread; here we just need the result.

Thanks. That's what I was hoping for (and includes the correct version of what you miswrote):

View attachment 38232

When I think of "half variable identity" I expect one that starts "cosh(x/2) = ..."; what this shows is the form you need to use.

Would this be the next step?

 

[HATLEY1997]

View attachment 38233
Would this be the next step?

I’ve had a look online and this seems to be the next step. I don’t want to just copy this out without understanding it though. Where is that top line coming from - cosh(7x)+2cosh(5x)+cosh(3x)

 

[BigBeachBanana]

Distribute [imath](\cosh(2x)+1)(\cosh(5x)) =\cosh(2x)\cosh(5x)+\cosh(5x)[/imath]
Then apply the [imath]\cosh(a)\cosh(b)[/imath] identity.

 

[Dr.Peterson]

View attachment 38234
I’ve had a look online and this seems to be the next step. I don’t want to just copy this out without understanding it though. Where is that top line coming from - cosh(7x)+2cosh(5x)+cosh(3x)

Definitely don't just copy this; you've got to learn to think for yourself. This is why students' dependency on "searching for an answer" worries me. (Luckily, you know better than to fall completely into the trap.)

But what they've done is to convert what you gave them back to the original form of your integral! They are not using the half-variable formula to carry out the integration, as you were told to do. Possibly the "alternative" they will give you when you tell them to "skip simplification" would be something like what you are asked to do.

What they show is what I probably would have done without the instructions, using identities like what you used for part (i). But they show so little detail that I can't be sure exactly what they did. (Possibly, since they show the same half-variable formula in their list, they actual do use that, followed by the others, as BBB just suggested.)

All you can do is try ...