one variable too many (for comfort)

[allegansveritatem]

Here is problem:

Here is what I did:
and then, in desperation:


How to deal with this problem? sinxcosx doesn't seem to have anything to offer in the way of a saving identity. And two variables multiplied doesn't lend itself to a solution involving presenting one variable in terms of another.

 

[lookagain]

I don't think you have one variable too many. You may feel you have
one function too many. I know you are dealing with theta for the angles
in the problem.
Aside from that, sin(x)cos(x) = (1/2)sin(2x).

Your line 5 could be: 3/8 = (1/2)sin(2x)

 

[allegansveritatem]

I don't think you have one variable too many. You may feel you have
one function too many. I know you are dealing with theta for the angles
in the problem.
Aside from that, sin(x)cos(x) = (1/2)sin(2x).

Your line 5 could be: 3/8 = (1/2)sin(2x)

really? How is that? I will have to think about this...is their some identity such that sinxcox=1/2 sin2x? Wow! I didn't know that....which is no big thing becasue I don't know a lot of things...well I will check this out and post back later. Thanks

 

[Dr.Peterson]

really? How is that? I will have to think about this...is their some identity such that sinxcox=1/2 sin2x? Wow! I didn't know that....which is no big thing becasue I don't know a lot of things...well I will check this out and post back later. Thanks

Yes, look up the double angle identities.

I would expect that this problem would only be given after teaching them, as that is the most natural way to solve this.

Are you not going through the book sequentially? Does it not have examples like this, showing the method of solution?

 

[allegansveritatem]

I don't think you have one variable too many. You may feel you have
one function too many. I know you are dealing with theta for the angles
in the problem.
Aside from that, sin(x)cos(x) = (1/2)sin(2x).

Your line 5 could be: 3/8 = (1/2)sin(2x)

I took your advice, after proving to myself that sinxcosx does indeed stand in for 1/2sin2x and here is what I got:

I had to use a calculator to get the 8,9, and 10. Not sure how else to do it and also not sure why the problem asks us for the angles (plural) when there seems to be only one.

 

[allegansveritatem]

Yes, look up the double angle identities.

I would expect that this problem would only be given after teaching them, as that is the most natural way to solve this.

Are you not going through the book sequentially? Does it not have examples like this, showing the method of solution?

I can say, and my words would ring with the ring of truth if you could hear them, that I NEVER skip anything. Yes, I have been introduced to double angle identities but there was nothing about 1/2sin2x. Only sin2x, cox2x and tan2x. But Dr. Stokowski seems to me to be one of those teachers who expects his students to put two and two together for themselves. Most of what I learn from his book I learn from doing the exercises at the end of sections and chapters and from the solutions manual. He probably threw that problem in with the expectation that if one knows that Sin2x = 2sinxcosx that one could hit on the fact that sinxcosx = 1/2 sin2x. Ain't always so, I'm afraid.

 

[Dr.Peterson]

I took your advice, after proving to myself that sinxcosx does indeed stand in for 1/2sin2x and here is what I got:
View attachment 18596
I had to use a calculator to get the 8,9, and 10. Not sure how else to do it and also not sure why the problem asks us for the angles (plural) when there seems to be only one.

There are two different angles whose sine is 0.75. One is 48.59°; what is the other?

That leads to two acute angles that could be x.

 

[allegansveritatem]

There are two different angles whose sine is 0.75. One is 48.59°; what is the other?

That leads to two acute angles that could be x.

Oh I see. We are looking for sin2x, not sinx! I am not in math mode right now but will post later with a reply. I could look it up on the net but that would be cheating. I will get back to this tonight

 

[allegansveritatem]

so, after consulting my encased-in-plastic laminate unit circle, I deduced that the only other candidate is approx 132 degrees. After the fact I found that I could have come to this directly by subtracting 48.5 from 180. Is there some nice rule of thumb that can be applied to cases like this?

 

[Dr.Peterson]

That is the rule, and it is not just a rule of thumb (that is, an approximation); it's a fact.

If the sine of an angle has a desired value, then its supplement has the same sine, so, as you found, you can just subtract from 180 degrees to get that other angle.

If you think about the unit-circle definition of the sine, it is the y-coordinate of a point on the circle. Two points on the circle have the same y-coordinate; the angles add up to 180 degrees.

 

[allegansveritatem]

That is the rule, and it is not just a rule of thumb (that is, an approximation); it's a fact.

If the sine of an angle has a desired value, then its supplement has the same sine, so, as you found, you can just subtract from 180 degrees to get that other angle.

If you think about the unit-circle definition of the sine, it is the y-coordinate of a point on the circle. Two points on the circle have the same y-coordinate; the angles add up to 180 degrees.

Good. I know that through knowledge and understanding of the unit circle is key in this subject. it's like the hub of the wheel of trigonometry. In that analogy the spokes would be those pesky identities in all their numerousness.

 

[Dr.Peterson]

Of course, this, too, is an identity! [MATH]\sin(\pi - \theta) = \sin(\theta)[/MATH] in radians, or [MATH]\sin(360° - \theta) = \sin(\theta)[/MATH] in degrees.

And many identities, like this one, can be remembered directly in terms of the definitions.

Using identities is like having a lot of friends, and knowing them well enough to say, "This is a situation where Fred could help me." The hard part of trig is simply that you have too many friends to keep track of at first! It takes time, and lots of adventures together, to get to know them well enough.

 

[JeffM]

Good. I know that through knowledge and understanding of the unit circle is key in this subject. it's like the hub of the wheel of trigonometry. In that analogy the spokes would be those pesky identities in all their numerousness.

I disagree with Dr. Peterson on this. I dislike burdening my mind with lots of memorization.

I used to contribute at a different forum, where some tutor posted a pdf on trig that I have always thought was excellent because it was so simple. His point was that most of what you need in trig can be derived fairly easily if you know what is on one and a half pages. The pdf is an attachment to this post

Trigonometry to Memorize, and Trigonometry to Derive

I have attached a pdf document containing the vast majority of trigonometry I have needed to know on a working basis. The first page consists of trigonometry I think everyone should have memorized. I have never needed to have anything more than the first sheet memorized for any application. The...

mathhelpforum.com

I gave an example of how to use the pdf in the same thread

Trigonometry to Memorize, and Trigonometry to Derive

I don't suppose anyone wants to take upon themselves the challenge/effort of writing up how each identity on the 2nd page can be derived from the stuff on the 1st page? (I'd try, but doubt I could get them all).

mathhelpforum.com

 

[Dr.Peterson]

I disagree with Dr. Peterson on this. I dislike burdening my mind with lots of memorization.

I can't see the sheet without signing up, but I suspect we don't disagree at all. The sheet I give my students is one page long, and centers on a few pictures that make those easy. Did I say to memorize hundreds of identities? Maybe you took "a lot" differently than I intended it.

I frequently say I prefer to minimize what I memorize. In fact, what I said there wasn't that I've memorized that particular fact, but that I remember it by referring to other facts (namely, the definition). Of course, when I've been using it recently it tends to stick, but it does so the way friends' names stick -- not by having made a burdensome effort to memorize, but by becoming familiar with them. I use the analogy of friends very deliberately.

On the other hand, I find that some people memorize things more easily than I do, and visualize (e.g. the unit circle definitions) less easily than I do. So what works for me may not be what works for them.

 

[JeffM]

I can't see the sheet without signing up, but I suspect we don't disagree at all. The sheet I give my students is one page long, and centers on a few pictures that make those easy. Did I say to memorize hundreds of identities? Maybe you took "a lot" differently than I intended it.

I frequently say I prefer to minimize what I memorize. In fact, what I said there wasn't that I've memorized that particular fact, but that I remember it by referring to other facts (namely, the definition). Of course, when I've been using it recently it tends to stick, but it does so the way friends' names stick -- not by having made a burdensome effort to memorize, but by becoming familiar with them. I use the analogy of friends very deliberately.

On the other hand, I find that some people memorize things more easily than I do, and visualize (e.g. the unit circle definitions) less easily than I do. So what works for me may not be what works for them.

I suspect what I misinterpreted was "many identities can be remembered." I recognize now that "can be" does not imply "should be," and I read more into your statement than was there.

I hate to copy someone else's work that is posted at another site, but since the work is on a free site open to all and downloadable from there, it seems like fair use to post it here. If the admins disagree, they can delete it.



Not my work as I said before.

 

[Deleted member 4993]

One more to add:

\(\displaystyle \int sec(x) dx = ln |sec(x) + tan(x)| +C \)

Impossible to derive without "divine intervention".

 

[allegansveritatem]

Of course, this, too, is an identity! [MATH]\sin(\pi - \theta) = \sin(\theta)[/MATH] in radians, or [MATH]\sin(360° - \theta) = \sin(\theta)[/MATH] in degrees.

And many identities, like this one, can be remembered directly in terms of the definitions.

Using identities is like having a lot of friends, and knowing them well enough to say, "This is a situation where Fred could help me." The hard part of trig is simply that you have too many friends to keep track of at first! It takes time, and lots of adventures together, to get to know them well enough.

I like that!

 

[allegansveritatem]

I suspect what I misinterpreted was "many identities can be remembered." I recognize now that "can be" does not imply "should be," and I read more into your statement than was there.

I hate to copy someone else's work that is posted at another site, but since the work is on a free site open to all and downloadable from there, it seems like fair use to post it here. If the admins disagree, they can delete it.

View attachment 18661
View attachment 18662
Not my work as I said before.

Thanks for this. I will copy it and keep it in a file for reference.

 

[lookagain]

One more to add:

\(\displaystyle \int sec(x) dx = ln |sec(x) + tan(x)| +C \)

Impossible to derive without "divine intervention".

There is a method that does not rely on "that trick."

Rewrite sec(x) as 1/cos(x).

Multiply top and bottom by cos(x).

Rewrite the denominator as \(\displaystyle \ 1 - sin^2(x)\).

Let u = sin(x)

Then du = cos(x)dx

You have 1/(1 - u^2) to separate using partial fractions.

Integrate using the natural logarithm function and use of the
absolute value symbols.

Substitute back for the x-variable (trigonometric).

Write the difference of two logs as a log of a quotient.

Multiply the argument of the log, top and bottom, by (1 + sin(x)).

Rewrite the denominator of the argument as \(\displaystyle \ cos^2(x)\).

There would be a coefficient of (1/2) for the logarithm. Transfer it to
the argument as an exponent and take the square root.

The new argument would look like (1 + sin(x))/(cos(x)).

Split these up, change the first to sec(x), the second to tan(x), and
get the "+ C" written down after the logarithm expression.