Only by abandoning math’s connection to realty, could we discover reality’s true nature.

But, because it discusses some really intriguing history (e.g. math duels, with you job on the line) and how folks went about certain problems totally differently than we do today (e.g. using geometry to solve quadratic equations, versus symbols).

Way to go Geralamo Cardano!

P.S. BTW, the “job on the line” is why folks also hid their solutions…

Cross posted on my Nibbles Ninja blog.

]]>Of course, this is true outside of trig too…so my recommended approaches work beyond just this one subset of mathematics.

Now, I used the word “maze” deliberately. If you go into a real world maze (or do a maze puzzle), sometimes you’ll go the wrong way…have to back up…and try again. That’s true with these problems too. Don’t feel defeated if your first, second, etc. attempt fails. Just back up and try again!

And, understand that it’ll become easier and easier over time, where just looking at a problem will give you an idea of what the best steps likely are.

What should you try (numbering does *not* imply order or importance):

- For a proof: Choose one side and work it to the other side (you can accomplish a proof working both sides, but there is a good chance your teacher will require choosing a single side, and you are more likely to get lost in the maze if you don’t)
- For a proof: Choose the most complex side (think of ${\sin ^2}x + {\cos ^2}x = 1$…would you really want to make the right side become the left?)
- Don’t forget you can use all the algebra tricks you’ve learned before; the scary trig functions don’t change how they work (e.g. if you can handle factoring ${x^2} – {y^2} = (x + y)(x – y)$, you can handle factoring $1 – 4{\sin ^2x})$
- If you are dealing with fractions, find the common denominator
- Multiply by one (which either means multiplying what you have by a value over itself or both sides of an equation by the same value)
- Multiply by the conjugate (e.g. if you have $1 + \sin x$, multiply by $\frac{1-\sin x}{1-\sin x}$); this is a form of “multiply by one”
- Convert the equation into $\sin$ and $\cos$; e.g. if you are given $\tan$ and $\csc$, see what happens if you use $\frac{{\sin }}{{\cos }}$ and $\frac{1}{{\sin }}$ instead
- Watch for patterns; e.g. if you see ${\sin ^2}x$ or ${\cos ^2}x$, you should ask yourself, “Will the Pythagorean Identity help?”…or if you see something squared minus something squared, will factoring it to the equivalent of $(x + y)(x – y)$ help?
- Use identities, they are tools for converting something you don’t know how to handle into something you do

Note that there is quite a bit of overlap between the approaches. #6 is a form of #5. #7 will often use #9. Some of the patterns you should look for in #8 are identities from #9. And so on…

So, what do you think? Any other advice you would give folks based on your experiences?

P.S. The picture is from an Indiana University Supplemental Instruction session where I was discussing this…and, as you can see, the list is growing…

]]>If you want to see how easy it is to memorize, here you go!:

Short version: All you have to do is memorize five values…

Warning: Unscripted!

]]>We have two straight lines, one bigger than the other. How do we cut a piece out of the longer one equal to the length of the shorter one? (Or as *Euclid’s Elements: All Thirteen Books Complete in One Volume* puts it: “Given two unequal straight lines, to cut off from the greater a straight line equal to the less.”

Ready to drive into the mathematical wild?

Hop on the Land Cruiser!

The GeoGerbra link is available here.

This is a relatively short proof. It needs more cowbell.

Thanks for watching!

P.S. The YouTube “Video Safari of Euclid’s Elements” is here, and Math.Promo’s page for it is here.

]]>We have a point and we have a line segment apart from that point. How do we draw a line segment from that point that is exactly the same length as the original line segment? (Or as *Euclid’s Elements: All Thirteen Books Complete in One Volume* puts it: “To place at a given point a straight line equal to a given straight line.”

Ready to go into the mathematical wild?

Hop on the Land Cruiser!

The GeoGerbra link is available here.

I had a heck of the time getting the green screen lighting right…and I may be noticeably tired (it’s headin’ toward midnight). But, I did get to use the smiley-face bell the first time!

Thanks for watching!

P.S. The YouTube “Video Safari of Euclid’s Elements” is here, and Math.Promo’s page for it is here.

]]>If we have a line, how do we create a true, equilateral triangle? (Or as *Euclid’s Elements: All Thirteen Books Complete in One Volume* puts it: “On a given finite straight line to construct an equilateral triangle.”

With no ruler! ‘Cause a ruler wouldn’t be exact regardless. Ready to climb on the Land Cruiser and go into the mathematical bush?

Hop on!

Okay, it was a little rough…but it’s my first time trying out a few things…and you didn’t get eaten by a lion…so, they’ll get smoother.

Thanks for watching!

P.S. The YouTube “Video Safari of Euclid’s Elements” is here, and Math.Promo’s page for it is here.

]]>Parametric equation behind it available here:

https://www.desmos.com/calculator/nbuivqi4uw

Credit for equation to Wolfram MathWorld.

Stay healthy everyone.

]]>I replied that when I lead my trig supplemental instruction session tomorrow and go over the area of a sector, I am going share it like this:

$Alan=\frac{1}{2}{r}^{2}\theta $🙂

]]>Come on. Admit it. You asked it before, didn’t you?

Ever used decibels? Then, my friend, you have used logarithms! From ”Understanding Decibels” on The SWLing Post:

dB = 10 log

_{10}(P2/P1)

Okay, you should read the article for more than just that.

However, just like how we measure earthquakes, hurricanes, and other items that vary by a large range, leveraging powers of 10 to measure signal or sound strength makes a lot of sense. Using earthquakes as an example, would you really want a Richter scale \(10^5\), that is, a Richter scale 100,000 earthquake? Richter scale 5 is so much easier for the brain, isn’t it?^{1}

Although, in the spirit of Spinal Tap’s guitar amp with a maximum volume of 11 versus 10, I suppose I’d love it if when I cranked Shinedown I could tell people I turned it up to 1,000,000.

Getting back to the mathematics…

Logs are just inverse powers (e.g. \(log_{10} 10^n=n\)). If you do not see a base with your log (e.g. \(log10^n\), then it is the “common log,” which is base 10.

Of course, a dB is a bit more complicated because (a) it is 10 times the common log; (b) it is the log of a ratio and (c) we don’t know what \(P1\) above is.

Well, “c” is what immediately came to mind for me.

And since I am most familiar with decibels for sound levels (in nature and in audio files), I wanted to find out what the reference value is for sound. “dB: What is a decibel?” from Physclips (University of South Whales) not only was a great discussion of decibels overall, but it provided the reference value for sound (and more):

What does 0 dB mean? This level occurs when the measured intensity is equal to the reference level. i.e., it is the sound level corresponding to 0.02 mPa. In this case we have

sound level = 20 log (p

_{measured}/p_{ref}) = 20 log 1 = 0 dBRemember that decibels measure a ratio. 0 dB occurs when you take the log of a ratio of 1 (log 1 = 0). So

0 dB does not mean no sound, it means a sound level where the sound pressure is equal to that of the reference level. This is a small pressure, but not zero. It is also possible to have negative sound levels: – 20 dB would mean a sound with pressure 10 times smaller than the reference pressure, i.e. 2 μPa.

Now, since I titled this “Shortwave Decibels,” I suppose it would be a bait and switch to leave it there.

However, I am not (yet) a Ham radio operator…so my knowledge is limited…and this neat article, “Decibel & S-Readings,” by Serge Stroobandt, ON4AA, convinced me it’s best to let you read it and decide which cases where decibels intersect with shortwave you care most about.

Either way, you now know, with logarithms, at least some cases you’ll use the math…even if you don’t realize you are.

P.S. Are there anything neater than good, old-fashioned analog stereo meters?

^{1}Yes, I am oversimplifying how the Richter scale works. However, when we hear a newsperson say “Richter scale,” they aren’t really using it anyway.

Oh, and don’t forget that what you are measuring is important. For instance, a category 5 hurricane doesn’t have 10 times faster winds then a category 4, it has 10 times the “damage potential multiplier.”

]]>So, yesterday…yes, Christmas…I proved (or perhaps, better said “derived”) the Quadratic Formula. The fun didn’t stop there.

My oldest son (visiting for the holiday), seemed to enjoy what I was doing…so I figured I’d blow his mind with how polar equations act in graphs, first showing him a simple \(r = 2\sin \theta\) one. After a bit discussing that, I thought I’d blow his mind more and create the equivalent of \(x = 1\), which is \(r = \frac{1}{{\cos \theta }}\).

Having a brain cramp, I thought I’d shift it one by adding 1 to it (\(r = \frac{1}{{\cos \theta }} + 1\)), but that results in something quite weird. You see, in a \(g(x) = f(x) + 1\) world, \(g(x)\) is just \(f(x)\) shifted up 1. Not so in the polar equation world.

The green line is \(r = \frac{1}{{\cos \theta }}\). The red lines are what happened when I added one to it (\(r = \frac{1}{{\cos \theta }} + 1\)):

But, why would we want to stop having fun there? Could we convert the polar equation to a Cartesian one? What would…

\[r = \frac{1}{{\cos \theta }} + 1\]

…be with \(x\) and \(y\)?

To help our transformation, let’s remember three things:

\[\begin{gathered}

{r^2} = {x^2} + {y^2} \hfill \\

x = r\cos \theta \hfill \\

y = r\sin \theta \hfill \\

\end{gathered} \]

Ready?!

Start with our original:

\[r = \frac{1}{{\cos \theta }} + 1\]

Get rid of the \(cos()\) in the denominator by multiplying both sides by \(\cos \theta\):

\[(r = \frac{1}{{\cos \theta }} + 1) \times \cos \theta \]

Which, multiplied out, is:

\[r\cos \theta = 1 + \cos \theta \]

Remember this?:

\[x = r\cos \theta \]

That means we can immediately transform what we have to:

\[x = 1 + \cos \theta \]

That’s great! However, how do we get rid of that pesky \(\cos \theta\) on the right?

Well, you may recall that:

\[\cos = \frac{{adjacent}}{{hypotenuse}}\]

What value is *adjacent* in an xy world? \(x\)

What value is *hypotenuse*? \(r\)

But \(r\) doesn’t do us much good…since it is a polar value. However, do you remember this?:

\[{r^2} = {x^2} + {y^2}\]

Therefore:

\[r = \sqrt {{x^2} + {y^2}} \]

(\(r\) is always positive.)

So…and it’s really about to get messy…replacing \(\cos \theta\) with *adjacent* divided by *hypotenuse* means:

\[x = 1 + \cos \theta \]

Becomes:

\[x = 1 + \frac{x}{{\sqrt {{x^2} + {y^2}} }}\]

Which is the same as:

\[x – 1 = \frac{x}{{\sqrt {{x^2} + {y^2}} }}\]

We want to end up with \(y\) alone on the left. To do that, we divide both sides by \(x-1\) and multiply both sides by \({\sqrt {{x^2} + {y^2}} }\), giving us:

\[\sqrt {{x^2} + {y^2}} = \frac{x}{{x – 1}}\]

Now we square both sides to get rid of the square root on the left:

\[{x^2} + {y^2} = {(\frac{x}{{x – 1}})^2}\]

Then, we move \(x^2\) to the right side of the equation by subtracting it from both sides:

\[{y^2} = {(\frac{x}{{x – 1}})^2} – x^2\]

And take the square root of each side so that it is \(y\) versus \(y^2\):

\[y = \pm \sqrt {{{(\frac{x}{{x – 1}})}^2} – {x^{^2}}} \]

Voila!

That was easy, eh?

*But, is it right?*

Let’s check it’s graph and see how it compares. The green line is \(x=1\) and the red is our new Cartesian equation:

Okay, to save you scrolling, here is our original beside it:

Did we do it?!

Well, it’s time to watch the rest of Die Hard 2. That’s a Christmas movie too, right?

P.S. You can see the polar graph here on Desmos, and the Cartesian one here.

]]>