A spring is attached at one end to support

*B*and at the other end to collar

*A*, as represented in the figure. Collar

*A*slides along the vertical bar between points

*C*and

*D*. In the figure, the angle theta is the angle created as the collar moves between points

*C*and

*D*.

(The figure shows that the distance directly from support

*B*to the vertical bar is 3 ft.)

Part A

When theta = 28 degrees, what is the distance from point

*A*to point

*B*to the nearest tenth of a foot?

Part B

When the spring is stretched and the distance from point

*A*to point

*B*is 5.2 feet, what is the value of theta to the nearest tenth of a degree?

(A) 35.2 degrees

(B) 45.1 degrees

(C) 54.8 degrees

(D) 60.0 degrees

To answer Part A, we notice that the known side of 3 feet is adjacent to the given angle, and we are asked to find the hypotenuse

*AB*. Adjacent and hypotenuse -- that sounds like a job for the cosine:

cos theta =

*BD*/

*AB*

cos 28 = 3 /

*AB*

*AB*= 3 / cos 28

*AB*= 3.398

Rounding off to the nearest tenth of a foot, the answer is 3.4 feet. Moving on to Part B, we now have that the hypotenuse is 5.2 feet and we wish to find the angle:

cos theta =

*BD*/

*AB*

cos theta = 3 / 5.2

theta = arccos(3 / 5.2)

theta = 54.77

Rounding off to the nearest tenth of a degree, the answer is 54.8 degrees, which is choice (C). Some of the wrong answers use sine or tangent instead of cosine.

There are several things that I wish to say about this problem. First of all, notice that the test actually does use the Greek letter theta to denote angle measures. Just as with interval notation mentioned earlier this week, I usually think of theta a symbol that doesn't appear until the trigonometry portion of Precalculus, which may appear in Honors Algebra II. I once briefly mentioned the theta symbol to my Geometry students when I was tutoring, but then I quickly said something like "cos theta means the cosine of

*t*degrees." But now we see that the Common Core uses theta on its Geometry test -- which means that I should as well. I mentioned yesterday that the PARCC and SBAC have priority over what anyone else says -- and that includes

*myself*. I say that theta should appear until Honors Algebra II or Precalc, while the PARCC says that it can appear in Geometry, and the PARCC wins.

Here's an interesting question: why do we use the letter theta to denote angles? I decided to do a Google search to find out why. I noticed that an old Mathforum post from 1996 appeared in my results -- and that's the same site I found the information about Dr. John Conway and the inclusive definition of trapezoid. I clicked on the search result -- and sure enough, it was Conway:

http://mathforum.org/kb/message.jspa?messageID=1080353

This raises another question : just WHY ARE

theta, phi, psi

the traditional angle-letters? Let's see where they are in the

Greek alphabet, which I usually write in tetrads, because it

"rhymes" best that way :

alpha beta gamma delta

epsilon zeta eta THETA

iota kappa lambda mu

nu xi omicron pi

rho sigma tau upsilon

PHI chi PSI omega

There doesn't seem to be much logic in that particular triple,

but it occurs to me that chi and omega are also reasonably

common angle-letters, so it might well be that it's just

"theta and the last few letters". On the other hand, it might

be that they (or at least some of them) are the initial letters

of some obvious Greek words - I'm not competent enough to know.

I remark that since phi and psi are both written and pronounced

similarly, once you've used one for some purpose, you're quite

likely to use the other for a similar purpose.

Can anyone fluent in Greek (maybe Andreas?) tell us why theta

is THE standard angle name?

John Conway

So Conway doesn't know the answer -- at least he didn't 19 years ago. Just to show how old this post is, I myself was in a Precalc class and just learning about theta when this was posted. Earlier in this post, a (Geometry!) student who knew that his/her teacher spoke to Conway on the Internet said:

"Why don't you write to John Conway. If he doesn't

know, then nobody in this world knows." I replied, "Either that, or if

Prof. Conway doesn't know, then it's probably not worth knowing."

Now in Part B, we must use the inverse cosine to solve the problem. The U of Chicago introduces the cosine (and sine) function in Section 14-4. But inverse trig functions are an afterthought. The inverse tangent is mentioned in Section 14-7 on vectors, because the only way to determine the direction (angle) of a vector given its

*x*- and

*y*-components is to use the inverse tangent. The inverse cosine (and sine) function don't appear in the U of Chicago text at all -- but I've seen it in other Geometry texts. I repeat that PARCC and SBAC have priority over the U of Chicago. Since inverse cosine appears on the PARCC, I should include inverse cosine and sine in my course next year.

One of my least favorite concepts regarding inverse trig is the notation. For some reason, we know that cos^2 (theta) = (cos theta)^2, yet cos^-1 (

*t*) =/= (cos

*t*)^-1. I absolutely

*hate*this notation because of the lack of consistency when the superscript is 2 vs. when it is -1. If I could, I would abolish the notation cos^-1 altogether, and replace it with "arccos," for

*arccosine*. The big problem is that calculators use the notation cos^-1, not arccos -- and as inverse trig functions are impractical to evaluate without a calculator, my priority must be the notation that appears on the calculator.

So now we ask, why does cos^2 denote an exponent of 2, yet cos^-1 denotes the inverse? Too bad that we don't have a Conway here to answer. I decided to look specifically for a Mathforum post from Conway to answer the question. I found this Mathforum post from 2005 -- but it's not Dr. Conway who answered, but someone named "Dr. Math" (and signed Dr. Vogler), describing why we use the notation

*f*^-1 to denote the inverse of a function

*f*:

http://mathforum.org/library/drmath/view/68250.html

So why do they use a power of -1 to mean inverse function? The main reason is that when people deal with *iterating* a function, they often write f^n or n f to mean f *composed* with itself n times (rather than f *multiplied* by itself n times). So then 2 f (x) = f(f(x)) and 3 f (x) = f(f(f(x))) and so on. And we also say that f^0(x) is the identity function x. This notation is convenient because it satisfies many of the well-known properties of exponents, such as n m n+m f (f (x)) = f (x). In this notation, it makes perfect sense to write f^-1(x) for the inverse function, because then -1 f (f(x)) = x is exactly the formula I wrote above, with n = -1 and m = 1.

So according to Dr. Vogler,

*f*^-1 really is an inverse of

*f*, but it's not the

*multiplicative*inverse of

*f*, but rather an inverse where the operation is "iteration," which involves the composite of a function with itself.

Let's think about what Vogler is saying if we let

*f*be a transformation from Geometry. Suppose that D is a dilation of scale factor

*k*centered at the origin. If

*P*is a point, then D(

*P*) denotes the dilation image of point

*P*. So what would D^2 be? It is the composite of D with itself, D o D. So D^2(

*P*) denotes a dilation of scale factor

*k*^2 centered at the origin. Likewise D^3(

*P*) denotes a dilation of scale factor

*k*^3 centered at the origin. So D^-1(

*P*) would be the inverse dilation -- it would be a dilation of scale factor 1/

*k*(that is,

*k*^-1) centered at the origin. Notice that because D^

*n*(

*P*) is a dilation of scale factor

*k*^

*n*centered at the origin, the superscript

*n*really does mean exponentiation, and D^-1 actually means multiplicative inverse.

But what if we considered a translation T, with translation vector

**v**? Then T^2(

*P*) translates a point by the vector 2

**v**, and T^3(

*P*) translates a point by the vector 3

**v**. Then T^-1(

*P*) is the inverse transformation -- a translation by the vector -

**v**. Notice that because T^

*n*(

*P*) is a translation by the vector

*n*

**v**, the superscript

*n*now means scalar

*multiplication*, and T^-1 now means

*additive*inverse.

And what if we considered a reflection r, with mirror

*m*? Then r^2(

*P*) reflects a point over

*m*and then reflects it back (Flip-Flop Theorem), so this is the identity. Likewise r^3(

*P*) reflects a point over the mirror thrice, so it's equivalent to the original reflection

*m*. So the set containing two elements -- the identity and a single reflection -- is in fact a group, and this group is isomorphic to the additive group known as

**Z**/2

**Z**. So r^-1(

*P*), the inverse of r, is also r itself. We conclude that T^

*n*(

*P*) is either reflection in

*m*or the identity depending on whether

*n*is odd or even, and r^-1 now means just r.

So we just saw that sometimes the -1 superscript means multiplicative inverse (of a scale factor

*k*), sometimes it means additive inverse (of a translation vector

**v**), and sometimes it means nothing at all (if the transformation is a reflection). In the notation

*f*^-1, the -1 is not the inverse of multiplication, but the inverse of

*composition*. Just as

*n** 1/

*n*= 1, the identity element for multiplication,

*f*^-1 o

*f*, as normally written, must be the identity function. It's tricky because numerical functions can be either multiplied or composed, but other functions, such as reflections, can't be multiplied -- they can only be composed.

Composition is an operation in its own right -- it's an operation on

*functions*. Just as addition has the plus sign, composition has "o" -- the circle sign. Back in 2009, I've seen some people propose that, in order to emphasize that

*f*^-1 is the composite inverse, a small circle be written near the -1. This looks something like f°[-1](x) in ASCII, or cos°[-1](x) for inverse cosine. I first read about this notation at the following link:

http://math.eretrandre.org/tetrationforum/showthread.php?tid=321

Let me explain what this link is exactly. This is the

*Tetration Forum*, a website that describes a new operation called "tetration." If we think about the fact that multiplication is repeated addition and exponentiation is repeated multiplication, we may wonder, what is repeated exponentiation? The 20th century British mathematician Reuben Goodstein named this operation "tetration," because it is the

*fourth*operation in the sequence addition, multiplication, exponentiation (and

*tetra-*is Greek for four, if we think about the four-faced

*tetrahedron*).

As tetration is repeated exponentiation, it can be defined using function iteration. In particular, we can define

*b*tetrated to the

*n*, written in ASCII as

*b*^^

*n*, by letting

*f*be exponentiation and then iterating it:

*f*(

*x*) =

*b*^

*x*

*b*^^

*n*=

*f*^

*n*(1)

Here are a few tetrations calculated out:

*f*(

*x*) = 2^

*x*

2^^0 =

*f*^0(1) = 1

2^^1 =

*f*^1(1) = 2

2^^2 =

*f*^2(1) = 2^2 = 4

2^^3 =

*f*^3(1) = 2^2^2 = 16

2^^4 =

*f*^4(1) = 2^(2^2^2) = 65,536

Yes, tetration increases very quickly -- notice that 2^^5 is 2^65,536, with nearly 20,000 digits! We can use inverses to determine one more value of tetration:

2^^-1 =

*f*^-1(1) = log_2(1) = 0

or written in our new notation,

2^^-1 = f°[-1](1) = log_2(1) = 0

Since tetration is so closely linked to function iteration, tetraters are very interested in a consistent notation for iteration -- hence the thread at the link above. For example, a tetrater may want to know what the value of 2^^(1/2) is. By our definition, this would be f°[1/2](1), but we don't know how to calculate such "half-iterates." The entire Tetration Forum website is devoted to finding such suitable half-iterate functions so that expressions like 2^^(1/2) can make sense.

Notice in the thread linked above, the original poster considered using the symbol inv(

*f*) for the inverse of

*f*, and there was even a brief mention of just using arc(

*f*), since we already have already established arccos to mean inverse cosine. The reason for introducing an "inv" or "arc" operator is because "inv" is a function in its own right -- it maps functions to other functions. (Another example of an operator that appears in Calculus class is the derivative.) And so "inv" can also be iterated -- inv^2 is just the identity, but inv°[1/2] would be an interesting operator in its own right.

Tetration is another one of my more esoteric interests -- along with calendar reform and musical reform, which I've already posted on the blog. I won't discuss tetration very often on the blog, but mentioning half-iterates raises the question, what are half-iterates for the Geometry transformations. I point out that some transformations are easy to half-iterate:

-- If T is a translation with vector

**v**, then T°[1/2] could be a translation with vector (1/2)

**v**.

-- If R is a rotation with angle theta, then R°[1/2] could be a rotation with angle theta/2.

-- If D is a dilation with scale factor

*k*, then D°[1/2] could be a dilation with scale factor sqrt(

*k*).

-- If I is the identity, then I°[1/2] could also be the identity.

But notice that half-iterates of functions are never uniquely determined:

-- If T is a translation with vector

**v**, then T°[1/2] could be a glide reflection with vector (1/2)

**v**and any mirror parallel to

**v**.

-- If R is a rotation with angle theta, then R°[1/2] could be a rotation with angle theta/2 + 180.

-- If D is a dilation with scale factor

*k*, then D°[1/2] could be a dilation with scale factor -sqrt(

*k*) -- that is, we dilate with scale factor sqrt(

*k*) and then rotate 180 around the center of dilation.

-- If I is the identity, then I°[1/2] could also be any reflection.

I almost have the transformation worked out, but I need to get back to the PARCC question. We've already completed it quickly, so I could spend extra time discussing the theta and ^-1 notation, but this post it getting long enough. Maybe I'll give the transformation next time -- or who knows, maybe you will be able to figure it out. Find a discontinous transformation T such that T o T gives reflection in the

*x*-axis.

**PARCC Practice EOY Exam Question 27**

**U of Chicago Correspondence:**

**Section 14-4, The Sine and Cosine Ratios**

**Key Theorem: Definition of cosine**

**In right triangle**

*ABC*with right angle*C*:**the cosine of Angle**

*A*, written cos*A*, is leg adjacent to angle*A*/ hypotenuse.

**Common Core Standard:**

CCSS.MATH.CONTENT.HSG.SRT.C.8

Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.

**Commentary: The U of Chicago gives more problems with sine than cosine. But some of the problems, particularly Question 14 in this section and Question 55 in SPUR under "Uses," could use either sine or cosine depending on which angle is used. There are no problems using inverse cosine anywhere in the text.**

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