"From Stargazers to Starships" tries to be self-contained, even in the mathematics it uses. The sections below are meant to refresh your memory about some basic ideas in algebra and trigonometry.

--Even if this material is completely new to you, with enough motivation and persistence you should be able to understand it. Care was taken to remove any potential stumbling blocks.

--And if algebra and trig are old hat to you, rest assured that you will still find here some unexpected goodies.

Here is what is covered:

Elements of algebra

(M-1) Basic ideas
(M-1A) Algebra Proficiency Drill
(M-2) How it all started
(M-3) Formulas
(M-4) Identities
(M-5) Deriving Approximate Results

(M-6) The Theorem of Pythagoras

Elements of trigonometry

(M-7) What is it good for?
(M-8) How to tell sines from cosines
(M-9) Deriving sines and cosines
(M-10) Going past 90 degrees
(M-11) Deriving sin(a+b), cos(a+b)
(M-11A) Trigonometry Proficiency Drill
(M-12) The Tangent

Don't expect here a full-blown course on high-school algebra; it cannot be done, not in such limited space. These are just the bare bones--just three basic ideas and rules for handling relations ("equations") involving unknown quantities whose values you are trying to find.

In most calculations you try to find a number. For instance, the area of a rectangular plot of land 25 meters long and 40 meters wide (or yards, or feet) is

25  x  40 = 1000 square meters

Until the multiplication is carried out, we may represent the answer by some letter, usually x, and write

25  x  40 = x

One can then say "x stands for the unknown quantity". The fundamental idea of algebra is very simple:

The unknown quantity x is a number like any other. It may be added, subtracted, divided or multiplied in any way appropriate for ordinary numbers.

A mathematical relationship involving known numbers (like 25 or 40) and unknown ones (like x) is known as an equation. Often x is not given as cleanly as above, but is buried inside some complicated expression. To get a solution, one must replace the given equation (or equations) with others, containing the same information but cleaner in appearance. The final goal is to isolate the unknown, to make it stand apart ("isola" is island in Italian), to bring the equation to the above form, namely

x = (expression containing only known numbers)

Once that form is reached, the number which x represents can be quickly calculated.

For instance:

"What is the number which, if you double it, then add 5 and divide the sum by 3, you get 3?"

Call that number x. The information stated here in words can also be written down in equation form:

Do not expect anything profound or interesting--this is just a drill, like the finger excercises you do if you wish to master a musical instrument, or the parallel parking practice before your driver's test. Do them all--do not leave any out!

The notation of the instructions is as follows. For operations applied to both sides:

(+2)     add 2
(-6)     subtract 6
(*3)     multiply by 3
(/5)     divide by 5
For other operations:
(+/-)     add or subtract terms wherever you can
(*)     multiply terms wherever you can

(x/2)+5 = (x/3)+6
5x - 20 = x+8
(x+6)/2 = 2x - 21
(2x-3)/(4x-3) = 1

2/(3-x) + 1/(2+x) = 0
(x+10)/(3x+5) = 2
(11x+1)/(6x-2) = 2
(x+2)(x+3) = (x+1)(x+7)

(a) expressing y of one equation in terms of x, then
(b) substituting the expression for y in the other equation, then
(c) deriving x, and finally
(d) putting that value in the substituted expression and getting y.

Here are some examples--the first is worked out, for the rest just the steps are given. In this notation, II always means the 2nd equation at this stage of the calculation--it need not be the original 2nd equation but could have been (say) multiplied by 6. If the instructions just name an operation, it is to be applied to the equation obtained in the preceding step.

                5x - 12y = 2 (I)
                -3x + 2y = 4 (II)

(II*6)
                -18x + 12y = 24
(I+II)
                5x - 18x = 26                 (12y and -12y cancel)
(+)
                -13x = 26
(*(-1))
                13x = -26
(/13)
                x = -2
To find y, put this in (I)

(-3)(-2) + 2(-1) = 4? (4 = 4, result OK)

In what follows, only the steps for obtaining one variable are given. On your own, derive also the other variable and check the result.

The word "algebra" comes from a phrase (in bold below) in the title of an Arab book "Kitab al muhtasar fi hisab al gabr w'al muqubalah." This has been translated as "A compact introduction (book) to calculation using rules of completion and reduction," but Solomon Gandz has suggested "al gabr" comes from Babylonian "gabru" meaning solution of an equation, and that "muqubalah" (q reads like k) was its equivalent in Arabic. The book covered simple equations like the one in the preceding section, also quadratic ones involving x², as well as other areas such as geometry and the division of inheritances.

Its author, Mukhammad ibn Musa Al-Khorezmi (lived about 780-850) was the chief mathematician in the "House of Wisdom", an academy of sciences established in Baghdad by the Caliph Al Ma'mun, son of Harun Al Rashid of "Arabian Nights" fame. The "House of Wisdom" was involved in Al Ma'mun's expedition to measure the size of the Earth, which Al-Khorezmi afterwards estimated to have a circumference of 21000 Arab miles. (We are not sure how big the Arab mile was, the actual figure is about 25000 of our miles; more about such estimates, here).

Al-Khorezmi's family (and possibly he as well) apparently came from the oasis of Khorazem, at the southern end of the Aral Sea, in what is now Uzbekistan. He is also credited with helping establish among the Arabs the Indian numbering system, using decimal notation and the zero. Previous systems of writing numbers used letters, like the Roman numeral systems or the cruder ones of the Greeks and Hebrews. When Al-Khorezmi's book on the new system reached Europe, the Europeans called its use "algorism" or "algorithm," a corruption of the author's name. Today "algorithm" means method of calculation, and the rise of computers has led to extensive work on developing efficient computer algorithms.

More about Al-Khorezmi's work, here.

Note

This material is taken from a long and charming article about Al-Khorezmi, his work and his times, by Heinz Zemanek of Vienna, Austria. In September 1979, in what was (give or take a year or two) the 1200th anniversary of Al-Khorezmi's birth, mathematicians marked the occasion by convening at the site of the Khorazem oasis a conference "Algorithms in Modern Mathematics and Computer Science." Dr. Zemanek's article opens the bound proceedings of that conference, edited by A.P. Ershov and D.E. Knuth and published in 1981 by Springer Verlag as volume 122 of "Lecture Notes in Computer Science."

Further Exploration

  However, different letters are also used, often hinting at the quantity they represent--v for an unknown velocity, T for an unknown time,m or M for unknown masses and F for an unknown force. (These will be used later in the calculation of Lagrangian points).

  "Unknown" may be broadened to include "unspecified"--letters standing for numbers which may actually be known, but with whose exact values we do not want to bother, either because (#1) we want to save writing them out, or (#2) because we have not yet chosen them.

  An example of case (#1) is the ratio between the length of a circle and its diameter, a universal constant appearing in many equations (for example, Kepler's equation in section (12a)), whose value to 11 decimals is 3.14159265359... Although the number is known, it is universally represented in all equations by the letter p (Greek p, pronounced "pi"). It is only replaced with the actual number when the unknown quantity is derived.

  As an example of case (#2), consider the distance s which a dropped ball covers in a time of t seconds, starting from rest. It is

s = (1/2) gt²

  The above is a typical formula. It looks like an equation, but the game is a little different: you are not asked to "dig out the unknown number," because it is already in plain sight. All you do is choose the value of t and you get the corresponding value of s.

  Still, the "digging out" skills you learn with equations are also useful here. Suppose you seek the inverse relationship--given s, what is t? One now views t as the unknown and proceeds to isolate it. Multiply both sides by 2

2s = gt²

and divide by g

2s/g = t²

To go from t² to t one must find the square root, an easy task for anyone with a calculator having a square root button (slower methods also exist, using pencil and paper). Mathematics has a sign for this, but since the web does not provide it, we write instead SQRT:

SQRT(2s/g) = t

Now, whatever the distance s is, one can put it in the equation and derive the appropriate time t, in seconds.

Substitution of Formulas

where T₁ and R₁ are variables (known or unknown) different from T and R. If both sides of equation (2) are divided by the left side of (1)--that is, by VT--what remains is still equal, because we have treated both sides of (2) equally:

However, because of the = sign in (1), we could replace the denominator on the left by
2 p R, giving

Canceling equal multipliers ("factors") on top and on bottom leaves

which turns out to be useful in the rest of the calculation. This is a general rule: given two equations, we may divide each side of one by one of the sides of the other. "After dividing equals by equals, the results remain equal".

In some algebraic manipulations entire expressions get multiplied. For instance, one can write

(a + b)c = ac + bc

This is not an equation but an identity, an expression true for any three numbers (a,b,c). For instance if a = 3, b = 7, c = 5, then

(3 + 7)(5) = (3)(5) + (7)(5) = 15 + 35 = 50

If the addition is performed first

(3 + 7)(5) = (10)(5) = 50

Identities do not add any information about the quantities which they contain, because they are true whatevers their values may be. They are however useful in reshuffling equations to new, cleaner forms. The identity written on top is actually one of the basic properties of numbers ("the distributive law"). From it one gets more generally

(a + b)(c + d) = (a + b)c + (a + b)d

which can be further broken up and which holds for any values of (a,b,c,d). In particular

(a + b)² = (a + b)(a + b) = (a + b)a + (a + b)b

= a² + ba + ab + b²

= a² + 2ab + b²

which is quite useful (you can try it out with some specific values for a and b). Similarly

(a - b) ² = (a - b)(a - b) = (a - b)(a) + (a - b)(-b)

= a² - ba - ab + b²

= a² - 2ab + b²

Again, the two last identities

(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²

hold for any values of a and b, and as will be seen, are very useful in proving Pythagoras' Theorem.

A Preliminary Derivation

where (remember?) the two letters ac stand for "a times c" and similarly for bc.

  That is so because (c/c) = 1, no matter what the value of c is, and multiplying anything by 1 does not change its value. In multiplying fractions, the rule is to multiply top with top, bottom with bottom, so we get

As for dividing top and bottom by the same number d

it follows at once from the preceding, if we choose the number c to equal 1/d.

Working with Small Quantities

  Many such calculations make use of the following observation. When we derive squares, 3rd powers, 4th powers etc. of numbers larger than 1, the results are always bigger, while for numbers smaller than 1, the results are always smaller. For example:

Since z² is much smaller than 1 or z, we can write, using the symbol ~ for "approximately equal"

and dividing both sides by (1 -z)

(Many texts use the symbol ~ not alone but placed above an equal sign; however, that combination is not available for web documents). For example (check with your calculator)

If           z = 0.01,     (1+z) = 1.01,      (1-z) = 0.99,

then      1/(1-z) = 1/0.99 = 1.010101...

  The basic rule is: one may neglect small quantities like z, z², z³ etc. when they are added to (or subtracted from) something much bigger. One may not do so if they are just multiplied or divided, because then, if they are removed, nothing is left of the expression containing them.

  Here z can be either positive or negative. If we write z = -y, where y is a small number of opposite sign, we get

which is another useful result, valid for any small number. If that small number is again renamed and is now called z (not the same z as before, of course), we get

which can also be obtained from the earlier equation

by dividing both sides by (1 + z).

  In section (34a) where the distance to the Lagrangian point L1 is derived, it turns out necessary to approximate 1/[1-z]³. You start from (1+z) ~ 1/(1-z) and raise both sides to their 3rd powers:

Multiply out the left side:

However, if z² and z³ are much smaller than z, then dropping the terms containing them only increases the error slightly, leaving

A step beyond: The Binomial Theorem

Pythagoras of Samos was a Greek philosopher who lived around 530 BC, mostly in the Greek colony of Crotona in southern Italy. According to tradition he was the first to prove the assertion (theorem) which today bears his name:

If a triangle has sides of length (a,b,c), with sides (a,b) enclosing an angle of 90 degrees ("right angle"), then

a² + b² = c²

A right angle can be defined here as the angle formed when two straight lines cross each other in such a way that all 4 angles produced are equal. The theorem also works the other way around: if the lengths of the three sides (a,b,c) of a triangle satisfy the above relation, then the angle between sides a and b must be of 90 degrees.

For instance, a triangle with sides a = 3, b = 4, c = 5 (inches, feet, meters--whatever) is right-angled, because

a² + b² = 3² + 4²

= 9 + 16 = 25 = c²

Ancient Egyptian builders may have known the (3,4,5) triangle and used it (with measured rods or strings) to construct right angles; even today builders may still nail together boards of those lengths to help align a corner.

Many proofs exist and the easiest ones are probably the ones based on algebra, using the elementary identities discussed in the preceding section, namely

(a + b)² = a² + 2ab + b²

15² = (10 + 5)²
   = 10² + (2)(10)(5) + 5²
   = 100 + 100 + 25    = 225

and

(a - b) ² = a2 - 2ab + b²

5² = (10 - 5)²
   = 10² - (2)(10)(5) + 5²
   = 100 - 100 + 25    = 25

It is also necessary to know some simple areas: the area of a rectangle is (length) times (width), so the area of the one drawn above is ab. A diagonal cut divides it into two right-angled triangles with short sides a and b, and the area of such a triangle is therefore (1/2) ab.

Now look at the square on the left constructed out of four (a,b,c) triangles. The length of each side is (a+b) and therefore the entire square has an area (a+b)².

However, the square can also be divided into four (a,b,c) triangles plus a square of side c in the middle (strictly speaking, we also ought to prove it is a square, but we will skip that). The area of each triangle, as shown earlier, is (1/2)ab, and the area of the square is c². Since the big square is equal to the sum of all its parts

(a + b) ² = (4)(1/2)(a)(b) + c²

Using the identity for (a + b)² and multiplying (4)(1/2) = 2

a² + 2ab + b² = 2ab + c²

Subtract 2ab from both sides and you are left with

a² + b² = c²

The same result can also be shown using a different square, of area c². As the drawing on the right shows, that area can be divided into 4 triangles like the ones before, plus a small square of side (a-b). We get

c² = (4)(1/2)(a)(b) + (a-b) ²

= 2ab + (a² - 2ab + b²)

= a² + b² Q.E.D.

Q.E.D. stands for "quod erat demonstrandum," Latin for "which was to be demonstrated," and in traditional geometry books those letters mark the end of a proof. The importance of the work of Pythagoras and of later masters of Greek geometry (especially Euclid) was not only in what they proved, but of the method they developed: start from some basic statements which are assumed to be valid ("axioms") and deduce by logic their more complicated consequences ("theorems"). Mathematics still follows the same pattern.

The basic problem of trigonometry runs somewhat like this:

You stand next to a wide river and need to know the distance across it--say to a tree on the other shore, marked on the drawing here by the letter C (for simplicity, let's ignore the 3rd dimension). How can this be done without actually crossing the river?

The usual prescription is as follows. Stick two poles in the ground at points A and B, and with a tape measure or surveyor's chain measure the distance c between them ("the baseline").

Then remove the pole at A and replace it with a surveyor's telescope like the one shown here ("theodolite"), having a plate divided into 360 degrees, marking the direction ("azimuth") in which the telescope is pointing. Sighting the telescope, first at the tree and then at pole B, you measure the angle A of the triangle ABC, equal to the difference between the numbers you have read from the azimuth plate.. Replace the pole, take your scope to point B and measure the angle B in the same way.

The length c of the baseline, and the two angles A and B, contain all there is to know about the triangle ABC--enough, for instance, to construct a triangle of the same size and shape on some convenient open field. Trigonometry (trigon = triangle) was originally the art of deriving the missing information by pure calculation. Given enough information to define a triangle, trigonometry lets you calculate its remaining dimensions and angles.

Why triangles? Because they are the basic building blocks from which any shape (with straight boundaries) can be constructed. A square, pentagon or another polygon can be divided into triangles, say by straight lines radiating from one corner to all others.

In mapping a country, surveyors divide it into triangles and mark each corner by a "benchmark", which nowadays is often a round brass plate set into the ground, with a dimple in its center, above which the surveyors place their rods and telescopes (George Washington did this sort of work as a teenager). After measuring a baseline--such as AB in the example of the river--the surveyor would measure (as described here) the angles it formed with lines to some point C, and use trigonometry to calculate the distances AC abd BC. These can serve as baselines for 2 more triangles, each of which provides baselines for two more... and so on, more and more triangles until the entire country is covered by a grid involving only known distances. Later a secondary grid may be added, subdividing the bigger triangles and marking its points with iron stakes, providing additional known distances on which any maps and plans can be based.

One large surveying project of the 1800s was the "Great Trigonometric Survey" of British India. The two largest-ever theodolites were built for the project, monsters with circular scales 36 inches wide, on which settings were read with exceptional accuracy by 5 microscopes. Each in its box weighed half a ton and needed 12 men for carrying it around. Using them, the project covered the country with multiple strings of triangles in the north-south and east-west directions (the areas between the strings were left for later) and it took decades to complete.

In 1843 Andrew Scott Waugh took charge of the project as Surveyor-General, and gave special attention to the Himalaya peaks north of India. Because of clouds and haze, those peaks are only rarely seen from the lowlands, and until 1847 few measured sightings were achieved. Even after they were made, the results had to be laboriously analyzed by "computers" in the survey's offices--not machines, but persons who performed the trigonometric calculations.

The story is told that in 1852 the chief computer, Radanath Sikhdar, came to the director of the survey and told him: "Sir, we have discovered the highest mountain in the world." From a distance of over 100 miles (160 km), the peak was observed from six different stations, and "on no occasion had the observer suspected that he was viewing through his telescope the highest point on Earth." Originally it was designated as "Peak XV" by the survey, but in 1856 Waugh named it after Sir George Everest, his predecessor in the office of chief surveyor. Everest was the one who commisioned and first used those giant theodolites; they are now on display in the Museum of the Survey of India in Dehra Dum.

Nowadays position on Earth can be found pretty accurately using the global positioning system (GPS) of 24 satellites in precise orbits, constantly broadcasting their position. A small hand-held electronic instrument receives their signals and gives one's position within 10-20 meters (even more accurately for the military, the sponsor of the system). A great deal of trigonometry is involved, but it is all done for you by the computer inside your gadget, all you need do is push the proper buttons.

Now that you know a bit about the uses of trigonometry, you are welcome to advance to the actual nitty-gritty.

Note: The details about the discovery of Mt. Everest and the survey of India are taken from "Who Discovered Mount Everest?" by Parke A. Dickey, Eos (Transactions of the American Geophysical Union), vol 66, p. 697-700, 8 October 1985. The article is reprinted on p. 54-59 of History of Geophysics, Vol. 4, edited by C. Stewart Gillmor, published by the American Geophysical Union, 1990.

A very readable account, including the story of William Lambton, who founded that survey, can be found in "The Great Arc : The Dramatic Tale of How India Was Mapped and Everest Was Named" by John Keay, 160 pages hard cover, Harper and Collins, Sept. 2000.

Triangles come in many shapes. It would be hard to classify triangles of arbitrary shapes, but we note that any triangle ABC can always be divided into two right angled triangles, triangles with one angle equal to 90°. Those are easier to handle.

The three angles inside a triangle add up to 180^o (we won't prove it here) and therefore, in a right-angled triangle with sharp angles A and B

A + B + 90° = 180°

Subtracting 90° from both sides

A + B = 90°

Given the value of just one angle A, the other angle B is completely determined (it equals 90°-A), and so is the shape of the triangle, though not its size.

Let the sides of the triangle be named (a,b,c), each matching the name of the angle across from it. The angle A does not determine the length of any side, but it does uniquely fix the ratios between sides. Those ratios have names, and a specific notation exists for writing them:

a/c = sin A -- "the sine (or sinus) of A"
b/c = cos A -- "the cosine (or cosinus) of A"

To tell them apart, just remember:

sin A has the side opposite to the angle A on top of its fraction
cos A has the side adjacent to the angle A on top of its fraction

A simple connection exists between the sine and cosine of any angle. For by the theorem of Pythagoras

a² + b² = c²

Therefore, for any angle A

(sin A)² + (cos A)² = (b²/c²) + (a²/c²) = (a² + b²)/c² = 1

This statement is usually written with the square written sin²A (not sinA², which might be taken to mean the sine of an angle equal to A²):

sin²A + cos²A = 1

Both sinA and cosA must be numbers smaller than 1, for the adjacent and opposite sides of a triangle are always shorter than the side across from the 90° angle (called the hypotenuse, a word dear to the lovers of jokes and puns).

As the angle A gets closer and closer to 90° (and B gets smaller and smaller), the triangle becomes increasingly narrow and skinny. The length of side a then approaches that of c, while the length of b becomes very small: therefore as A approaches 90°, sinA approaches 1 and cosA approaches zero. For the derivation of some other values, see the next section.

Deriving the sine or cosine of an arbitrary angle takes a bit more math than is covered here. However, deriving them for a few special angles is relatively straightforward.

Complementary angles

Note first of all that a right-angled triangle contains two angles. Since all three angles (in any triangle) add up to 180°, the two sharp angles add up to 90°. It follows that if one of the angles is of A degrees, the other one (its "complementary angle") is (90°- A).

The sine and cosine were defined as the following ratios:

sin A = (side opposite to A)/(long side)
cos A = (side adjacent to A)/(long side)

Because the side opposite to A is the one adjacent to (90°- A), it follows that the sine of one angle is the cosine of the other, and vice versa:

sin A = a/c = cos (90° - A)
cos A = b/c = sin (90° - A)

This is a great help: deriving (for instance) the sine and cosine of 30°also gives us, as a bonus, the sine and cosine of 60°.

sin 45° = cos 45°

sin² 45° = cos² 45°

sin²A + cos²A = 1

2 sin² 45° = 1

sin² 45° = 1/2

sin 45° = SQRT(1/2)

sin 45° = 0.707107... = cos 45°

sin² 45° = 1/2 = 2/4

sin 45° = SQRT(2)/SQRT(4) = SQRT(2)/2

(2) A = 30°, (90° - A) = 60°

SR = (1/2) PR

In the notation of the drawing

a = (1/2) c

a/c = 1/2 = sin 30° = cos 60°

sin² 30° = 1/4

sin² 30° + cos² 30° = 1

1/ 4 + cos² 30° = 1

Subtract 1/4 from both sides

cos² 30° = 3/4

cos 30° = SQRT(3)/ SQRT(4) = SQRT(3)/2  =  1.7320508/2

cos 30° = 0.8660254 = sin 60°

(3) A = 90° , (90° - A) = 0

cos A = b/c = 0

and since

1 = sin²A + cos²A = sin²A + 0

it follows that

sin²A = 1     sin A = 1

Therefore

cos 90° = sin 0° = 0
sin 90° = cos 0° = 1

You should be able to draw a fairly good graph of sinA and cosA using the above points

(4) Postgraduate: A = 15°, (90° - A) = 75°

The preceding derivations and table are standard fare in practically any course or text on trigonometry. You notice however the gaps between 0° and 30°, and between 60° and 90°. If we want the angle A to grow in equal steps of 15^o, we still need the sine and cosine of 15° and 75°.

Are you interested? Here is how it can be done; hold on to your calculator!

Draw a triangle ABC, with the angle A equal to 30° and the two bottom angles each equal to 75°. Then draw line BD perpendicular to AC (see drawing on the right). By symmetry, the sides AB and AC are of the same length; let that length be denoted by the letter a.

The triangle ABD has angles of 90, 60 and 30 degrees, and is therefore of the kind examined earlier. We get

BD = a sin 30° = 0.5 a
AD = a cos 30° = 0.866025 a

Then

DC = AC - AD = a - 0.866025 a = 0.133975 a

BD² + CD² = c² = (0.5 a)² + (0.133975 a)²
= 0.25 a² + 0.0179493 a² = 0.2679493 a²

Taking the square root

c = 0.517638 a

From that, to 5 decimals (and involving the complementary angle of 75° as well)

sin 15° = 0.133975/0.517638 = 0.25882 = cos 75°
cos 15° = 0.500000/0.517638 = 0.96593 = sin 75°

Now go and draw your graph!

"Stargazers" introduced two ways of describing the position of a point P on a flat plane (e.g. a sheet of paper): cartesian coordinates (x,y) and polar coordinates (r,f).

Both used for reference a point O ("origin") and some straight line through it ("x-axis"). In cartesian coordinates a second "y-axis" is drawn through O, perpendicular to the first, and lines parallel to the axes are then dropped from P, cutting the axes at the points A and B on the drawing. The distances OA and OB then give the two numbers which define P, the x and y coordinates of the point.

In polar coordinates, the point P is defined by its distance r from the origin O (see drawing) and by its polar angle ("azimuth" on a map) between the x-axis and the "radius" r = OA, measured counter-clockwise.

Since the figure OAPB is a rectangle, the distance AP also equals y. Therefore

sinf = y/r
cosf = x/r

Multiplying everything by r gives the relation between the two systems of coordinates (symbols standing next to each other are understood to be multiplied):

x = r cosf
y = r sinf

These relations allow (x,y) to be calculated when (r,f) are given. To go in the opposite direction--given (x,y), find (r,f)--one notes that in the triangle OAP, by Pythagoras

x² + y² = r²

Therefore, given (x,y), r can be calculated, and then (sinf, cosf) can be derived as before by

sinf = y/r
cosf = x/r

(except at the origin point O, where (x, y, r) are all zero and the above fractions become 0/0; any value can then be chosen for the angle f).

However, there remains a problem. The angle f as defined above can go from 0 to 360°, but (sinf, cosf) are only defined for 0 to 90°, covering only the part of the plane where both x and y are positive. When one or both are negative, the angle f is larger than 90 degrees, and such angles never appears in any right-angled triangle. What sort of meaning can (sinf, cosf) have for f larger than 90 degrees?

There is a simple solution, though: use the above equations to re-define sinf and cosf for such larger angles! The equations are

sinf = y/r
cosf = x/r

They are now viewed as new definitions of the sine and cosine, for the polar angle f given by x and y (a slightly different way of formulating this definition is described further below). If (x,y) are both positive, the result is exactly the same as for angles inside a right-angled triangle. But it also works for larger angles. The sine and cosine can now be negative (just like x and y) but their magnitude still cannot exceed 1, because the magnitude of x and y is never larger than r. Here are those signs:

Allowing the line OP to go around the origin more than once allows the angle f to grow past 360°; the sine and cosine are still defined as y/r and x/r, and repeat their previous values. Similarly, turning OA in the opposite direction--clockwise--can define negative values of f. Together, these extensions define (sinf, cosf) for any angle f, positive or negative, of any size.

The relation derived from Pythagoras' theorem

sin²f + cos²f = 1

holds for any of those angles. If either the sine or the cosine is zero, the other function must be +1 or -1, depending on the sign of the coordinate (x or y) that defines them. At 90° and 270°, x = 0 and therefore cosf = 0, while at 0° and 180° y = 0 and therefore sinf = 0. We then get

Of course, f = 0° and f = 360° represent the same position of r, namely, along the positive branch of the x-axis. Below is the actual plot of cosf:

A Slightly Different Definition: the Unit Circle

Many trigonometry texts define the sine and cosine slightly differently, using the so-called unit circle. That is a circle whose radius is chosen to equal 1 unit (in whatever units we measure). We draw the circle so that its center is at the origin O of an (x,y) set of coordinates, and imagine a movable radius OB making an angle a with the x-axis.

Then the distance AB ("the sine line") equals sin a, and the distance OB ("the cosine line") equals cos a. The line CD, cut off the tangent to the circle by the extension of OB, is the "tangent line," it equals tan a and explains why the name "tangent" was given to this quantity.

Suppose the radius rotates to position OB', so that its angle with the x-axis is b, larger than 90°. Then the sine line A'B' still has positive length, since it is above the horizontal axis. However the cosine line OA' is to the left of the origin, so its length--which gives the cosine--must be counted as negative. The slice cut off the tangent is now the line CD' produced by the extension OD' of the rotating radius, and its length is counted as negative, too.

By letting the radius rotate all around the circle and measuring distance--with the understanding that anything extensing below the horizontal axis or to the left of the vertical one is negative--we find that the sine line, the cosine line and the tangent line always give the correct functions. It is really not different from the previous definition: but if you ever wondered, how the term "tangent" entered trigonometry, now you know.

A triangle ABC has a right angle C and two sharp angles A and B. The triangle's sides AC and BC on both sides of the right angle C are given as:
(a) AC = 3     BC = 4
(b) AC = 5     BC = 12
(c) AC = 8     BC = 15


In each case, use the theorem of Pythagoras to find the third side and then find the sine and cosine of angles A and B.

1. You are traveling uphill on a road and see a sign telling you this is a 5% grade, i.e. rising 5 meters for every 100 meters of road. What is the angle between the road and the horizontal direction?

   

2. An airplane is flying at 170 km/s towards the north-east, in a direction making an angle of 52° with the eastward direction.

   The wind is blowing at 30 km/s towards the north west, making an angle 20° with the northward direction. What is the actual "ground speed" of the airplane, and what is the angle A between the airplane's actual path and the eastward direction?

(The solution follows below: read it only after you have worked out the problem yourself. Teachers in class may substitute different numbers and directions)

Let us denote the velocity vector of the airplane relative to the air as V, that of the wind relative to the ground as W, and the velocity of the airplane relative to the ground U=V+W, where the addition is one of vectors. Draw a diagram with the given speeds and angles labeled appropriately.

To perform the actual addition each vector must be resolved into its components. We get

Add:

Therefore

4. In a triangle ABC, we name the angles (A,B,C) according to their corners ("vertices," singular vertex), and name the sides (a,b,c), so that side a is facing angle A, b faces angle B, and c faces angle C. Prove the "law of sines"
   sinA/a = sinB/b

   Hint: From C draw a line CD perpendicular to side c. The line CD is a "height" of the triangle and will therefore be denoted by the letter h. Use h in your proof.

5. Before attempting the next problem, note two points:

• If in the above proof we had used a perpendicular "height" from A or B, rather than from C, we would have got a relation like
  sinA/a = sinC/c

  Therefore the "law of sines" is completely symmetric:

  sinA/a = sinB/b = sinC/c

  

• The sum of angles in a triangle is always 180 degrees. A rigorous proof takes some work, but the statement can be made plausible by the following argument. In the triangle ABC, draw a line through point C that is parallel to AB. This creates two additional angles, A' and B'.

  The three angles (A',C,B') add up to 180 degrees, because they are adjacent to each other and backed off against a straight line. However, by the properties of parallel lines, the angles (A,A') are equal, as are (B,B'). Therefore (A,C,B) also add up to 180 degrees.

Don't read any further until you have tried to solve it. Teachers in class may substitute different numbers and directions.

Multiply both sides by sin(55) to obtain the length b = AC.

Further question: what is the perpendicular distance from C to the line c = AB? (hint: it equals the height h in the derivation of problem (4).)

6. (after section M-10) Find the sine and cosine of
   (1) 145°         (2) 210°         (3) 300°

   

7. (a) When a beam of light hits the surface of a flat piece of glass, it is generally bent by some angle. Draw a line perpendicular to the point on the surface where the beam enters. Then if the beam reaches the surface along a path making an angle A with the surface, it continues inside the glass with an angle making an angle B, where
   sin B = (sin A)/n
   The number n ("refractive index") is a property of the glass and is larger than one.

   The problem: given values of A=0, 20, 40, 60, and 80 degrees, and n = 1.45, what is B in each case?

   (b) the same rule holds when the light leaves the glass, except that now the angle B (inside the glass) is given and the angle A (in air) must be found. We then use

The problem:Given B = 0, 20, 90, 60, 80 --what are the angles A? Does any of these beams fail to leave the glass?

(By the way: the reason a prism splits light is that the value of n depends slightly on color of the light--or more accurately, on its wavelength).