Trigonometry is an extremely useful branch of elementary mathematics, and besides can be distinctly entertaining. It is pleasant exercise for the mind to consider its many ramifications, though this would not be the opinion of a high school student suffering through it for the first time. The mature mind finds it logical and connected, and can grasp it as a whole, not as a collection of rules. In fact, only very bad trigonometry can be done relying only on rules and formulas. It is much more powerful if understood, and if its beautiful power is appreciated. High school trigonometry was largely concerned with the solution of triangles, a very practical and concrete target, and one of great use, rather than with the mathematical properties of the trigonometric functions. Questions of precision and checking should be given more attention than is usually devoted to them. I really do not know what has happened to it in schools in the last few decades, but it has probably suffered like geometry.
In this article I will review what we both know very well, as I recently have done, trying for utility as well as conciseness. For trigonometry to be fun, one should be prepared in geometry and algebra, at least in the properties of triangles, circles and parallels, and in the manipulation of algebraic formulas. This is not really much preparation, but it is essential. The trigonometry text itself introduced the student to logarithms, once the invariable accompaniment to trigonometry, but now much less important that electronic pocket calculators are available. In omitting logarithms in trigonometry, it should be kept in view that logarithms are very important in mathematics, and the proper and efficient use of tables is still a valuable skill. These subjects belong somewhere, even if not in trigonometry. Now, come review with me!
Trigonometric functions are also called circular functions because of their simple relation to various lengths defined by a radius of a circle making an angle with a reference direction, as shown in the figure at the right. This is not the only way of defining trigonometric functions; they can be defined as analytic functions of a complex variable z by power series, for example.
The trigonometric functions of the angle θ are defined in the figure, as the lengths of certain lines when the radius is unity. The lengths of these lines for any radius are simply these values times the radius, and that is how they are used to solve trianglesz. The angle φ = 90° - θ, the complement of θ, is at the right end of the horizontal line. Each of the "co-" functions is the corresponding functions of φ. For example, sin φ = cos θ. Using this fact, and the properties of similar triangles, all the relations shown in the figure can be obtained. Note especially the three Pythagorean relations at the bottom. That the sum of the squares of the sine and cosine is equal to unity is easy to see and the most used trigonometric identity. Also, it is clear that all the functions of an angle can be found if only one of them is known.
As the radius rotates, the trigonometric functions go through a regular cycle of variations with period 2π, the sine and cosine bounded by ±1, the tangent and cotangent going to ±∞. We see that sin(-x) = -sin x and cos(-x) = cos x by looking at the corresponding lines. Functions of supplementary angles are the same, up to a sign. For example, cos(180° - θ) = -cos θ and sin (180° - θ) = +sin θ. Functions of complementary angles are the co-functions, as sin(90° - θ) = cos θ. A table of sines from 0° to 90° is also a table of cosines for the same range, and equivalent to a table of sines and cosines from 0° to 45°, the usual arrangement of tables. One should be able to express the trigonometric functions of any positive or negative angle in terms of the functions for an angle in the first quadrant, between 0° and 90°.
Review the available sources of functional values. The pocket calculator is an excellent resource, but it is instructive as well to consider the slide rule (which has sine and tangent scales, from which any function can be read off at once). The slide rule shows clearly that the sine and tangent tend to equality as the angle becomes small, and eventually θ = sin θ = tan θ (if θ is expressed in radians). This is easily seen from the figure. My Brunton compass has a three-place table of sines for integral degrees from 0° to 45°. It is easy to interpolate to find functions of tenths of a degree. For example, sin 25.7° = 0.423 + 0.7(0.438 - 0.423) = 0.433. It is not easy to find other functions with such a limited table. The cosine can be found from √(1 - sin2θ), and this used for the sine of the complementary angles greater than 45°, as well as for the tangent and cotangent by division. In the present case, cos 25.7° = 0.901 = sin 64.3° and tan 25.7° = 0.481. Of course, if you have the right calculator, you have all the functions anyway, but this could be of use if the only thing available is a four-function calculator. Things are much easier with a full set of tables, such as appeared in Wentworth and Smith. It was so common to use logarithms in trigonometric calculations that a table of sines, for example, was labelled natural sines, to distinguish it from logarithmic sines.
Before the invention of logarithms in the 17th century, trigonometric calculations were so tedious and error-prone that they were performed graphically or by ingenious mechanical devices (such as the gunner's quadrant).
For serious computation, tables such as Vega's (see References) were used. The seven-place tables give tabulated values for 5 significant figures. Logarithms with 7 significant figures can be obtained by interpolation, using the tables of proportional parts that are supplied on each page. These save the arithmetic of finding decimal fractions of the difference between two tabular values. For example, log 5470.612 is log 5470.6, or 3.7380350, plus 0.12 of the difference 79 to the next tabulated value for 5470.7. The proportional parts for 1 is 7.9, for 2 15.8, so we must add 7.9 + 1.58 = 9, dropping the decimals, and the final result is log 5470.612 = 3.7380359. This is precisely the result given by my calculator. The digital computer has made the preparation of accurate and extensive tables very easy, but it has also supplanted them. Nevertheless, much data is still presented in the form of tables.
Sexagesimal notation was retained for angles, as for time, since it has advantages in tabular organization. However, electronic calculators are decimal (some can use sexagesimal, however), so decimal degrees are now overwhelmingly employed. For an accuracy equivalent to seconds of arc, four decimal places should be kept. The natural unit for angles in trigonometry is the radian, which is always divided decimally. Only when radians are used is the sine of an angle approximately equal to the angle for small angles. However, the radian is usually inconvenient in practical problems. Angles are often measured in sexagesimal time units in Astronomy, where 1h = 15°, 1m = 15' and 1s = 15".
The mil is an angle unit in military service. The British mil is 0.001 radian. The NATO mil is defined by 1600 mils = 90°, so it equals 0.0009817 radian, close enough to 0.001 radian for military work. The mil used by the U.S. in WWII was defined as 1000 mils = 90°, and was 0.00157 radian. It was a tenth of the French grad, 100 grad = 90°, just an example of decimania, and rather useless, as was the US mil as well. Using s = rθ, an object of width s metres observed to subtend m UK or NATO mils is at a range of r = 1000(s/m) metres. This works with yards, and indeed with any length unit.
There exist a couple of now unfamiliar functions, the versed sine and the exsecant. The versed sine is 1 - cos θ, so it approaches zero as the angle does, but more rapidly. It gives the sagitta distance v, which for small angles is the difference between two nearly equal quantities, 1 and cos θ. For small angles, 1 - cos θ = θ2/2, so if the circle radius is r, we immediately get the sagitta formula v = h2/2r, where h = rθ is small. It is often seen in the form of the haversine, or (1 - cos θ)/2, since this is equal to sin2(θ/2). There is a formula for solving a triangle that gives sin (θ/2) as a square root of a quantity. Therefore, the haversine will be equal to the quantity, and if a table of the haversine is available, a square root operation is avoided. Pocket calculators have reduced the value of such ingenuity. The exsecant, sec θ - 1, was useful to surveyors, who are often interested in this distance. Such functions were convenient when tables had to be used; now we can easily calculate them directly. These functions were tabulated to simplify the evaluation of small corrections, which will be illustrated below under Applications.
The solution of general, or oblique, triangles was the climax of the high-school trigonometry course. By "solution" is meant the determination of all the parts of a triangle when enough parts are given to determine the triangle. It is really rather wonderful that this can be done so easily, and is a remarkable triumph of mathematics. The most basic fact about a plane triangle is that the sum of its angles is a straight angle, or 180°. This is very easily proved by drawing a parallel to one side through the opposite vertex, as in the figure at the left. The properties of a parallel make the angles A and B equal as shown, so the sum is 180°. This is a fundamental property of a Euclidean space. If we draw a triangle on a convex surface, such as the surface of the earth, the sum of the angles is greater than 180°. If the triangle is drawn on a concave surface, then the sum is less than 180°. When surveying on the earth, the spherical excess δ of the sum of the angles above 180° is easily observed for large triangles, and is related to the area S of the triangle by S = R2δ, where R is the radius of the sphere (check this for an octant of a sphere).
The parts of a triangle will be denoted as in the diagram at the right. The letters A, B, C refer to the vertices of the triangle, as well as to the magnitude of the internal angles at each vertex. The lengths of the sides opposite a vertex are denoted by the corresponding small letters. When we speak of the angle A or the side a, we also mean their magnitudes.Of the six parts, only five are independent, since A + B + C = 180°. The sides must be such that the sum of any two is greater than the third, or a + b > c, and so on for any pair of sides. This is the familiiar triangle inequality. Relations between the sides and the angles fall into classes called "laws" of trigonometry. There are three familiar laws, all of which may be easily derived by considering the figure shown.
From the definitions of the trigonometric functions, it is easy to see that h = a sin B = b sin A, so that a/sin A = b/sin B. By considering the altitude to another side, we see that this also means that a/sin A = b/sin B = c/sin C. These relations are called the Law of Sines. If we use the Pythagorean theorem in the same diagram, we find that b2 = h2 + (c - s)2, as well as a2 = h2 + s2. Subtracting the second equation from the first, we get b2 - a2 = (c - s)2 - s2, from which b2 = a2 + c2 - 2ac cos B, since s = a cos B. This, and the other two similar relations involving C and A, are called the Law of Cosines. For the third law, we begin with a/b = sin A/sin B. Subtracting 1 from both sides, we find (a - b)/b = (sin A/sin B) - 1. Adding 1, the result is (a + b)/b = (sin A/sin B) + 1. Dividing the first equation by the second, we find that (a - b)/(a + b) = (sin A - sin B)/(sin A + sin B). Expressing the difference and the sum of the sines by sin A - sin B = 2 sin[(A - B)/2]cos[(A + B)/2] and sin A + sin B = 2 cos[(A - B)/2]sin[(A + B)/2], and forming the tangents, we finally get (a - b)/(a + b) = tan[(A - B)/2]/tan[(A + B)/2]. This equation, and its two companions for other pairs of sides, are the Law of Tangents.
Suppose we know the angle C and the sides adjacent to it, a and b. These parts determine the triangle uniquely (as a sketch reveals). We can use trigonometry to find A, B and c. This case is abbreviated SAS, for side-angle-side, and is perhaps the most frequently encountered in practice. We see first that the Law of Sines cannot be applied at once, but the Law of Cosines gives us immediately c2 = a2 + b2 - 2ab cos C. This equation is easily evaluated on a pocket calculator for any value of C from 0° to 180°. However, if C is close to 0° or 180°, the cosine does not vary rapidly with the angle, so that the calculations must be carried out to a precision greater than that of the final answer. When logarithms are used, the Law of Cosines is very inconvenient.
The Law of Tangents was derived just for this case. Since we know C, we also know (A + B)/2 = (180° - C)/2. This makes the Law of Tangents convenient, for then tan[(A - B)/2] = [(a - b)/(a + b)] tan[(A + B)/2]. The sum and difference of the sides is easily found, and then a logarithmic solution can do all the hard work. This formula does not suffer from the disadvantages of the Law of Cosines for C near 0° or 180°, and so is still useful for this case with the pocket calculator when the triangle is long and skinny. Then, we can find A = (A + B)/2 + (A - B)/2 and B = (A + B)/2 - (A - B)/2, and can determine c by c = (a/sin A)sin C or c = (b/sin B)sin C. It is easy to remember to work fat triangles by the Law of Cosines, and skinny triangles by the Law of Tangents. The parts of a triangle may be relabeled, and any formula involving them is still valid.
A second frequently-met case is when we know the three sides a, b and c, and are to determine the angles, a case called SSS. This is a job for the Law of Cosines again, solved for the cosine: cos A = (a2 - b2 - c2)/2bc. This suffers again from loss of precision, and incompatibility with logarithms. To overcome this, add 1 to both sides, and use 1 + cos A = 2 sin2(A/2) and 1 - cos A = 2 cos2(A/2). Using the abbreviation s = (a + b + c)/2 for the semiperimeter of the triangle, a little algebra gives tan(A/2) = √[(s - b)(s - c)/s(s-a)]. These relations can be called half-angle formulas, and considered as related to the Law of Tangents.
A third case is ASA, where the angles A and B at the ends of side c are given. Of course, we know all three angles, since C = 180° - A - B. Then the other sides are easily determined by the Law of Sines: a = (c/sin C) sin A and b = (c/sin C) sin B. A related case is AAS, where we are given two angles and the side opposite one of them. This is not frequently encountered, but it is also easy to solve by the Law of Sines, since we know one angle and the side opposite it. Note that in these two cases, the Law of Sines is used to find a side, not the angle.
There is one more case, and it is peculiar. This is SSA, where we know two sides and the angle opposite one of them, not the angle between them. Since we know a side and its opposite angle, we can use the Law of Sines to determine the other angle. The problem here is that we are considering angles to go from 0° to 180°, and for such angles the sine is positive and repeats itself. That is, sin(180° - θ) = sin θ. Therefore, unless the angle happens to be 90°, the Law of Sines gives us two possible values, which are supplementary. For that reason, this case is called the ambiguous case. However, there is really nothing wrong here, just that SSA may not specify a triangle uniquely, as a sketch easily reveals. If you know which triangle to choose, and that is part of the statement of the problem, then the problem is as easily solved as any other. In the diagram, ABC and AB'C' are the two possible triangles, one acute, the other obtuse. Since triangle BCB' is isosceles, it is easy to see that B and B' are supplementary.
Mollweide's Equations, which involve all six parts of a triangle, are an easy and valuable check on their consistency, and so on the correctness of a solution. They follow from the laws of sines, and are: (a - b)/c = sin (A - B)/2 / cos C/2 and (a + b)/c = cos (A - B)/2 / sin C/2. The triangle should be relabeled to avoid angles near 0° or 90°. Mollweide's contribution was to publish them in 1808; they were known considerably earlier.
Three noncollinear points determine a unique circle. The three vertices of a triangle, therefore, determine a circle called the circumscribed circle, of which the sides of the triangle are chords. The centre of this circle must lie on the perpendicular bisectors of the chords, which must meet at one point. It is not hard to see that the diameter of this circle is 2R = a/sin A = b/sin B = c/sin C, so that we have an interpretation of the meaning of this ratio. We have used the fact that the angle subtended by a chord at the circumference is half the angle subtended at the centre.
The largest circle that will fit inside the triangle, the inscribed circle, must be tangent to the three sides. Its centre, then, is the point common to the three bisectors of the angles. Note that the radius r must subtend equal angles at any vertex, which proves the property. It is easy to see that the area of the triangle is the sum of the areas of three triangles with the same altitude r and bases equal to the sides. The radius of the inscribed circle is r = √[(s - a)(s - b)(s - c)/s] in terms of the sides. This formula can be found by combining the formula for the area, S = rs, and Heron's Formula (see below) for the area.
The area of a triangle is given by ch/2, where c is the length of a side and h is the corresponding altitude. This is easily seen, since the area of the triangle is half the area of a parallelogram of the same base c and height h, and the area of the parallelogram is equal to the area of a rectangle of that base and altitude, or ch. Since h = a sin B, for example, we have formulas like S = (1/2)ac sin B = (1/2)bc sin A for the area S (S is used to avoid confusion with angle A). This is generally the easiest formula to use, after solving the triangle.
A median of a triangle is a line drawn from a vertex to the centre of the opposite side. The medians meet at one point, which is the centre of gravity of the triangle. An altitude is a line drawn from a vertex perpendicular to the opposite side. The altitudes meet at one point, called the orthocentre of the triangle. We won't use these definitions here, but these lines should be compared with the ones we do use with the circumscribed and inscribed circles.
If we use sin B = 2 sin(B/2)cos(B/2), and then the expressions we found above while deriving the half-angle formulas, we can find the remarkable expression S = √[s(s - a)(s - b)(s - c)], called Heron's Formula. If R is the radius of the circumscribed circle, then sin B = b/2R and S = abc/4R = 2R2sin A sin B sin C. If r is the radius of the inscribed circle, then the circle is divided into three triangles with the same altitude r by the bisectors of the angles, so the area becomes S = r(a + b + c)/2 = rs. Combining this with Heron's Formula gives the expression for r that we quoted above. Heron's formula is convenient for the SSS case, and is easily evaluated with logarithms.
Some relations between trigonometric functions are an immediate consequence of the definitions, such as cos2x + sin2x = 1, but many are not quite so easy to derive, though very useful. Although I know most of them, I usually look them up in a reference in case I don't remember correctly. I discovered that this habit has led me to forget the derivations. Since the knowledge of a logical path through them lends confidence to trigonometric algebra, I'll show the way into the most useful relations here. Incidentally, an identity is an equation that is true for any value of the variables.
Surely one of the most remarkable relations in mathematics is Euler's Formula that connects trigonometric functions with exponential functions: eix = cos x + i sin x. At the cost of introducing complex numbers and the exponential, many derivations can proceed algebraically instead of geometrically. For example, the starting point for many identities is the expression of functions of x + y and x - y, the sum and difference of two angles. Euler's Formula gives ei(x + y) = cos (x + y) + i sin (x + y) = eixeiy = (cos x + i sin x)(cos y + i sin y) = (cos x cos y - sin x sin y) + i(sin x cos y + cos x sin y), and we can read off the results. For x - y, we simply use the fact that sin (-y) = sin y and cos (-y) = cos y. Now we can use the identities we already know to find tan (x + y) and all the other functions we need.
It follows from Euler's Formula that sin x = (eix - e-ix)/2 and cos x = (eix + e-ix)/2. Identities are easily derived algebraically from these expressions. For example, squaring sin x gives sin2x = (1 - cos 2x)/2 without difficulty. Also, the relation to hyperbolic functions is easily seen: sin ix = i sinh x, cos ix = cosh x and tan ix = i tanh x. We can go on from this to find sin (x + iy) and cos (x + iy). If we put x = π in Euler's Formula, we find eiπ = -1, surely a remarkable expression containing the transcendental numbers e and π, as well as the imaginary unit i and -1.
A geometrical proof for the functions of x + y is not much harder. The key is writing down the diagram, which is shown at the right. Once you have the correct diagram, the results just fall out. We see that sin (x + y) = CF and cos (x + y) = OF. CF = DE + CG and OF = OE - DG. In the small triangle CGD, the angles x and GCD have corresponding sides perpendicular, so the angle at C is x. The hypotenuse CD = sin y, while side OD = cos y. CG, then, is sin y cos x, while DG is sin y sin x; DE = sin x cos y, and OE = cos x cos y. Now we see that sin (x + y) = sin x cos y + cos x sin y, and cos (x + y) = cos x cos y - sin x sin y. These formulas for the functions of x + y and x - y are the basic formulas for deriving all the others.
It would be customary in geometric proofs like this to draw figures for all the possible cases (such as x + y > 90°) to show that the results are valid in all cases. However, we know that the trigonometric functions are analytic functions of a complex variable, so any proof in a restricted region can be extended to everywhere in the region of definition. The fact that we can express them algebraically in terms of the analytic function ex shows that they are analytic.
Now the double- and half-angle formulas can be found immediately. For example, we have cos 2x = cos2x - sin2x. We can add and subtract 1 = cos2x + sin2x to find 1 + cos 2x = 2 cos2x and 1 - cos 2x = 2 sin2x. These can be written in terms of half-angles if desired, as 1 - cos x = 2 sin2(x/2) = vers x.
The sums and differences of sines and cosines are also easily found from the basic formulas. If we let u = x + y and v = x - y, then sin u + sin v = 2 sin[(u + v)/2] cos [(u - v)/2] and sin u - sin v = 2 cos[(u + v)/2] sin[(u - v)/2]. Similarly, cos u + cos v = 2 cos[(u + v)/2] cos [(u - v)/2] and cos u - cos v = -2 sin[(u + v)/2] sin [(u - v)/2]. Corresponding formulas for the tangent and cotangent are derived from their definitions as ratios of the sine and cosine. We have now derived all of the trigonometric relations commonly given in handbooks.
It is not profitable to try to memorize all the trigonometric identities that you will use. There is a vast forest of them, and the memory can be inaccurate. Far better is to learn the paths of deriving them from a few basic relations, since this will give skill in deriving any relation that may be required. The only basic relations that are needed are sin2x + cos2x = 1 and the sum of angles formulas sin(A + B) = sinAcosB + cosAsinB and cos(A + B) = cosAcosB - sinAsinB, and these are easy to remember, and easy to derive with diagrams. Of course, the definitions of the functions, and their oddness or evenness are also necessary. If you practice deriving identities a little, you will find that the skills are easy to learn.
A spherical triangle is formed by three great circles on the surface of a sphere, as shown in the diagram. A great circle is the intersection of a plane through the centre O with the sphere, and is analogous to a straight line in the plane. We consider a sphere of unit radius without loss of generality. The vectors a, b, c of unit length locate the vertices A, B, C of the triangle. The angles of the triangle are denoted A, B, C and the sides opposite each angle by a, b, c. Both the angles and the sides are measured in angular measure. The angle A is the dihedral angle between the planes defined by the vectors a, b and a, c. The side a is the angle between the vectors b, c. As the length of the largest side approaches zero, the spherical triangle approaches a plane triangle. The angles cannot remain constant as this limit is approached. The sum of the angles of a spherical triangle is greater than 180°.
The scalar product a·b = cos c (et cycl.). The product (c x b)·(a x c) can be expanded in two ways. First, it is equal to sin a sin b cos C from evaluating the cross products. It is also equal to (c·a)(b·c) - (c·c)(a·b), which is cos b cos a - cos c. Equating these, we find cos c = cos a cos b + sin a sin b cosC, which is the Law of Cosines for the side c. There are similar formulas for the other two sides. In a similar way, but using cross products of the vectors a, b, c, we find the laws of cosines for the angles: cos A = -cos B cos C + sin B sin C cos c, and so forth. Note the minus sign, which may be unexpected.
For the angle A, we have sin A = (axb)x(axc) / |axb||axc| = [axb·c]/sin b sin c. Therefore, sin A / sin a = [axb·c]/sin a sin b sin c. The same result is obtained for the other angles, so we have the Law of Sines sin A / sin a = sin B / sin b = sin C / sin c.
As the largest side becomes small, we may use the approximations sin a = a and cos a = 1 - a2/2, and similarly for the other sides. Then, the law of cosines becomes c2 = a2 + b2 - 2ab cos C, and the law of sines sin A / a = sin B / b = sin C / c, the analogous formulas for a plane triangle.
These relations are sufficient to solve any spherical triangle for the unknown parts, but another relation is quite useful. To derive it, we substitute cos c = cos a cos b + sin a sin b cos C in cos b = cos a cos c + sin a sin c cos B, use cos2 b = 1 - sin2 b, and eliminate the common factor sin a to get sin a cos b = cos a sin b cos C + sin c cos B. Dividing by sin b, we get sin a cot b = cos a cos C + sin c cos B / sin b. The last term is equal to sin C cot B, using the law of sines. Therefore, we have sin a cot b = cos a cos C + sin C cot B. If we work with the angles instead of the sides, we find sin A cot B = sin c cot b - cos c cos A. Interchanging a, b or A, B (which is just a relabeling) gives valid formulas. For small sides, these formulas reduce to the law of sines when A + B + C = 180° is taken into account.
Another identity involving two angles and three sides can also be obtained from the Law of Cosines. If we write sin b sin c cos A = cos a - cos b cos c by rearranging the formula for cos a, and then substitute the formula for cos c in this, we can use cos2b = 1 - sin2b and then divide by sin b to find sin c cos A = cos a sin b - cos b sin a cos C. This formula can be used to find c after using the cot formula to find A in an SAS case, instead of using the Law of Sines directly. There does not appear to be any advantage in doing this, however.
There are several formulas involving all six parts of a triangle that may be used to check a solution, due to Gauss and Napier. Two of Gauss's formulas, sin (a - b)/2 / sin c/2 = sin (A - B)/2 / cos C/2, and sin (a + b)/2 / sin c/2 = cos (A - B)/2 / sin C/2, reduce to Mollweide's equations for small sides, which are used to check plane triangle solutions.
There are half-angle formulas for finding the angles when three sides are known, and for finding the sides when three angles are known. For example, tan A/2 = k / sin(s - a), where k2 = sin (s - a) sin (s - b) sin (s - c) / sin s, where s is the semiperimeter s = (a + b + c)/2. Also, tan a/2 = K cos(s - A), where K2 = cos S / cos (S - A) cos (S - B> cos (S - C), and S = (A + B + C)/2.
The haversine (introduced above under Trigonometric Functions) is the function hav θ = (1 - cos θ)/2, or sin2(θ/2). Its use makes some computations simpler. For example, the law of cosines becomes hav c = hav(a - b) + sin a sin b hav C. The haversine is also better for small angles, when the cosine has precision difficulties. The haversine appears to have fallen into obscurity; the HP-48 does not recognize it.
To solve a spherical triangle when two sides and the included angle are known, the law of cosines can be used to find the third side, and then the law of sines gives the other angles. Another method that is recommended is to use the cotangent formula to find either angle A or B, then the law of sines to find side c. A third method is to use the haversine, as just explained.
If one of the angles is 90°, say C, then sin C = 1, cos C = 0 and the formulas simplify. Any unknown part can be expressed as the product of functions of two known parts. The Law of Sines gives sin a = sin A sin c and sin b = sin B sin c. The Law of Cosines gives cos c = cos a cos b. The relation between two angles and three sides gives sin c cos A = cos a sin b. If we substitute for sin b and then cancel sin c, we find cos A = cos a sin B, which also gives cos B = cos b sin A. Five more identities are found involving tangents and cotangents from these. For example, from a relation just quoted, sin b = sin c cos A /cos a. Replacing sin c by sin a / sin A from the Law of Sines, sin b = tan a cot A. Also, cos c = cos a cos b = cot B cot A, where we have used cos A = cos a sin B and its twin on interchanging A and B, a and b. There are 10 identities in all.
These identities can be summarized by Napier's Rules of Circular Parts. In the diagram above, the parts on either side of a certain part, called the middle part, are the adjacent parts, while the others are the opposite parts. The sine of a middle part is the product of the cosines of the opposite parts, or the product of the tangents of the adjacent parts. Check this rule with the identities in the preceding paragraph.
Vector calculations can be made graphically, but trigonometry is required for numerical solutions. Finding components of a vector in a given direction, and finding the resultant of vectors, are the most common problems. A vector A has a magnitude A = |A| and a direction given by an angle with respect to a reference direction. The component of vector B in the direction of vector A is AB cos θ, where θ is the angle between the positive directions of the vectors. If A were reversed, then the angle would be 180° - θ, and the cosine would change sign. If A is a unit vector (that is, A = 1), then the component of B is just B cos θ. We may take three unit vectors i,j,k in the directions of a right-handed coordinate system x,y,z. The three components Bx = i·B, and so on are the rectangular components of the vector B, and represent it completely. Often, it is much more convenient to work with the components than with magnitude and direction. If we wish to find the resultant of two vectors, we can use either the law of cosines, or alternatively can add corresponding components to find the components of the resultant.
It is no surprise that trigonometry is of great use in plane surveying. A typical surveying text (see References) includes 5-place tables of logarithms and natural and logarithmic functions (which, of course, are no longer necessary). A problem that can be solved trigonometrically is that of an inaccessible distance, such as across a river. With electronic distance measuring, this is no longer much of an obstacle, since the line need not be accessible to chainmen. Without such assistance, the problem is a significant one that can be solved by trigonometry. Let's assume, then, that we can measure the angles between points that can be sighted from a given point with a transit, both horizontally and vertically, and distances are measured by "chaining" with a 100-foot steel tape.
Suppose A and C are two points on a line, on opposite banks of a river or across a lake, and the distance between them is required. Three solutions are illustrated at the right. One solution is for the chainmen at point C to give a "swinging offset" of one chain, or 100 ft, to the transitman at A. The point B need not be precisely determined, since the transitman can get an accurate measure of the angle by setting the crosshairs on the farthest excursion of the 100 ft point. Then, the right triangle gives AC = 100/sin θ, where θ is the angle CAB. Alternatively, a line at right angles AB of length s can be laid off, and the angle ACB can be measured with the transit at C. Then, AC = s/tan ACB. If neither of these methods is satisfactory, a line AB of length s can be laid off on one bank, and the angle CAB measured. The transit is moved to C, and the angle ACB measured. Then, by the Law of Sines, AC = (s/sin ACB) sin(180° - CAB - ACB).
The area of a field can be found by setting the transit at some point within it where all the corners can be seen. The distances from the transit to each corner are measured, as well as the angles between the corners. Then, application of S = (1/2)ab sin C to each of the triangles gives the area of the field. Any area with straight-line boundaries can be subdivided into triangles, and the area can be determined.
Quite commonly, the corners of an area are specified by rectangular coordinates, determined by means of a closed traverse around its perimeter. The y-coordinate is the northing, measured positive northwards along grid north, while the x-coordinate is the easting, measured at right angles positive towards the east. The corners are numbered consecutively around the traverse. The best method for finding the area if the coordinates of the corners are known finds the algebraic sum of the areas of the trapezoids between the sides of the traverse and a coordinate axis, usually the y-axis. The figure shows point 2 of a traverse, with the points on either side. Point 2 enters in the two trapezoidal areas shown, whose area is (1/2)[(y3 - y2)(x3 + x2) + (y2 - y1)(x2 - x1)]. Multiplying out, and keeping only the terms containing y2, we have (1/2)y2(x1 - x3). The whole area is the sum of such terms for each corner of the traverse. Each value of y is multiplied by the difference between the preceding and following x-values. To find the area, a table is made with the first two columns the northing and easting (y,x), the third column the differences in eastings corresponding to each corner, and the fourth and fifth column the positive and negative products (these, of course, may be accumulated in a calculator). The net value is found and divided by 2 to get the area. The sign of the area is not significant, and reverses if the traverse is described in the opposite direction. With our formula, the area is positive for a clockwise tour around the traverse.
An application of the versed sine is shown in the figure. A and B are two points on a line, and s is the measured distance between them on the surface of the earth. However, surveys are always projected on a horizontal plane, and distances are always horizontal distances. It is easy to see that the horizontal distance is s cos θ, where θ is the angle of depression. However, it may be more accurate to subtract the correction s vers θ, since this is usually a small number and need not be calculated to the full number of significant figures required. If the difference in elevation h is small compared to s, then vers θ = h2/2s2, but one supposes that a table of versines is available. Suppose s = 100 ft, and h = 10 ft (a rather steep incline). then θ = 5.7106° and vers θ = 0.004963. The correction is then 0.49 ft. Only two significant figures in the correction gives five significant figures in the distance, 99.51 ft. Even the approximate sagitta formula gives a correction of 0.50 ft. Considerations of precision are really very important in practical problems, as well as the existence of checks on the results.
One can construct practice problems for oneself, but it is more challenging to solve ingenious problems constructed by others. For example, work this problem from Wentworth and Smith: A man in a balloon observes the angle of depression of an object on the ground, bearing south, to be 35° 30'; the balloon drifts 2.5 miles east at the same height, when the angle of depression of the same object is 23° 14'. Find the height of the balloon (Ans. 7096 ft).
Navigation can provide interesting vector problems. For an aircraft or a ship, the absolute velocity is the sum of the velocity relative to the wind or current plus the velocity of the wind or current. There are some esoteric aspects to ship navigation, but air navigation is simple. For example, suppose you fly from over point A at an airspeed of 115 mph for one hour on a heading of N 60° E, and find yourself over point B that is 130 miles from A in the direction N 75° E. What is the wind velocity? Ans. 33.65 mph from N 47.44° W. This is a good problem for the Law of Tangents. Check with the Law of Cosines. A problem that must often be solved is to find the direction in which you must fly to reach a certain destination, knowing the direction and speed of the wind, as well as your own airspeed. In this case, you must fly into the wind at an angle of φ from your intended course, where sin φ = w sin θ/a, where w sin θ is the wind component perpendicularly across your course, and a is your airspeed. If a = 220 mph and w = 40 mph at an angle of 45° to your intended course, then sin φ = (40 sin 45°/220), or φ = 7° 23.4'.
Two important applications involving spherical triangles are: (1) finding the distance and bearing of the great circle path between two points of known geographical coordinates, and (2) finding the altitude and azimuth of a celestial body whose declination and hour angle are known. Both are SAS cases, so there are several methods of solution. To illustrate the first problem, let the two points be Denver (40° N, 105° W) and Mecca (21° N, 40° E). C = 145°, a = 50°, b = 69°. The cot formula gives B = 34.95°, about NE by N. The law of sines then gives c = 110.81°. Note that a calculator will give the angle less than 90° with the given sine; we want the one greater than 90°. 110.81° is 1.934 radians, so the distance is 7657 miles (Earth's mean radius 3959 miles or 6371 km). For the second problem, let us find the altitude and azimuth of Fomalhaut (-29.65°, RA 22h57m) when its hour angle is 2h34m before the meridian (this is the angle at the celestial pole between the hour circle of Fomalhaut and the meridian). In general, the hour angle is the sidereal time minus the right ascension. Now C = 38.5005°, a = 50° b = 119.65° (90° - δ). We again use the cot formula, getting Az = 146.46°. The calculator gives a negative angle less than 90°, so we subtract it from 180° to get the angle we want. The law of sines then gives c, which is 90° - altitude, as 78.2854°. The altitude is then 11.7145°.
E. Maor, Trigonometric Delights (Princeton, NJ: Princeton U. P., 1998).
G. Wentworth and D. E. Smith, Plane Trigonometry and Tables, 4th ed. (Boston: Ginn and Company, 1943). A standard high-school trigonometry text of 1950. Answers to most exercises are included, a great help to self-learners.
J. J. Corliss and W. V. Berglund, Plane Trigonometry (Boston: Houghton Mifflin, 1950). A college trigonometry text, also of 1950. Illustrates the difference in emphasis, but covers much of the same ground as Wentworth and Smith. Logarithms are deferred until later, and complex numbers are treated in more detail. The solution of triangles is not ignored. Answers to odd-numbered problems are included.
G. Vega, Seven-Place Logarithmic Tables (New York: Hafner, 1960).
C. B. Breed and G. L. Hosmer, The Principles and Practice of Surveying, Vol I, 10th ed. (New York: John Wiley & Sons, 1970).
Composed by J. B. Calvert
Created 10 January 2004
Last revised 14 September 2007