Advanced and Spherical Trigonometry

The first trigonometry lesson gave you the right-triangle ratios and the laws of sines and cosines. This lesson stacks the identities the board leans on for cleaner algebra, shows how to hand back every solution of a trigonometric equation rather than just one, collects the several ways to get the area of a triangle, and then leaves the flat plane entirely for the spherical triangle that surveying and geodesy questions ride on. Every result is carried to a number with units so you can check your own habits against it.

Sum and difference identities

These six identities let you split an awkward angle into two friendly ones, or fold two angles into one. They are the parents of every double-angle and half-angle rule further down.

Identity	Formula
sin of a sum	sin(A + B) = sin A cos B + cos A sin B
sin of a difference	sin(A - B) = sin A cos B - cos A sin B
cos of a sum	cos(A + B) = cos A cos B - sin A sin B
cos of a difference	cos(A - B) = cos A cos B + sin A sin B
tan of a sum	tan(A + B) = (tan A + tan B) / (1 - tan A tan B)
tan of a difference	tan(A - B) = (tan A - tan B) / (1 + tan A tan B)

Watch the sign flip: the cosine formulas carry the opposite sign to the one in the parentheses, and the tangent denominators do too. That crossed sign is the single most common slip.

Worked example: Find sin 75 degrees exactly. Write 75 as 45 + 30, so sin 75 = sin 45 cos 30 + cos 45 sin 30 = (sqrt(2)/2)(sqrt(3)/2) + (sqrt(2)/2)(1/2) = (sqrt(6) + sqrt(2))/4. Numerically that is (2.449 + 1.414)/4 = 3.863/4 = 0.9659. As a check, cos 15 = cos(45 - 30) = cos 45 cos 30 + sin 45 sin 30 = 0.6124 + 0.3536 = 0.9659, and cos 15 must equal sin 75, which it does.

Double-angle and half-angle identities

Set B = A in the sum formulas and the double-angle identities fall out. The cosine one has three interchangeable forms, and picking the right form is often what makes an equation factor.

• Double angle: sin(2A) = 2 sin A cos A; cos(2A) = cos^2 A - sin^2 A = 2 cos^2 A - 1 = 1 - 2 sin^2 A; tan(2A) = 2 tan A / (1 - tan^2 A).
• Half angle: sin(A/2) = +/- sqrt((1 - cos A)/2); cos(A/2) = +/- sqrt((1 + cos A)/2); tan(A/2) = (1 - cos A) / sin A = sin A / (1 + cos A).

The plus-or-minus on a half angle is settled by the quadrant that A/2 lands in, not the quadrant of A. The two tangent half-angle forms never need that sign because they carry it automatically.

Worked example: Given sin A = 3/5 with A in Quadrant I, so cos A = 4/5. Then sin(2A) = 2(3/5)(4/5) = 24/25 = 0.96, and cos(2A) = 1 - 2 sin^2 A = 1 - 2(9/25) = 1 - 18/25 = 7/25 = 0.28. For the half angle, using cos A = 4/5 would be the wrong value here, so instead take a case where cos A = 3/5 in Quadrant I: cos(A/2) = sqrt((1 + 3/5)/2) = sqrt((8/5)/2) = sqrt(0.8) = 0.8944, while sin(A/2) = sqrt((1 - 3/5)/2) = sqrt(0.2) = 0.4472, and tan(A/2) = 0.4472/0.8944 = 0.5.

Solving general trigonometric equations

A calculator hands you one principal value; the exam usually wants every solution in a stated interval, or the full general solution. Reduce the equation to a single trig function equal to a constant, read the principal value, then add the period.

Equation form	General solution (n any integer)
sin x = k	x = arcsin k + 360n degrees, or x = 180 - arcsin k + 360n degrees
cos x = k	x = +/- arccos k + 360n degrees
tan x = k	x = arctan k + 180n degrees

Worked example (linear): Solve 2 sin x = 1 for 0 <= x < 360 degrees. Then sin x = 0.5, whose principal value is 30 degrees; sine is also positive in Quadrant II, giving 180 - 30 = 150 degrees. So x = 30 degrees and 150 degrees. Worked example (quadratic in cosine): Solve 2 cos^2 x - cos x - 1 = 0 on the same interval. Factor as (2 cos x + 1)(cos x - 1) = 0, so cos x = -1/2 or cos x = 1. From cos x = -1/2 come 120 degrees and 240 degrees; from cos x = 1 comes 0 degrees. The full solution set is x = 0, 120, and 240 degrees. Dropping the cos x = 1 branch is the usual miss.

The area of a triangle, several ways

The board picks the givens so that exactly one area formula is short. Recognize the shape of the data and reach straight for the matching row.

Given	Area formula
base b and height h	(1/2) b h
two sides a, b and the included angle C	(1/2) a b sin C
three sides a, b, c (Heron)	sqrt(s(s - a)(s - b)(s - c)), s = (a + b + c)/2
one side a and its two adjacent angles B, C	a^2 sin B sin C / (2 sin(B + C))
three sides and circumradius R	a b c / (4R)
semiperimeter s and inradius r	r s

Worked example (Heron and the radii): A triangle has sides 5, 6, and 7. The semiperimeter is s = (5 + 6 + 7)/2 = 9, so the area is sqrt(9(9 - 5)(9 - 6)(9 - 7)) = sqrt(9 * 4 * 3 * 2) = sqrt(216) = 14.70 square units. The same triangle then hands you both radii: the circumradius R = a b c / (4 * area) = (5)(6)(7)/(4 * 14.70) = 210/58.79 = 3.57 units, and the inradius r = area/s = 14.70/9 = 1.633 units. Worked example (two sides and the included angle): With a = 8, b = 11, and C = 40 degrees, the area is (1/2)(8)(11) sin 40 = 44 * 0.6428 = 28.28 square units. Forgetting the 1/2 and reporting 56.57 is the classic error.

Spherical triangles: sides and angles as arcs

Cut a sphere with three planes through its center and the great circles they trace bound a spherical triangle. Everything here is an angle: the three sides a, b, c are the angles the arcs subtend at the center (not chord lengths), and the three vertex angles A, B, C are the dihedral angles between the arcs. Because the surface curves, the vertex angles always sum to more than 180 degrees.

The law of sines and cosines on a sphere

The plane laws have direct spherical cousins. The spherical law of sines pairs the sine of each side with the sine of its opposite angle, and the spherical law of cosines for sides links one side to the other two and their included angle. Sines and cosines act on both sides and angles here, since both are measured as angles.

• Spherical law of sines: sin A / sin a = sin B / sin b = sin C / sin c.
• Law of cosines for sides: cos a = cos b cos c + sin b sin c cos A.
• Law of cosines for angles: cos A = -cos B cos C + sin B sin C cos a.

Worked example (a side from two sides and the included angle): A spherical triangle has b = 60 degrees, c = 75 degrees, and included angle A = 40 degrees. Then cos a = cos 60 cos 75 + sin 60 sin 75 cos 40 = (0.5)(0.2588) + (0.8660)(0.9659)(0.7660) = 0.1294 + 0.6408 = 0.7702, so a = arccos(0.7702) = 39.6 degrees. Worked example (an angle by the law of sines): Using that a = 39.6 degrees with A = 40 degrees and b = 60 degrees, sin B = sin b sin A / sin a = (0.8660)(0.6428)/(0.6381) = 0.5567/0.6381 = 0.8724, so B = arcsin(0.8724) = 60.7 degrees.

A common application is the great-circle distance between two points on the earth. Form the terrestrial triangle whose vertices are the north pole and the two points; the two sides from the pole are the co-latitudes and the angle at the pole is the difference in longitude, so cos d = sin(lat1) sin(lat2) + cos(lat1) cos(lat2) cos(dLong).

Worked example (great-circle distance): Two places sit at latitudes 40 degrees N and 50 degrees N with a longitude difference of 60 degrees. Then cos d = sin 40 sin 50 + cos 40 cos 50 cos 60 = (0.6428)(0.7660) + (0.7660)(0.6428)(0.5) = 0.4924 + 0.2462 = 0.7386, so d = arccos(0.7386) = 42.4 degrees. Since one degree of arc on the earth is 60 nautical miles, the distance is 42.4 * 60 = 2543 nautical miles.

Spherical excess and the area of a spherical triangle

The amount by which the vertex angles overshoot 180 degrees is the spherical excess, E = (A + B + C) - 180 degrees, and it is not a defect to correct but the very thing that measures area. Girard's theorem says the area of a spherical triangle is proportional to its excess:

Area = pi R^2 E / 180, with E in degrees, or equivalently Area = R^2 E, with E in radians.

Worked example: An octant of a sphere of radius R = 10 is a triangle with three right angles, A = B = C = 90 degrees. The excess is E = (90 + 90 + 90) - 180 = 90 degrees, so the area is pi (10^2)(90)/180 = pi (100)(0.5) = 50 pi = 157.1 square units. Check it against the whole sphere: one eighth of 4 pi R^2 is (1/8)(4 pi)(100) = 50 pi, the same 157.1 square units, which confirms the excess formula.

Quantity	Plane triangle	Spherical triangle
Angle sum	exactly 180 degrees	more than 180 degrees, by the excess E
Sides	lengths	angles a, b, c subtended at the center
Law of cosines	c^2 = a^2 + b^2 - 2 a b cos C	cos a = cos b cos c + sin b sin c cos A
Law of sines	a / sin A = b / sin B = c / sin C	sin a / sin A = sin b / sin B = sin c / sin C
Area	(1/2) a b sin C, Heron, and more	pi R^2 E / 180, from the excess E

Exam-day strategy

• On the cosine sum and difference identities, copy the sign flip carefully: cos(A + B) uses a minus and cos(A - B) uses a plus, the opposite of the sign inside the parentheses.
• Pick the cos 2A form that matches what you already know: use 2 cos^2 A - 1 when you have cos A, and 1 - 2 sin^2 A when you have sin A, so the equation collapses to one function.
• After solving a trig equation, sweep the whole stated interval for the second and third angles; the principal value from the calculator is only one of them, and the quadratic forms usually have three.
• Match the area formula to the givens: three sides means Heron, two sides plus the included angle means (1/2) a b sin C, and once you have the area both radii follow from R = a b c / (4 area) and r = area / s.
• In a spherical triangle remember that the sides are angles, so cosines and sines act on the sides too; a chord length or a straight distance has no place inside the spherical laws.
• For area on a sphere, do not fight the angle sum exceeding 180 degrees, use it: compute the excess E = (A + B + C) - 180 and feed it into Area = pi R^2 E / 180.