Introductory Spherical Trigonometry

In this notes, we'll cover -
Part 1. Foundations

• Section of a sphere by a plane
• Great Circle and Small Circle
• Properties of Poles
• Spherical Radius, Angular Distance
• Secondaries
• Spherical Angle
• Poles of Great Circles
• Angle between Great Circles
• Length of Arc of Small Circles
• Spherical Triangles
• Polar Triangles and Reciprocity of Polarity
• Relations between sides and angles of Polar Triangles
• Properties of Spherical Triangles

Part 2. Fundamental Formulae

• Cosine Formula
• Supplemental Cosine Formula
• Sine Formula
• Formula for Half Angle
• Formula for Half a Side
• Sine-Cosine Formulae
• Relation between Three Sides and Two Angles
• Supplemental Sine Cosine Formula
• Cotangent Formula
• Napier's Analogies
• De-Almbret's Analogies

Part 3. Right-Angled Triangles

• Spherical Triangle (right-angled case)
• Napier's Rule of Circular Parts

Part 1. Foundations

The Sphere & Section by a Plane

Definition 1. Sphere:
A sphere is the locus of all points in space that are at a constant distance from a fixed point.

• The fixed point is called the centre
• The constant distance is called the radius

When a plane intersects a sphere, the cross-section is always a circle. There are two cases: great circle and small circle.

Defintion 2. Great Circle
A great circle is the intersection of a sphere with a plane that passes through the centre of the sphere.
Its properties:

• Its radius is the radius of the sphere.
• It divides the sphere into two equal halves.
• It's the largest possible circle on a sphere.
• Through any two points on a sphere (not diametrically opposite), exactly one great circle passes.
• Through two diametrically opposite points (antipodal points), infinitely many great circles pass.

E.g., on Earth, the equator, all lines of longitude (meridians) are great circles. Lines of latitude (except equator) are not great circles.

Definition 3. Small Circle:
A small circle is the intersection of a sphere with a plane that does not pass through the centre.
Its properties:

• Its radius is less than the radius of the sphere.
• It does not divide the sphere into two equal halves.

E.g., on Earth, all parallels of latitude except the equator (like 30°N, 60°S) are small circles.

Poles and Their Properties

Definition 4. Poles of a Circle:
The extremeties of the axis of a circle on a sphere are called the poles of that circle. And the axis of a circle is the diameter of the sphere that is perpendicular to the plane of that circle.

For a Great Circle-
Since the plane of a great circle passes through the centre of the sphere, the axis also passes through the centre. Therefore both lies symmetrically, they are diametrically opposite (antipodal) points, equidistant from the plane.

For a Small Circle-
Since the plane of a small circle does not pass through the centre, the two poles lie at unequal distances from the plane:

• Nearer Pole - The pole on the same side as centre
• Farther Pole - the pole on the opposite side

Theorem 1

A pole of a circle is equidistant from every point on the circumference of that circle.

Proof:

Let a plane cut the sphere with centre O and radius R, forming a circle with centre C and radius r. Let P and P' be the poles of this circle, so PP' is the axis, perpendicular to the plane of the circle at C.

Let A and B be any two points on the circumference of the circle.

Since A and B lie on the circle with centre C:
$CA = CB = r$

Since PC $\perp$ plane of the circle, triangle PCA and triangle PCB are both right-angled at C.

By Pythagoras' theorem:
$PA^2 = PC^2 + CA^2 = PC^2 + r^2$
$PB^2 = PC^2 + CB^2 = PC^2 + r^2$

Therefore: $PA = PB$

Since A and B are arbitrary points on the circle, P is equidistant from every point on the circumference. The same holds for P'.

Theorem 2

The angular distance from a pole of a great circle to any point on its circumference is a quadrant (90°).

Proof:

Let the great circle have centre O (which is also the centre of the sphere) and radius R. Let P be a pole of this great circle.

Since P is a pole, OP is perpendicular to the plane of the great circle.

Let A be any point on the great circle. Then OA = R (radius of sphere).

In triangle OPA:
$\angle OAP = 90° \quad \text{(since OP } \perp \text{ plane, OP } \perp \text{ OA)}$

Since OA = OP = R (both radii of the sphere):

The arc PA subtends angle AOП at the centre O.

$\cos(\angle POA) = \frac{OP \cdot \hat{n} \cdot OA}{|OP||OA|}$

Since OP is along the axis (normal to the plane) and OA lies in the plane:
$\vec{OP} \cdot \vec{OA} = 0 \implies \angle POA = 90°$

Therefore the arc PA = 90°, i.e., a quadrant. Since A is arbitrary, every point on the great circle is at 90° from P.

Theorem 3 (Converse)

If a point P on a sphere is at a quadrant's distance from two distinct points A and B on a great circle, then P is a pole of that great circle.

Proof:

Let the sphere have centre O. Let A and B be two distinct points on a great circle, and let P be a point on the sphere such that:
$\text{arc } PA = 90° \quad \text{and} \quad \text{arc } PB = 90°$

This means:
$\angle POA = 90° \quad \text{and} \quad \angle POB = 90°$

Therefore:
$\vec{OP} \perp \vec{OA} \quad \text{and} \quad \vec{OP} \perp \vec{OB}$

Since $\vec{OA}$ and $\vec{OB}$ are two distinct vectors lying in the plane of the great circle, and $\vec{OP}$ is perpendicular to both, it follows that:

$\vec{OP} \perp \text{plane of the great circle}$

Hence OP is the axis of the great circle, and P is an extremity of this axis on the sphere.

Therefore, P is a pole of the great circle.

Spherical Radius, Angular Distance

Definition 5. Spherical Distance/Angular Distance:
The spherical distance (or angular distance) between any two points on the surface of a sphere is the length of the shorter arc of the great circle passing through them.

• Units: it's measured in degrees, minutes, and seconds, or in radians.
• Formula: If $\theta$ is the central angle in radians and $R$ is the linear radius of the sphere, the actual linear arc length $s$ is given by: $s = R \cdot \theta$

Definition 6. Spherical Radius:
The spherical radius of a small circle on a sphere is the spherical arc distance from the pole of the circle to any point on its circumference.
For a Great Circle, the spherical radius is always a quadrant ($90\degree$ or $\frac{\pi}{2}$ radians).
For a Small Circle, the spherical radius (say $\alpha$) is always less than $90\degree$. If the linear radius of the small circle is $r$ and the linear radius of the sphere is $R$, they are related by:

Theorem 4:

If $r$ is the linear radius of a small circle, $R$ is the linear radius of the sphere, and $\alpha$ is the spherical radius of the small circle, then $r = R \sin \alpha$.
Proof:
Let $O$ be the center of the sphere with linear radius $R$.
Let a plane intersect the sphere to form a small circle with center $C$ and linear radius $r$.
Let $P$ be the pole of this small circle, so the line $OP$ is the axis of the circle, intersecting the plane perpendicularly at $C$ ($\angle OCD = 90^\circ$).
Let $D$ be any point on the circumference of the small circle.

In the right-angled plane triangle $\triangle OCD$:
$\sin(\angle COD) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{CD}{OD}$

By definition, $CD = r$ (linear radius of the small circle) and $OD = R$ (linear radius of the sphere).

The central angle $\angle COD$ subtends the spherical arc ${PD}$ on the surface.
By definition, the arc length of this milestone is the spherical radius $\alpha$.
Therefore, $\angle COD = \alpha$.

Substituting these values into the trigonometric ratio:
$\sin \alpha = \frac{r}{R}$

$\implies \boxed{r = R \sin \alpha}$

Secondaries to a Great Circle

Definition 7. Secondaries: A secondary to a given great circle is any great circle that passes through the poles of that given great circle.

• Perpendicularity: Every secondary intersects its primary great circle at right angles ($90^\circ$).
• Real-World Analogy: If the Earth's Equator is the primary great circle, all meridians of longitude are its secondaries because they all meet and intersect at the geographic North and South Poles.

Spherical Angle && Angle Between Two Great Circles

Definition 8. Spherical Angle:
A spherical angle is the angle between two great circles at their point of intersection. It is measured by the angle between the tangents to the two great circles at that point.

Example: On Earth, two meridians meet at the North Pole. The angle between them at the pole equals the difference in their longitudes.

Definition 9. Angle Betwwen Two Great Circles:
The angle between two great circles at their point of intersection is equal to the arc intercepted by them on a great circle which is a secondary to both.

Key Properties:

• Two great circles always intersect at exactly two points
• The angles at these two points are supplementary (sum = 180°)
• If two great circles are perpendicular, each passes through the poles of the other

Length of Arc of Small Circles

Definition 10. Length of Arc of Small Circles:
Let a small circle have spherical radius θ (the arc of a secondary from the pole to the small circle). If an arc of the small circle subtends an angle φ at the pole of the small circle, then the length of that arc is:

$l = \sin\theta \cdot \phi \cdot r$

where r is the radius of the sphere. In angular terms (r = 1):

$l = \phi \sin\theta$

Theorem: The arc of a small circle subtending an angle φ at the pole is equal to φ sinθ, where θ is the spherical radius of the small circle.

Proof:

Let the sphere have centre O and radius R. Let CD be a small circle with pole P and spherical radius θ.

Let the plane of the small circle cut the axis OP at point C'. Then:

$OC' = R\cos\theta \quad \text{and} \quad C'D = R\sin\theta$

So the radius of the small circle is: $\rho = R\sin\theta$

Now, let arc AB be an arc of the small circle subtending angle φ at the pole P (and also at C', the centre of the small circle, since PC' is the axis).

The arc of a circle = radius × angle subtended at centre:

$\text{arc } AB = \rho \cdot \phi = R\sin\theta \cdot \phi$

For a unit sphere (R = 1):
$\boxed{\text{arc } AB = \phi\sin\theta}$

Key Notes

• Radius of a small circle with spherical radius θ = R sinθ
• Arc of small circle = φ sinθ (unit sphere)
• When θ = 90°, sinθ = 1, then small circle becomes a great circle

Spherical Triangle & Its Properties

Definition 11. Spherical Triangle: A spherical triangle is a figure formed on the surface of a sphere by three arcs of great circles, each arc being less than a semicircle (less than 180°).

The three arcs are called the sides and the three points of intersection are called the vertices of the spherical triangle.

If the vertices are A, B, C then:

• Side a = arc BC (opposite to vertex A)
• Side b = arc CA (opposite to vertex B)
• Side c = arc AB (opposite to vertex C)

Note: The sides a, b, c are measured in degrees or radians (not length). They are the angles subtended at the centre O of the sphere.

Properties of Spherical Triangles

1. Each side is less than 180° (a semicircle).

2. The sum of any two sides is greater than the third side:
$a + b > c, \quad b + c > a, \quad c + a > b$

3. The sum of all three sides:
$0° < a + b + c < 360°$

4. Each angle is less than 180°.

5. The sum of all three angles A + B + C satisfies: $180° < A + B + C < 540°$

This is the key difference from plane geometry where A + B + C = 180° always. The excess over 180° is called the spherical excess E = A + B + C − 180°.

6. The greater side has the greater angle opposite to it.

Polar Triangles

Definition 12. Polar Triangle:
Let ABC be a spherical triangle. For each vertex, construct the pole of the opposite side that lies on the same side as the vertex:

• A' = pole of arc BC, on the same side as A
• B' = pole of arc CA, on the same side as B
• C' = pole of arc AB, on the same side as C

The triangle A'B'C' formed by these three poles is called the polar triangle and A'B'C' is its polar triangle.

Theorem (Reciporcity of Polarity)

If A'B'C' is the polar triangle of ABC, then ABC is also the polar triangle of A'B'C'.

Proof:
Since A' is the pole of BC, every point on BC is 90° from A'. In particular:

$\text{arc } A'B = 90° \quad \text{and} \quad \text{arc } A'C = 90°$

Similarly, since B' is the pole of CA:
$\text{arc } B'C = 90° \quad \text{and} \quad \text{arc } B'A = 90°$

Now consider arc B'C'. Since B is 90° from both A' and C', B is the pole of arc A'C'.

Similarly C is the pole of A'B', and A is the pole of B'C'. Therefore ABC is the polar triangle of A'B'C'.

Relations Between Sides and Angles of Polar Triangles

If the sides and angles of ABC are a, b, c and A, B, C, and the sides and angles of A'B'C' are a', b', c' and A', B', C', then:
$a' = 180° - A, \quad b' = 180° - B, \quad c' = 180° - C$
$A' = 180° - a, \quad B' = 180° - b, \quad C' = 180° - c$

Part 2. Fundamental Formulae

Cosine Formula

Statement: In a spherical triangle ABC with sides a, b, c and angles A, B, C:

$\cos a = \cos b \cos c + \sin b \sin c \cos A$

Similarly:
$\cos b = \cos c \cos a + \sin c \sin a \cos B$
$\cos c = \cos a \cos b + \sin a \sin b \cos C$

Proof:

Let O be the centre of the sphere of unit radius.
Let ABC be a spherical triangle with sides a, b, c opposite to vertices A, B, C respectively.

At vertex A, draw tangents to the two sides AB and AC in the planes of the great circles AB and AC respectively. Let these tangents meet the radii OB and OC extended at points P and Q respectively.

Since OA ⊥ AP and OA ⊥ AQ (tangent ⊥ radius):

In triangle OAP:
$OP^2 = OA^2 + AP^2 = 1 + AP^2$
$\cos c = \frac{OA}{OP}$
$\implies OP = \frac{1}{\cos c}, \quad AP = \tan c$

In triangle OAQ:
$\cos b = \frac{OA}{OQ} \implies OQ = \frac{1}{\cos b}, \quad AQ = \tan b$

The angle PAQ = A (the spherical angle at A, measured by angle between tangents).

Now applying the plane cosine rule in triangle OPQ:
$PQ^2 = OP^2 + OQ^2 - 2 \cdot OP \cdot OQ \cdot \cos a$
$PQ^2 = \frac{1}{\cos^2 c} + \frac{1}{\cos^2 b} - \frac{2\cos a}{\cos b \cos c} \quad \cdots (1)$

Applying the plane cosine rule in triangle APQ:
$PQ^2 = AP^2 + AQ^2 - 2 \cdot AP \cdot AQ \cdot \cos A$
$PQ^2 = \tan^2 c + \tan^2 b - 2\tan b \tan c \cos A \quad \cdots (2)$

Equating (1) and (2):
$\frac{1}{\cos^2 c} + \frac{1}{\cos^2 b} - \frac{2\cos a}{\cos b \cos c} = \tan^2 c + \tan^2 b - 2\tan b \tan c \cos A$

Using $\frac{1}{\cos^2\theta} = 1 + \tan^2\theta$:

$1 + \tan^2 c + 1 + \tan^2 b - \frac{2\cos a}{\cos b \cos c} = \tan^2 c + \tan^2 b - 2\tan b \tan c \cos A$

$2 - \frac{2\cos a}{\cos b \cos c} = -2\tan b \tan c \cos A$

$\frac{\cos a}{\cos b \cos c} = 1 + \tan b \tan c \cos A$

$\boxed{\cos a = \cos b \cos c + \sin b \sin c \cos A}$

Supplemental Cosine Formula

How it's derived: Apply the cosine formula to the polar triangle A'B'C' of ABC.

In the polar triangle, sides and angles satisfy: $a' = 180° - A, \quad b' = 180° - B, \quad c' = 180° - C$
$A' = 180° - a, \quad B' = 180° - b, \quad C' = 180° - c$

Applying the cosine formula to the polar triangle:
$\cos a' = \cos b' \cos c' + \sin b' \sin c' \cos A'$

Substituting:
$\cos(180° - A) = \cos(180° - B)\cos(180° - C) + \sin(180° - B)\sin(180° - C)\cos(180° - a)$

Using $\cos(180° - x) = -\cos x$ and $\sin(180° - x) = \sin x$:

$-\cos A = (-\cos B)(-\cos C) + \sin B \sin C(-\cos a)$

$-\cos A = \cos B \cos C - \sin B \sin C \cos a$

$\boxed{\cos A = -\cos B \cos C + \sin B \sin C \cos a}$

Therefore In a spherical triangle ABC:

$\cos A = -\cos B \cos C + \sin B \sin C \cos a$

Similarly:
$\cos B = -\cos C \cos A + \sin C \sin A \cos b$
$\cos C = -\cos A \cos B + \sin A \sin B \cos c$

Sine Formula

Statement: In a spherical triangle ABC:

$\frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C}$

Proof:

From the cosine formula:
$\cos a = \cos b \cos c + \sin b \sin c \cos A$

Rearranging:
$\cos A = \frac{\cos a - \cos b \cos c}{\sin b \sin c}$

Now using the identity $\sin^2 A = 1 - \cos^2 A$:

$\sin^2 A = 1 - \left(\frac{\cos a - \cos b \cos c}{\sin b \sin c}\right)^2$

$\sin^2 A = \frac{\sin^2 b \sin^2 c - (\cos a - \cos b \cos c)^2}{\sin^2 b \sin^2 c}$

Expanding the numerator:
$= \sin^2 b \sin^2 c - \cos^2 a + 2\cos a \cos b \cos c - \cos^2 b \cos^2 c$

$= (1 - \cos^2 b)(1 - \cos^2 c) - \cos^2 a + 2\cos a \cos b \cos c - \cos^2 b \cos^2 c$

$= 1 - \cos^2 b - \cos^2 c - \cos^2 a + 2\cos a \cos b \cos c$

So:
$\sin^2 A = \frac{1 - \cos^2 a - \cos^2 b - \cos^2 c + 2\cos a \cos b \cos c}{\sin^2 b \sin^2 c}$

Therefore:
$\frac{\sin^2 A}{\sin^2 a} = \frac{1 - \cos^2 a - \cos^2 b - \cos^2 c + 2\cos a \cos b \cos c}{\sin^2 a \sin^2 b \sin^2 c}$

The right side is symmetric in a, b, c. It does not change if we replace a with b or c. Therefore:

$\frac{\sin^2 A}{\sin^2 a} = \frac{\sin^2 B}{\sin^2 b} = \frac{\sin^2 C}{\sin^2 c}$

Taking square roots (all values positive in a spherical triangle):

$\frac{\sin A}{\sin a} = \frac{\sin B}{\sin b} = \frac{\sin C}{\sin c}$

Which gives:

$\boxed{\frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C}}$

Formula for Half Angle

Notation: Let $s = \dfrac{a + b + c}{2}$ (semi-perimeter of the spherical triangle).

Derivation:

From the cosine formula:
$\cos A = \frac{\cos a - \cos b \cos c}{\sin b \sin c}$

Using the identity $\cos a = 1 - 2\sin^2\dfrac{a}{2}$ and $\cos a = 2\cos^2\dfrac{a}{2} - 1$:

For $\sin\dfrac{A}{2}$:

Using $1 - \cos A = 2\sin^2\dfrac{A}{2}$:

$2\sin^2\frac{A}{2} = 1 - \frac{\cos a - \cos b \cos c}{\sin b \sin c}$

$= \frac{\sin b \sin c - \cos a + \cos b \cos c}{\sin b \sin c}$

$= \frac{\cos(b - c) - \cos a}{\sin b \sin c}$

Using sum-to-product: $\cos P - \cos Q = 2\sin\dfrac{P+Q}{2}\sin\dfrac{Q-P}{2}$:

$\cos(b-c) - \cos a = 2\sin\frac{a+b-c}{2}\sin\frac{a-b+c}{2}$

$= 2\sin(s-c)\sin(s-b)$

where $s - c = \dfrac{a+b-c}{2}$ and $s - b = \dfrac{a+c-b}{2}$.

Therefore:
$\sin^2\frac{A}{2} = \frac{\sin(s-b)\sin(s-c)}{\sin b \sin c}$

$\boxed{\sin\frac{A}{2} = \sqrt{\frac{\sin(s-b)\sin(s-c)}{\sin b \sin c}}}$

For $\cos\dfrac{A}{2}$:

Using $1 + \cos A = 2\cos^2\dfrac{A}{2}$:

$2\cos^2\frac{A}{2} = 1 + \frac{\cos a - \cos b \cos c}{\sin b \sin c}$

$= \frac{\sin b \sin c + \cos a - \cos b \cos c}{\sin b \sin c}$

$= \frac{\cos a - \cos(b+c)}{\sin b \sin c}$

$= \frac{2\sin\frac{a+b+c}{2}\sin\frac{b+c-a}{2}}{\sin b \sin c}$

$= \frac{2\sin s \sin(s-a)}{\sin b \sin c}$

Therefore:
$\boxed{\cos\frac{A}{2} = \sqrt{\frac{\sin s \sin(s-a)}{\sin b \sin c}}}$

For $\tan\dfrac{A}{2}$:

Dividing:
$\boxed{\tan\frac{A}{2} = \sqrt{\frac{\sin(s-b)\sin(s-c)}{\sin s \sin(s-a)}}}$

All three together:

$\sin\frac{A}{2} = \sqrt{\frac{\sin(s-b)\sin(s-c)}{\sin b \sin c}}$

$\cos\frac{A}{2} = \sqrt{\frac{\sin s \cdot \sin(s-a)}{\sin b \sin c}}$

$\tan\frac{A}{2} = \sqrt{\frac{\sin(s-b)\sin(s-c)}{\sin s \cdot \sin(s-a)}}$

where $s = \dfrac{a+b+c}{2}$.

By symmetry, similar formulas hold for $\dfrac{B}{2}$ and $\dfrac{C}{2}$ by cyclic permutation.

Formula for Half a Side

Notation: Let $S = \dfrac{A + B + C}{2}$ (semi-sum of angles of the spherical triangle).

Derivation:

These follow directly by applying the half angle formulas to the polar triangle A'B'C', exactly as the supplemental cosine formula was derived from the cosine formula.

In the polar triangle:
$a' = 180° - A, \quad b' = 180° - B, \quad c' = 180° - C$
$A' = 180° - a, \quad s' = \frac{a'+b'+c'}{2} = \frac{(180°-A)+(180°-B)+(180°-C)}{2}$
$s' = 270° - S$

Therefore:
$s' - a' = 270° - S - (180° - A) = 90° - (S - A)$
$s' - b' = 90° - (S - B)$
$s' - c' = 90° - (S - C)$

Applying $\sin\dfrac{A'}{2}$ formula to the polar triangle:

$\sin\frac{A'}{2} = \sqrt{\frac{\sin(s'-b')\sin(s'-c')}{\sin b' \sin c'}}$

Substituting $A' = 180° - a$, so $\dfrac{A'}{2} = 90° - \dfrac{a}{2}$, giving $\sin\dfrac{A'}{2} = \cos\dfrac{a}{2}$:

$\cos\frac{a}{2} = \sqrt{\frac{\sin(90°-(S-B))\sin(90°-(S-C))}{\sin(180°-B)\sin(180°-C)}}$

$\boxed{\cos\frac{a}{2} = \sqrt{\frac{\cos(S-B)\cos(S-C)}{\sin B \sin C}}}$

Similarly applying $\cos\dfrac{A'}{2}$ formula:

$\boxed{\sin\frac{a}{2} = \sqrt{\frac{-\cos S \cdot \cos(S-A)}{\sin B \sin C}}}$

And dividing:

$\boxed{\tan\frac{a}{2} = \sqrt{\frac{-\cos S \cdot \cos(S-A)}{\cos(S-B)\cos(S-C)}}}$

All three together:

$\sin\frac{a}{2} = \sqrt{\frac{-\cos S \cdot \cos(S-A)}{\sin B \sin C}}$

$\cos\frac{a}{2} = \sqrt{\frac{\cos(S-B)\cos(S-C)}{\sin B \sin C}}$

$\tan\frac{a}{2} = \sqrt{\frac{-\cos S \cdot \cos(S-A)}{\cos(S-B)\cos(S-C)}}$

where $S = \dfrac{A+B+C}{2}$.

By symmetry, similar formulas hold for $\dfrac{b}{2}$ and $\dfrac{c}{2}$ by cyclic permutation.

Sine-Cosine Formula

Statement: In a spherical triangle ABC: $\sin a \cos B = \cos b \sin c - \sin b \cos c \cos A$ $\sin a \cos C = \cos c \sin b - \sin c \cos b \cos A$

By symmetry, similar formulas hold for the other corresponding parts by cyclic permutation of $a, b, c$ and $A, B, C$.

Proof: From the cosine formula, we know: $\cos b = \cos c \cos a + \sin c \sin a \cos B$

Substitute the value of $\cos a$ from its cosine formula ($\cos a = \cos b \cos c + \sin b \sin c \cos A$): $\cos b = \cos c (\cos b \cos c + \sin b \sin c \cos A) + \sin c \sin a \cos B$ $\cos b = \cos b \cos^2 c + \sin b \sin c \cos c \cos A + \sin c \sin a \cos B$

Transposing $\cos b \cos^2 c$ to the left side: $\cos b (1 - \cos^2 c) = \sin b \sin c \cos c \cos A + \sin c \sin a \cos B$ $\cos b \sin^2 c = \sin b \sin c \cos c \cos A + \sin c \sin a \cos B$

Dividing the entire equation by $\sin c$ (which is strictly positive in a proper spherical triangle, so $\sin c \neq 0$): $\cos b \sin c = \sin b \cos c \cos A + \sin a \cos B$

Rearranging the terms yields the final formula: $\boxed{\sin a \cos B = \cos b \sin c - \sin b \cos c \cos A}$

Supplemental Sine-Cosine Formula

Statement: Also known as the Relation between Three Angles and Two Sides. $\sin A \cos b = \cos B \sin C + \sin B \cos C \cos a$

Derivation: Applying the Sine-Cosine formula to the polar triangle $A'B'C'$ of $ABC$: The sides and angles of the polar triangle satisfy: $a' = 180^\circ - A$, $b' = 180^\circ - B$, $c' = 180^\circ - C$ $A' = 180^\circ - a$, $B' = 180^\circ - b$, $C' = 180^\circ - c$

Apply $\sin a' \cos B' = \cos b' \sin c' - \sin b' \cos c' \cos A'$: $\sin(180^\circ - A) \cos(180^\circ - b) = \cos(180^\circ - B) \sin(180^\circ - C) - \sin(180^\circ - B) \cos(180^\circ - C) \cos(180^\circ - a)$

Using the basic trigonometric identities $\sin(180^\circ - \theta) = \sin \theta$ and $\cos(180^\circ - \theta) = -\cos \theta$: $\sin A (-\cos b) = (-\cos B) \sin C - \sin B (-\cos C) (-\cos a)$ $-\sin A \cos b = -\cos B \sin C - \sin B \cos C \cos a$

Multiplying both sides by $-1$: $\boxed{\sin A \cos b = \cos B \sin C + \sin B \cos C \cos a}$

Cotangent Formula (Four Parts Rule)

The Cotangent formula connects four consecutive parts (two sides and two angles) of a spherical triangle when read in a cyclical order along the perimeter.

Let the four consecutive parts be $(Side_1, Angle_{inner}, Side_{inner}, Angle_2)$.

Statement: $\cos(\text{inner side}) \cos(\text{inner angle}) = \sin(\text{inner side}) \cot(\text{other side}) - \sin(\text{inner angle}) \cot(\text{other angle})$

For example, considering the sequence of consecutive parts $a, C, b, A$:

• Inner side = $b$
• Inner angle = $C$
• Other side = $a$
• Other angle = $A$

Applying the rule: $\cos b \cos C = \sin b \cot a - \sin C \cot A$

Note: This is one of the most powerful formulas for practical calculations in astronomy and navigation as it relates four adjacent variables without requiring an auxiliary triangle.

Napier's Analogies

These formulas beautifully relate the semi-sums and semi-differences of the angles to the semi-sums and semi-differences of the sides. They are particularly useful for solving oblique spherical triangles.

1. $\tan\frac{A-B}{2} = \frac{\sin\frac{a-b}{2}}{\sin\frac{a+b}{2}} \cot\frac{C}{2}$

2. $\tan\frac{A+B}{2} = \frac{\cos\frac{a-b}{2}}{\cos\frac{a+b}{2}} \cot\frac{C}{2}$

3. $\tan\frac{a-b}{2} = \frac{\sin\frac{A-B}{2}}{\sin\frac{A+B}{2}} \tan\frac{c}{2}$

4. $\tan\frac{a+b}{2} = \frac{\cos\frac{A-B}{2}}{\cos\frac{A+B}{2}} \tan\frac{c}{2}$

De-Almbret's Analogies (Gauss's Formulas)

Similarly, these equations relate trigonometric functions of half-angles to half-sides. They differ from Napier's primarily by using sines and cosines instead of tangents.

1. $\frac{\sin\frac{a-b}{2}}{\sin\frac{c}{2}} = \frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}}$

2. $\frac{\sin\frac{a+b}{2}}{\sin\frac{c}{2}} = \frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}$

3. $\frac{\cos\frac{a-b}{2}}{\cos\frac{c}{2}} = \frac{\sin\frac{A+B}{2}}{\cos\frac{C}{2}}$

4. $\frac{\cos\frac{a+b}{2}}{\cos\frac{c}{2}} = \frac{\cos\frac{A+B}{2}}{\sin\frac{C}{2}}$

Part 3. Right-Angled Triangles

Spherical Triangle (Right-Angled Case)

Definition: A right-angled spherical triangle is a spherical triangle in which at least one of its angles is exactly $90^\circ$ (a quadrant).

Suppose $C = 90^\circ$. Then $\sin C = 1$ and $\cos C = 0$. Applying these values to the fundamental general formulas drastically simplifies them. For example:

• From Cosine Law: $\cos c = \cos a \cos b + \sin a \sin b \cos(90^\circ) \implies \cos c = \cos a \cos b$
• From Sine Law: $\frac{\sin c}{\sin 90^\circ} = \frac{\sin a}{\sin A} \implies \sin a = \sin c \sin A$

However, deriving these continuously can be tedious. To synthesize these simplified identities systematically, we use Napier's Rules.

Napier's Rule of Circular Parts

John Napier formulated an elegant mnemonic mechanism to easily reconstruct any right-angled spherical triangle identity without algebraic derivation from scratch.

Let $C = 90^\circ$. We ignore $C$ and arrange the other five parts in a circular wheel. When forming this circle, the sides adjacent to the right angle ($a$ and $b$) are taken as they are, but the remaining three parts ($A, c, B$) are replaced by their complements ($90^\circ - A$, $90^\circ - c$, $90^\circ - B$).

The five Circular Parts are: $a, b, (\frac{\pi}{2} - A), (\frac{\pi}{2} - c), (\frac{\pi}{2} - B)$

Often annotated as: $a, b, A', c', B'$ (where the prime denotes the $90^\circ$ complement).

The Rules: Pick any of the five parts to be the Middle Part.

• The two parts situated immediately next to it on the circle are the Adjacent Parts.
• The two remaining parts on the circle are the Opposite Parts.

1. Rule 1: The sine of the middle part is equal to the product of the tangents of the adjacent parts.

   • mnemonic: sine = product of tangents (Adjacents) $\sin(\text{Mid}) = \tan(\text{Adj}_1) \times \tan(\text{Adj}_2)$

2. Rule 2: The sine of the middle part is equal to the product of the cosines of the opposite parts.

   • mnemonic: sine = product of cosines (Opposites) $\sin(\text{Mid}) = \cos(\text{Opp}_1) \times \cos(\text{Opp}_2)$

Question Bank

1-Mark Questions

Definitions:

1. Define a great circle.
2. Define a small circle.
3. Define the pole of a circle on a sphere.
4. Define the axis of a circle on a sphere.
5. Define a secondary to a great circle.
6. Define spherical radius of a small circle.
7. Define a spherical angle.
8. Define a spherical triangle.
9. Define the polar triangle of a spherical triangle.
10. Define spherical excess.
11. Define a right-angled spherical triangle.
12. What are Napier's Circular Parts?

Short facts:

1. What is the angular distance from a pole of a great circle to any point on it?
2. How many great circles pass through two antipodal points?
3. What is the range of A + B + C in a spherical triangle?
4. What is the range of a + b + c in a spherical triangle?
5. State the relation between sides and angles of a polar triangle.
6. Write the arc length of a small circle subtending angle φ at its pole with spherical radius θ.
7. If C = 90° in a spherical triangle, write the value of cos c using Napier's Rule.
8. Write $\tan\dfrac{A}{2}$ in terms of sides.
9. Write $\tan\dfrac{a}{2}$ in terms of angles.
10. State the sine formula for a spherical triangle.
11. State the cosine formula for a spherical triangle.
12. What is the supplemental cosine formula?
13. Write the cotangent formula (four parts formula).
14. How many secondaries can be drawn to a great circle?
15. If the angles of a spherical triangle are A, B, C, write the sides of its polar triangle.

Some other:

1. Can the sum of two sides of a spherical triangle be greater than 360°? Why?
2. Is every diameter of a sphere the axis of infinitely many circles? Justify.
3. If A + B + C = 270° in a spherical triangle, what is the spherical excess?
4. Can all three angles of a spherical triangle be obtuse?
5. If a = b = c = 90°, what are the angles of the spherical triangle?
6. What happens to the cosine formula when the triangle is very small (plane limit)?
7. In a right-angled spherical triangle with C = 90°, if c = 90°, what can you say about A and B?
8. If the polar triangle of ABC is equilateral, what can you say about ABC?

5-Mark Questions

Standard theorems:

1. Prove that a pole of a circle is equidistant from every point on its circumference.
2. Prove that the angular distance from a pole of a great circle to any point on it is 90°.
3. Prove that if a point is at 90° from two distinct points on a great circle, it is a pole of that great circle.
4. Prove that all secondaries to a great circle are perpendicular to it.
5. Prove that the arc of a secondary intercepted between a primary great circle and any parallel small circle equals the spherical radius of the small circle.
6. Prove the reciprocity of polarity.
7. Prove that 180° < A + B + C < 540° for a spherical triangle.
8. Prove that 0° < a + b + c < 360° for a spherical triangle.
9. Prove that the greater side has the greater angle opposite to it.
10. State and prove the supplemental cosine formula using the polar triangle.

Formula derivations:

1. Derive $\sin\dfrac{A}{2}$ and $\cos\dfrac{A}{2}$ in terms of sides.
2. Derive $\tan\dfrac{A}{2}$ in terms of sides.
3. Derive $\tan\dfrac{a}{2}$ in terms of angles.
4. Derive the sine-cosine formula.
5. Derive the cotangent formula.
6. Verify any one of Napier's analogies.
7. Verify any one of De-Almbret's analogies.
8. Derive the relation between three sides and two angles.

Numerical:

1. In a spherical triangle, $a = 60°$, $b = 90°$, $C = 90°$. Find c.
2. If $a = b = c = 60°$, find angle A using the cosine formula.
3. In a spherical triangle, $A = B = C = 120°$. Find side a using the supplemental cosine formula.
4. Find $\tan\dfrac{A}{2}$ if $a = 90°$, $b = 60°$, $c = 60°$.
5. In a right-angled spherical triangle with $C = 90°$, $a = 60°$, $b = 45°$, find c using Napier's Rule.
6. Using the sine formula, find B if $a = 60°$, $A = 90°$, $b = 45°$.

Additional:

1. If two sides of a spherical triangle are equal, prove that the angles opposite to them are also equal.
2. Prove that in a spherical triangle, $\cos a = \cos b \cos c$ when $A = 90°$.
3. If A + B + C = 360°, what can you say about the triangle? Is it possible?
4. Show that the polar triangle of an equilateral spherical triangle is also equilateral.
5. Prove that if each side of a spherical triangle is a quadrant (90°), each angle is also 90°.
6. Two secondaries to a great circle intercept an arc of 60° on it. What is the angle between the secondaries?

7 or 8-Mark Questions

Derivations:

1. Derive the cosine formula for a spherical triangle from first principles.
2. Derive the sine formula for a spherical triangle.
3. Derive all three half angle formulas ($\sin\dfrac{A}{2}$, $\cos\dfrac{A}{2}$, $\tan\dfrac{A}{2}$).
4. Derive all three half side formulas ($\sin\dfrac{a}{2}$, $\cos\dfrac{a}{2}$, $\tan\dfrac{a}{2}$).
5. State and prove all four of Napier's analogies.
6. State and prove De-Almbret's analogies.
7. Derive the cotangent formula and state its use.
8. State and prove Napier's Rule of Circular Parts for right-angled spherical triangles.
9. Derive the supplemental sine-cosine formula.
10. Starting from the cosine formula alone, derive the sine formula and the supplemental cosine formula.

Combined derivation + application:

1. Derive the cosine formula and use it to find side a when $b = 75°$, $c = 60°$, $A = 45°$.
2. State and prove the sine formula. Hence find C when $a = 90°$, $A = 90°$, $c = 60°$.
3. Derive half angle formulas and hence find all angles when $a = 5°$, $b = 7°$, $c = 8°$ (small triangle approximation).
4. Derive Napier's analogies and use them to solve a triangle with $a = 70°$, $b = 50°$, $C = 80°$.
5. Prove Napier's Rule and solve the right-angled triangle with $C = 90°$, $a = 50°$, $B = 65°$.

Additional:

1. In a spherical triangle ABC, prove that $\cos a - \cos b \cos c = \sin b \sin c \cos A$. Hence deduce the sine formula.
2. If the angles of a spherical triangle are in the ratio 1:2:3, find the sides using appropriate formulas.
3. Prove that in any spherical triangle, $\dfrac{\cos A + \cos B \cos C}{\sin B \sin C} = \dfrac{\cos a - \cos b \cos c}{\sin b \sin c}$.
4. A spherical triangle has $A = B$. Using Napier's analogies, prove that $a = b$.
5. Derive the relation between three sides and two angles, and hence find angle B when $a = 90°$, $b = 60°$, $c = 90°$, $A = 90°$.
6. Show that for a spherical triangle on a unit sphere, as the sides tend to zero, the cosine formula reduces to the plane cosine rule.
7. Using De-Almbret's analogies, solve a spherical triangle completely given $A = 80°$, $B = 70°$, $c = 50°$.
8. If in a spherical triangle $a + b + c = 180°$, prove that $\tan\dfrac{A}{2}\tan\dfrac{B}{2} + \tan\dfrac{B}{2}\tan\dfrac{C}{2} + \tan\dfrac{C}{2}\tan\dfrac{A}{2} = 1$.