1.6: Trigonometry and Radians

1.6.1: Sine Rule

School textbooks tend to state the Sine Rule for a triangle \(ABC\) without worrying why it is true. So they often fail to give the result in its full form:

Theorem \(\PageIndex{1}\)

If \(R\) is the radius of the circumcircle of the triangle \(ABC\), then

\(\dfrac{a}{\sin{A}} = \dfrac{b}{\sin{B}} = \dfrac{c}{\sin{C}} = 2R\)

This full form explains that the three ratios

\(\dfrac{a}{\sin{A}}, \dfrac{b}{\sin{B}}, \dfrac{c}{\sin{C}}\)

are all equal because they are all equal to the diameter \(2R\) of the circumcircle of \(\triangle ABC\) – an additional observation which may well suggest how to prove the result (see Problem 32).

Problem 32

Given any triangle \(ABC\), construct the perpendicular bisectors of the two sides \(AB\) and \(BC\). Let these two perpendicular bisectors meet at \(O\).

(a) Explain why \(OA = OB = OC\).

(b) Draw the circle with centre \(O\) and with radius \(OA\). There are three possibilities:

(i) The centre \(O\) lies on one of the sides of triangle \(ABC\).

(ii) The centre \(O\) lies inside triangle \(ABC\).

(iii) The centre \(O\) lies outside triangle \(ABC\).

Case (i) leads directly to the Sine Rule for a right angled triangle \(ABC\) (remembering that \(\sin{90} = 1\)). We address case (ii), and leave case (iii) to the reader.

(ii) Extend the line \(BO\) to meet the circle again at the point \(A^{\prime}\). Explain why \(\angle BA^{\prime}C = \angle BAC = \angle A\), and why \(\angle A^{\prime}CB\) is a right angle. Conclude that

\(\sin{A} = \dfrac{BC}{A^{\prime}B} = \dfrac{a}{2R}\)

and hence that

\(\dfrac{a}{\sin{A}} = 2R \ \ \ \ \left( = \dfrac{b}{\sin{B}} = \dfrac{c}{\sin{C}}\right) \)

Problem 33

Let \(\triangle = \text{area}(\triangle ABC)\).

(a) Prove that

\(\Delta=\dfrac{1}{2} \cdot a b \cdot \sin C\)

(b) Prove that \(4R\triangle = abc\).

1.6.2: Radians and Spherical Triangles

There is no God-given unit for measuring distance; different choices of unit give rise to answers that are related by scaling. However the situation is different for angles. In primary and secondary school we measure turn in degrees – where a half turn is \(180^{\circ}\), a right angle is \(90^{\circ}\), and a complete turn is \(360^{\circ}\). This angle unit dates from the ancient Babylonians (~ 2000 BC). We are not sure why they chose \(360\) units in a full turn, but it seems to be related to the approximate number of days in a year (the time required for the heavens to make a complete rotation in the night sky), and to the fact that they wrote their numbers in “base \(60\)”. However the choice is no more objectively mathematical than measuring distance in inches or in centimetres.

After growing up with the idea that angles are measured in degrees, we discover towards the end of secondary school that:

there is another unit of measure for angles – namely radians.

It may not at first be clear that this is an entirely natural, God-given unit. The size of, or amount of turn in, an angle at the point \(A\) can be captured in an absolute way by drawing a circle of radius \(r\) centred at the point \(A\), and measuring the arc length which the angle cuts off on this circle. The angle size (in radians) of the angle at \(A\) is then defined to be the ratio

\(\dfrac{\text{arc length}}{\text{radius}}\).

That is,

\(\text{size of angle at the point} \ A = \text{arc length cut off on a circle of radius 1 centred at the apex} \ A\)

Hence a right angle is of size \(\dfrac{\pi}{2}\) radians; a half turn is equal to \(\pi\) (radians); a full turn is equal to \(2\pi\) (radians); each angle in an equilateral triangle is equal to \(\dfrac{\pi}{3}\) (radians); the three angles of a triangle have sum \(\pi\); and the angles of a polygon with \(n\) sides have sum \((n - 2)\pi\) (see Problem 230 in Chapter 6).

For a while after the introduction of radians we continue to emphasise the word radians each time we give the measure of an angle in order to stress that we are no longer using degrees. But this is not really a switch to a new unit: this new way of measuring angles is in some sense objective – so we soon drop all mention of the word “radians” and simply refer to the size of an angle (in radians) as if it were a pure number.

This switch affects the meaning of the familiar trigonometric functions. And though we continue to use the same names (\(\sin, \cos, \tan\), etc.), they become slightly different as functions, since the inputs are now always assumed to be in radians.

The real payoff for making this change stems from the way it recognizes the connection between angles and circles. This certainly makes calculating circular arc lengths and areas of sectors easy (an arc with angle \(\theta\) on a circle of radius \(r\) now has length \(\theta r\); and a circular sector with angle \(2\theta\) now has area \(\theta r^{2}\)). But the main benefit – which one hopes all students appreciate eventually – is that this change of perspective highlights the fundamental link between \(\sin{x}, \cos{x}\), and \(e^{x}\):

• "\(\cos{x}\)" becomes the derivative of \(\sin{x}\)
• "\(-\sin{x}\)" becomes the derivative of \(\cos{x}\), and
• the three functions are related by the totally unexpected identity

\(e^{i\theta} = \cos{\theta} + i\sin{\theta}\)

The next problem draws attention to a beautiful result which reveals, in a pre-calculus, pre-complex number setting, a beautiful consequence of thinking about angles in terms of radians. The goal is to discover a formula for the area of a spherical triangle in terms of its angles and \(\pi\), which links the formula for the circumference of a circle with that for the surface area of a sphere.

Suppose we wish to do geometry on the sphere. There is no problem deciding how to make sense of points. But it is less clear what we mean by (straight) lines, or line segments.

Before making due allowance for the winds and the tides, an airline pilot and a ship’s Captain both need to know how to find the shortest path joining two given points \(A\), \(B\) on a sphere. If the two points both lie on the equator, it is plausible (and correct) that the shortest route is to travel from \(A\) to \(B\) along the equator. If we think of the equator as being in a horizontal plane through the centre \(O\) of the sphere, then we may notice that we can change the equator into a circle of longitude by rotating the sphere so that the “horizontal” plane (through \(O\)) becomes a “vertical” plane (through \(O\)). So we may view two points \(A\) and \(B\) which both lie on the same circle of longitude as lying on a “vertical equator” passing through \(A\), \(B\) and the North and South poles: the shortest distance from \(A\) to \(B\) must therefore lie along that circle of longitude.

If we now rotate the sphere through some other angle, we get a “tilted equator” passing through the images of the (suitably tilted) points \(A\) and \(B\): these “tilted equators” are called great circles. Each great circle is the intersection of the sphere with a plane through the centre \(O\) of the sphere.

So

to find the shortest path from \(A\) to \(B\):

• take the plane determined by the points \(A\), \(B\) and the centre of the sphere \(O\);
• find the great circle where this plane cuts the sphere;
• then follow the arc from \(A\) to \(B\) along this great circle.

Once we have points and line segments (i.e. arcs of great circles) on the sphere, we can think about triangles, and about the angles in such a triangle. In a triangle \(ABC\) on the sphere, the sides \(AB\) and \(AC\) are arcs of great circles meeting at \(A\). By rotating the sphere we can imagine \(A\) as being at the North pole; so the two sides \(AB\) and \(AC\) behave just like arcs of two circles of longitude emanating from the North pole. In particular, we can measure the angle between them (this is exactly how we measure longitude): the two arcs \(AB\), \(AC\) of circles of longitude set off from the North pole \(A\) in different horizontal directions before curving southwards, and the angle between them is the angle between these two initial horizontal directions (that is, the angle between the plane determined by \(O\), \(A\), \(B\) and the plane determined by \(O\), \(A\), \(C\)).

Problem 34

Imagine a triangle \(ABC\) on the unit sphere (with radius \(r = 1\)), with angle \(\alpha\) between \(AB\) and \(AC\), angle \(\beta\) between \(BC\) and \(BA\), and angle \(\gamma\) between \(CA\) and \(CB\). You are now in a position to derive the remarkable formula for the area of such a spherical triangle.

Figure \(\PageIndex{1}\): Angles on a sphere

(a) Let the two great circles containing the sides \(AB\) and \(AC\) meet again at \(A^{\prime}\). If we imagine \(A\) as being at the North pole, then \(A^{\prime}\) will be at the South pole, and the angle between the two great circles at \(A^{\prime}\) will also be \(\alpha\). The slice contained between these two great circles is called a lune with angle \(\alpha\).

(i) What fraction of the surface area of the whole sphere is contained in this lune of angle \(\alpha\)? Write an expression for the actual area of this lune.

(ii) If the sides \(AB\) and \(AC\) are extended backwards through \(A\), these backward extensions define another lune with the same angle \(\alpha\), and the same surface area. Write down the total area of these two lunes with angle \(\alpha\).

(b) (i) Repeat part (a) for the two sides \(BA\), \(BC\) meeting at the vertex \(B\), to find the total area of the two lunes meeting at \(B\) and \(B^{\prime}\) with angle \(\beta\).

(ii) Do the same for the two sides \(CA\), \(CB\) meeting at the vertex \(C\), to find the total area of the two lunes meeting at \(C\) and \(C^{\prime}\) with angle \(\gamma\).

(c) (i) Add up the areas of these six lunes (two with angle \(\alpha\), two with angle \(\beta\), and two with angle \(\gamma\)). Check that this total includes every part of the sphere at least once.

(ii) Which parts of the sphere have been covered more than once? How many times have you covered the area of the original triangle \(ABC\)? And how many times have you covered the area of its sister triangle \(A^{\prime}B^{\prime}C^{\prime}\)?

(iii) Hence find a formula for the area of the triangle \(ABC\) in terms of its angles – \(\alpha\) at \(A\), \(\beta\) at \(B\), and \(\gamma\) at \(C\)

1.6.3: Polar Form and sin(A +B)

The next problem is less elementary than most of Chapter 1, but is included here to draw attention to the ease with which the addition formulae in trigonometry can be reconstructed once one knows about the polar form representation of a complex number. Those who are as yet unfamiliar with this material may skip the problem – but should perhaps remember the underlying message (namely that, once one is familiar with this material, there is no need ever again to get confused about the trig addition formulae).

Problem 35

(a) You may know that any complex number \(z = \cos{\theta} + i\sin{\theta}\) of modulus \(1\) (that is, which lies on the unit circle centred at the origin) can be written in the modulus form \(z = e^{i\theta}\). Use this fact to reconstruct in your head the trigonometric identities for \(\sin{(A + B)}\) and for \(\cos{(A + B)}\). Use these to derive the identity for \(\tan{(A + B)}\).

(b) By choosing \(X, Y\) so that \(A = \dfrac{X + Y}{2}\), and \(B = \dfrac{X-Y}{2}\), use part (a) to reconstruct the standard trigonometric identities for

\(\sin{X} + \sin{Y}, \sin{X} - \sin{Y}, \cos{X} + \cos{Y}, \cos{X} - \cos{Y}\)

(c) (i) Check your answer to (a) for \(\sin{(A + B)}\) by substituting \(A = 30^{\circ}\), and \(B = 60^{\circ}\).

(ii) Check your answer to (b) for \(\cos{X} - \cos{Y}\) by substituting \(X = 60^{\circ}\), and \(Y = 0^{\circ}\).

(d) (i) If \(A + B + C + D = \pi\), prove that

\(\sin{A}\sin{B} + \sin{C}\sin{D} = \sin{(B+ C)}\sin{(B+D)}\)

(ii) Given a cyclic quadrilateral \(WXYZ\), with \(\lt XWY = A\), \(\lt WXZ = B\), \(\lt YXZ = C\), \(\lt WYX = D\), deduce Ptolemy’s Theorem:

\(WX \times YZ + WZ \times XY = WY \times XZ\)