Skip to main content
SearchLoginLogin or Signup

Gamma Trigonometry : Applications of Extended Sine and Cosine Functions to Engineering

Published onSep 13, 2022
Gamma Trigonometry : Applications of Extended Sine and Cosine Functions to Engineering

Ever since sailors aligned quadrants with stars to guide themselves through the perils of the oceans (as illustrated in Figure 1), the need to project angles into distances (and vice versa) has been an inevitability for humanity in its endeavor for territorial and technological expansion. In modern times, this need still exists and is again present during the navigation of the SpaceX Dragon “Axiom-1” capsule as it made its final approach (with its four astronauts) into the International Space Station on April 9th, 2022.

Figure 1. Sailors determined angles using Quadrants to pinpoint their location on oceans

Trigonometric functions such as sin(α)\sin(\alpha) and cos(α)\cos(\alpha) are tools used to determine side lengths of a right triangle resulting from the orthogonal projection of the hypotenuse about a given internal reference angle α\alpha. Being inherently connected to the Pythagoras theorem, they only function encompassed in right-angle triangles (or in triangles where one of its internal angles is  90 degrees). If the right-angle  transformed into an angle of 60 or 120 degrees, what would these functions look like (Figure 2)? The projected lengths of the hypotenuse given by the traditional sine and cosine would change ― this time dependent on both angles α\alpha  and γ\gamma ― or simply put, into sin(α,γ)\sin^*(\alpha,\gamma)  and cos(α,γ)\cos^*(\alpha,\gamma)  [where the asterisk sign * implies that they are different from the original sine and cosine].

A typical approach to finding the side lengths and angles of a scalene triangle is to subdivide it into two right triangles and employ trigonometric operations. The resorting to orthogonality to solve a non-orthogonal problem is not an organic approach, and a more direct trigonometric and mathematical path is desired. If a mathematician or scientist is faced with a problem where he/she needs to know the (normalized) sides of a scalene triangle based on its angles, then the functions sin(α)\sin(\alpha) and cos(α)\cos(\alpha) [valid only for right triangles] need to be replaced by more generic expressions, herein termed the gamma or extended sine and cosine functions — denoted as sin(α,γ)\sin^*(\alpha,\gamma) and cos(α,γ)\cos^*(\alpha,\gamma), respectively.

Figure 2. Extending the notion of sine and cosine functions into scalene triangles

These extended functions  expand the usefulness of application of the already existing trigonometric functions in a variety of scientific fields, ranging from orbital mechanics [1], electronics [2], chemistry [3] and design [4]. Both the Pythagoras’ theorem and trigonometry (in general) form part of most secondary education curricula around the world, including in the American, Canadian and Australian Curriculum [5-7], which makes this paper of interest to both students and professionals.

1. Theory

1.1 Governing Equation

The Law of Cosines [written below in Eq.(1)] is a broad expression that relates the lengths of the three sides 𝑥, 𝑦 and 𝑧 of any triangle [8,9], which not only covers the particular case of the Pythagoras theorem 𝑥2+𝑦2=𝑧2𝑥^2+𝑦^2=𝑧^2 (where orthogonality defines the internal reference angle as γ=π/2\gamma=\pi/2 resulting in a right-angled triangle), but also all other possible values of 𝛾 that result in angle and distance relation within a scalene triangle.

z2=y2+x22xycos(γ)(1)z^2=y^2+x^2-2xycos(\gamma) \qquad(1)

The particular case of the Pythagoras theorem is satisfied by replacing γ=90\gamma=90 deg, x=cos(α)x=\cos(\alpha), y=sin(α)y=\sin(\alpha) and z=1z=1. Similarly, it was proven in [10] that the general case of the Law of Cosines is fulfilled by the extended expressions (defined below), implying a scalene triangle with sides 𝑥, 𝑦 and 𝑧 (as in Figure 3).

Figure 3. Angles and sides of a scalene triangle

1.2 Extended Functions

It has been proven [10] that the expressions for the extended (or also called gamma) sine function sin(α,γ)\sin^*(\alpha,\gamma) and the extended (or gamma) cosine function cos(α,γ)\cos^*(\alpha,\gamma) applicable to scalene triangles (Figure 2) [where the extended hypotenuse is normalized (i.e., z=1z=1)] are given as

y=sin(α,γ)=sin(α)sin(γ)(2)y=\sin^*(\alpha,\gamma)=\frac{\sin(\alpha)}{\sin(\gamma)} \qquad(2)

x=cos(α,γ)=cos(α)+cos(γ)sin(γ)sin(α)=sin(α+γ)sin(γ)(3)x=\cos^*(\alpha,\gamma)=\cos(\alpha)+\frac{\cos(\gamma)}{\sin(\gamma)}\sin(\alpha)=\frac{\sin(\alpha+\gamma)}{\sin(\gamma)} \qquad (3)

1.3 Identity Rules

In trigonometry, the angle sum and difference identity rules establish a link between additive or subtractive operations in angles and its impact on the lengths of their respective right triangles, and are commonly defined as

sin(A±B)=sin(A)cos(B)±cos(A)sin(B)(4)cos(A±B)=cos(A)cos(B)sin(A)sin(B)(5)\sin(A \pm B)=\sin(A)\cos(B) \pm \cos(A)\sin(B) \qquad (4)\\ \cos(A \pm B)=\cos(A)\cos(B) \mp \sin(A)\sin(B) \qquad (5)

This is only applicable to a triangle with an obtuse angle γ=90\gamma=90 deg — i.e. a right-angled triangle. In a recent publication [11], the definition of these rules has been extended to encompass scalene triangles, which accommodate a broader application including scalene triangles with any obtuse angle. For the summing of angles of the identity rule (where α=A+B\alpha=A+B)[Figure 4], these were proven to be

sin(A+B,γ)=sin(A,γ)cos(B,γ)+cos(A,πγ)sin(B,γ)(6)cos(A+B,γ)=cos(A,γ)cos(B,γ)sin(A,πγ)sin(B,γ)(7)\sin^*(A+B,\gamma)=\sin^*(A,\gamma)\cos^*(B,\gamma)+\cos^*(A,\pi-\gamma)\sin^*(B,\gamma)\quad (6) \\ \cos^*(A+B,\gamma)=\cos^*(A,\gamma)\cos^*(B,\gamma)-\sin^*(A,\pi-\gamma)\sin^*(B,\gamma)\quad (7)

Figure 4. An angle summation within a scalene triangle and associated projections [11]

For the angle difference identity rule (where α=AB\alpha=A-B)[Figure 5], these were proven to be

sin(AB,γ)=sin(A,γ)cos(B,γ)cos(Aπ+2γ,πγ)sin(B,γ)(8)cos(AB,γ)=cos(A,γ)cos(B,γ)+sin(Aπ+2γ,πγ)sin(B,γ)(9)\sin^*(A-B,\gamma)=\sin^*(A,\gamma)\cos^*(B,\gamma)-\cos^*(A-\pi+2\gamma,\pi-\gamma)\sin^*(B,\gamma)\quad (8) \\ \cos^*(A-B,\gamma)=\cos^*(A,\gamma)\cos^*(B,\gamma)+\sin^*(A-\pi+2\gamma,\pi-\gamma)\sin^*(B,\gamma) \quad (9)

Note that the drawings in Figure 4 and 5 are practically the same, except that the scalene triangle ABC\triangle ABC is upwards in Figure 4 (adding angle B to A), and it is downwards in Figure 5 (subtracting angle B from A).

Figure 5. An angle subtraction within a scalene triangle and associated projections [11]

This article concerns itself in the application of these equations (in Chapter 3); further details on their construction and proofs can be found in the respective publication.

2. Purpose

There are several ways to compute problems in trigonometry, and the point of this article is to solve the problems using the explicitly proposed (above) extended sine and cosine functions (section 1.2) and their identity rules (section 1.3), which have a broad application to scalene triangles, much in the same manner as conventional sine and cosine functions are applicable to right triangles (hence enabling a scientist or mathematician the possibility to, when the orthogonal condition fails, to replace the conventional sine and cosine by their extended versions immediately and effortlessly — culminating in a more flexible mathematical solution). Moreover, the new extended trigonometric functions open doors for possibilities of expanding scientific fields in which they (the functions) are the foundations. Some possibilities will be presented and discussed briefly.

2. Exercises & Possibilities

In order to assist the confirmation of the solutions of the following trigonometric problems, the open-source program Geogebra [12] can be used to draw the involved scalene triangles, as well as, in determining their lengths and angles.

2.1 Civil Engineering

2.1.1 Inclined Bridge

Problem. A suspension bridge has a series of main suspension cables connecting the horizontal road deck (through which the automobiles transit) to the vertical tower, making an angle γ=90\gamma=90 deg. The main suspension cables transmit tension forces and connect from anchors (concrete blocks at shore — located out of the photo to the left beyond point A) at each of their ends to the top of the intermediate towers [13]. The two connection points A and B have an original average diagonal cable length of 2km, where the cables are not completely stretched, making an angle of α=30\alpha=30 degrees with the horizontal. Over the years, the foundations yielded and the tip of the tower at point B is now a half a degree inclined in the direction away from the cables (the change of inclination from B to B’ was measured at point C by an engineer with a theodolite). How much have the cables stretched due to the inclination?

Figure 6. Assessment of cable elongation due to foundation inclination

Solution. The horizontal projection AC is assumed unchanged by the inclination. Hence, this is the starting point, as it is common to both cases straight ABC\triangle ABC and inclined ABC\triangle AB'C. At the beginning, we have a right triangle, hence the relation applies straightforward

AC=ABcos(α)=2cos(30)=1.732kmAC = AB\cos(\alpha) = 2\cos(30) = 1.732km

However, the inclination changes the right triangle (ACBγ=90\angle ACB \equiv \gamma=90 deg) to a scalene triangle ( with an angle change to ACBγ=90+0.5=90.5\angle ACB' \equiv \gamma'=90+0.5=90.5 deg), hence sine and cosine no longer apply directly. The extended versions given by Eq.(2) and Eq.(3) are applied in this case. The diagonal of the scalene triangle is given directly by the horizontal projection AC of the longest side AB´ giving

AB=ACsin(α+γ)sin(γ)=1.732sin(30+90.5)sin(90.5)=2.010kmAB' = \frac {AC} { \frac{\sin(\alpha+\gamma') }{ \sin(\gamma')} } = \frac {1.732} {\frac{\sin(30+90.5)}{\sin(90.5)} } = 2.010km

Therefore, the inclination of half a degree by the tower has stretched the main suspension cables by 10 meters.

2.1.2 Theodolites

Problem. Imagine that civil engineers with theodolites devices (Figure 7 left) are placed at either side of a river (located at points A and B), and wish to measure their distance to a location on the bridge (located at point C), which is inaccessible due to constructions. Among other things, Theodolites devices measure vertical and horizontal angles between visual reference points with great accuracy [14]. Point C is located on the bridge and directly over water, so it is not possible to measure the horizontal positions of point A and B to point C, and neither the height. However, the engineers can measure the distance between them, resulting in 200m. Engineer A rotates the theodolite from point B to C and measures a vertical angular distance of 38 degrees, while engineer B does the same from point A to C and measures 24 degrees. What are the distances of both engineers to point C on the bridge? What is the height from point C to the horizontal line connecting engineer A to B? And if the same line AB is inclined by ϵ=10\epsilon =10 degrees (α\alpha and β\beta remaining the same), what is the height of point C to line AB?

Figure 7. Finding distances from the shores to a point on an inaccessible bridge

Solution. The angle γ=ACB\gamma=\angle ACB formed by the two theodolites at point C is 180-38-24=118 degrees. The distance of the first engineer A to point C on the bridge is given by the direct application of the extended cosine function as

AC=200cos(38,118)=200[sin(38+118)sin(118)]=92.1mAC = 200 \cos^*(38,118) = 200 \Big[ \frac{\sin(38+118)}{\sin(118)} \Big] = 92.1m

On the other hand, the distance of the second engineer B to point C on the bridge is given by direct application of the extended sine function as

BC=200sin(38,118)=200[sin(38)sin(118)]=139.5mBC = 200 \sin^*(38,118) = 200 \Big[ \frac{ \sin(38) }{ \sin(118) } \Big] = 139.5m

Knowing the distance and angle from A to C, it is simply a matter of applying conventional sine function to obtain the vertical projection in the right triangle AOC\triangle AOC as

CO=92.1sin(38)=56.7mCO=92.1 \sin(38) =56.7m

If the line AB were to be inclined, making an angle ϵ=10\epsilon=10 degrees to the horizontal (here we assumed it is inclined to the left, as per Figure 8) — α=38\alpha=38 degrees and β=24\beta=24 degrees remain the same then the height of point C to line AB is (whilst remembering that the obtuse angle BOC\angle BO'C of triangle BOC\triangle BO'C in now 90+ϵ=10090+\epsilon=100 degrees) given by the direct application of the extended sine function, as

CO=BCsin(β,100)=139.5[sin(24)sin(100)]=57.6mCO'=BC\sin^*(\beta,100)=139.5 \Big[ \frac{ \sin(24) }{ \sin(100) } \Big] = 57.6m

Note that the added inclination (of 10 degrees) between the two theodolites A and B increased the vertical distance of point C to the line AB by almost 1 meter.

Figure 8. Same case as in Figure 6, but with line AB inclined by 10 degrees

2.2 Aerospace & Aeronautics

2.2.1 Turbine Velocity Diagrams

Problem. Velocity vector diagrams are tools used by aerospace and aeronautics engineers to design and understand the loading of a turbine stage on an aircraft engine (Figure 9 left) [15]. It links the magnitude and direction of air velocity vectors (both absolute VABSV_{ABS} and relative VRELV_{REL} at the vane exit to rotor exit) with blade tangential speed UU (Figure 9 right). Imagine that the flow exiting a row of stator vanes has been measured to be 250 m/s at an absolute angle to the engine axis of θ=70\theta=70 deg. The engineer selects a rotational speed for the rotor of 7,000 rpm. The mean blade diameter is 0.5m. What is the relative velocity VRELV_{REL} seen by the rotor as the flow enters it?

Figure 9. Velocity vector diagram for a turbine blade in a turbofan engine

Solution. The analysis revolves around the scalene triangle ABC\triangle ABC. At the stator exit plane, the air absolute velocity vector VABSV_{ABS} makes an angle with the blade speed vector UU of ACBα=1809070=20\angle ACB \equiv \alpha=180-90-70=20 deg. The rotational speed 7,000 rpm is converted to mean blade tangential velocity UU as

U=7000602πD=157.1m/sU=\frac{7000}{60}*2 \pi D =157.1 m/s

Finding the angle γABC\gamma \equiv \angle ABC

VABScos(α,γ)=VABS[cos(α)+cos(γ)sin(γ)sin(α)]V_{ABS}\cos^*(\alpha,\gamma)=V_{ABS} \Big[ \cos(\alpha)+\frac{\cos(\gamma)}{\sin(\gamma)}\sin(\alpha) \Big]

which, by replacing the values above, results in


That re-arranged gives


Note that, computing the arctangent of angle γ gives -47.7 degrees. Since we know that the angle γ needs to be obtuse, we add 180 degrees to give γ=132.3\gamma=132.3 degrees (the tangent is the same for both angles). To find the relative velocity VRELV_{REL}, one simply applies the sin(α,γ)\sin^*(\alpha,\gamma) expression (i.e., project side AC onto AB), which gives

VREL=VABSsin(20,132.3)=250[sin(20)sin(132.3)]=115.6m/sV_{REL}=V_{ABS}\sin^*(20,132.3)= 250\Big[ \frac{\sin(20)}{\sin(132.3)} \Big] = 115.6m/s

If the absolute angle of VABSV_{ABS} to the engine axis θ\theta is in fact 2 degrees more than anticipated, what is the new relative velocity VRELV_{REL} (use the angle difference identity rule)?

The new angle is θ=70+2=72\theta=70+2=72 deg, which based on the internal angle sum of the triangle ADC\triangle ADC, gives a angle α=1809072=202=18\alpha=180-90-72=20-2=18 deg. For the same mean blade tangential velocity UU, the obtuse angle of triangle ABC\triangle ABC becomes


As before the resulting obtuse angle is γ=136.2\gamma=136.2 deg. To find the relative velocity VRELV_{REL}, one applies the angle sum identity difference rule for sin(AB,γ)\sin^*(A-B,\gamma) [given by Eq.(8)]

VREL=VABSsin(202,136.2)=157.1[sin(202,136.2)]V_{REL}=V_{ABS}\sin^*(20-2,136.2)=157.1 \big[ \sin^*(20-2,136.2) \big]

which expands to

sin(202,136.2)=sin(20,136.2)cos(2,136.2)cos(2180+2×136.2,180136.2)sin(2,136.2)\sin^*(20-2,136.2)=\sin^*(20,136.2)\cos^*(2,136.2)- \\ \qquad \qquad -\cos^*(2-180+2\times 136.2,180-136.2)\sin^*(2,136.2)

By applying the extended sine and cosine functions given by Eq.(2) and Eq.(3), the end result is VREL=104.4m/sV_{REL}=104.4m/s. The two degree increase of the inlet absolute angle represents a reduction of the perceived relative velocity for the rotor of approximately 11m/s11m/s.

2.2.2 Aircraft Flight Path

Problem. Two engineers placed theodolites (height 1.7m) at a distance of 400m apart over a runway to record the flight path of aircraft taking off. One aircraft departs and climbs directly over them. As it crosses the sky, the engineers measure at either side the vertical angle for different synchronized times. The usage of electro-optical system to track the position of aircraft is common in the world of aviation [16]. The result are a succession of scalene triangles, as illustrated in Figure 10 (with the recorded angles presented on the right). What is the distance of the aircraft to each of the points A and B in the ground? Subsequently, what is the height of the aircraft at the different points in time (with respect to the theodolites measuring position)?

Figure 10. Measurement of aircraft flight trajectory during climb

Solution. Drawing lines from the three elements — engineers A, B and the aircraft — created a series of scalene triangles that change based on the time of flight of the aircraft. Using the three ground angles for both α\alpha and β\beta measured with the theodolites, the third internal angle γ\gamma of each scalene triangle was computed to be 116 degrees for T1, 109 degrees for T2 and 79 degrees for T3. The solution for each aircraft position with associated time is computed directly from the extended formula. For time T1, the distance from engineer A to the aircraft is

AT1=400cos(19,116)=400[sin(19+116)sin(116)]=191.3mAT_1 = 400 \cos^*(19,116) = 400 \Big[ \frac{sin(19+116)}{sin(116)} \Big] = 191.3m

and the distance to engineer B is

BT1=400sin(19,116)=400[sin(19)sin(116)]=144.9mBT_1 = 400 \sin^*(19,116) = 400 \Big[ \frac{ \sin(19) }{ \sin(116) } \Big] = 144.9m

and the aircraft is at a vertical distance of

h1=314.7.9sin(19)=102.5mh_1 = 314.7.9 \sin(19) = 102.5m

For time T2, the aircraft is at a distance from engineer A of

AT2=400cos(32,109)=400[sin(32+109)sin(109)]=224.2mAT_2 = 400 \cos^*(32,109) = 400 \Big[ \frac{\sin(32+109)}{\sin(109)} \Big] = 224.2m

and a distance from engineer B of

BT2=400sin(32,109)=400[sin(32)sin(109)]=266.2mBT_2 = 400 \sin^*(32,109) = 400 \Big[ \frac{ \sin(32) }{ \sin(109) } \Big] = 266.2m

while the aircraft is at a vertical distance of

h2=224.2sin(32)=141.1mh_2 = 224.2 \sin(32) = 141.1m

For time T3, the distance from engineer A to the aircraft is

AT3=400cos(73,79)=400[sin(73+79)sin(79)]=191.3mAT_3 = 400 \cos^*(73,79) = 400 \Big[ \frac{\sin(73+79)}{\sin(79)} \Big] = 191.3m

and the distance from engineer B is

BT3=400sin(73,79)=400[sin(73)sin(79)]=389.7mBT_3 = 400 \sin^*(73,79) = 400 \Big[ \frac{ \sin(73) }{ \sin(79) } \Big] = 389.7m

and the aircraft is at a height of

h3=191.3sin(73)=182.9mh_3 = 191.3 \sin(73) = 182.9m

Now, if the aircraft projects a shadow on the ground at an angle of 20 degrees to the vertical, what is the distance from engineer A to the projection at point O (Figure 11)? If the reading of the theodolite at A increases by an angle of dα=10d\alpha=10 degrees as the aircraft climbs (here, we assume that the aircraft rotation or flight path is such that its relative distance to point A remains the same, i.e. AT2=AT2AT_2=AT_2'), what is the new distance from engineer at point A to the new shadow at point O’ (use the angle sum identity rule)?

Figure 11. Measurement of the shadow location of the aircraft

The angle from the horizontal to the projected direction of the shadow is AOT2γ=90+20=110\angle AOT_2 \equiv \gamma=90+20=110 deg. For position T2, the extended cosine function gives directly the answer for the distance from point A to the projected shadow on the ground to be

AO=AT2cos(32,110)=224.2[sin(32+110)sin(110)]=146.9mAO = AT_2 \cos^*(32,110) = 224.2 \Big[ \frac{ \sin(32+110) }{ \sin(110) } \Big] = 146.9m

The fact that the relative distance of the aircraft does not change relative to point A means that, as the aircraft rotates in its climb, the distance AT2=AT2=224.2mAT_2=AT_2'=224.2m (in Figure 11) remains fixed. The extended angle sum identity rule [given by Eq.(7)] is used here to determine the new distance from engineer at point A to the new shadow at point O’. With the rotation, the original angle Aα2=32A\equiv\alpha_2=32 deg becomes A+B=α2+dα=32+10=42A+B=\alpha_2+d\alpha=32+10=42 deg. The new horizontal projection of the shadow becomes thus


The projected direction of the shadow is still AOT2γ=110\angle AOT_2' \equiv\gamma=110 deg (as the Sun’s rays are assumed parallel when they arrive at Earth). Expanding the above using Eq.(7) gives

AO=224.2[cos(32,110)cos(10,110)sin(32,180110)sin(10,110)]=AO'=224.2 \big[\cos^*(32,110)\cos^*(10,110)-\sin^*(32,180-110)\sin^*(10,110)\big]=

=224.2[sin(32+110)sin(110)sin(10+110)sin(110)sin(32)sin(180110)sin(10)sin(110)]=112m=224.2 \big[\frac{ \sin(32+110) }{ \sin(110) } \frac{ \sin(10+110) }{ \sin(110) }-\frac{ \sin(32) }{ \sin(180-110) } \frac{ \sin(10) }{ \sin(110) }\big]=112m

That is, the shadow moved from 146.9m146.9m to a distance from engineer A of 112m112m meters, a total distance covered of 34.9m34.9m. The advantage of Eq.(7) becomes evident when considering dαd\alpha as a variable (possibly part of a larger equation). Imagine that you would like to know where the shadow was (for both positions) for a different time of the day (say when the Sun was at an angle to the vertical of 10 deg, instead of 20 deg). Then, it is only required to re-apply the above equations, updating with the new Sun incidence angle of γ=90+10=100\gamma=90+10=100 degrees (here it is assumed that the aircraft flight path would remain the same).

2.2.3 Satellite Constellation Trajectories

Problem. Imagine three GPS satellites trailing each other on the same orbit. It is assumed that they operate best at a given (normalized) relative position to each other (as shown in Figure 12). It is worth noting that this is an abstraction, as in reality such a constellation of properly geometrically-spaced GPS satellites orbiting the Earth typically has 24 satellites disposed in a 3D configuration [17] — so this is a very simplified example. Here, each satellite has an optical sensor onboard that measures the satellite´s angle between the others two. What angles do the sensors of all three need to measure to guarantee that they are flying at the target (normalized) orbital distance to each other?

Figure 12. Satellite constellation flying in a specific target formation

Solution. Since the sum of the internal angles of a triangle is 180 degrees, all is required is to know two of them. This implies the need for two equations to solve for two variables. The first equation results from the cosine projection of the longest side of the scalene triangle as

BC=0.61=sin(α,γ)=sin(α)sin(γ)BC = 0.61 = \sin^*( \alpha,\gamma ) = \frac{\sin( \alpha )}{\sin( \gamma )}

which re-arranged gives

sin(γ)=sin(α)0.61(10)\sin( \gamma ) = \frac{\sin( \alpha )}{0.61} \qquad(10)

Similarly, the cosine projection of the longest side of the scalene triangles gives

AB=0.43=cos(α,γ)=cos(α)+sin(α)sin(γ)cos(γ)AB = 0.43 = \cos^*( \alpha, \gamma ) = \cos( \alpha )+ \frac{\sin( \alpha )}{\sin( \gamma )} \cos( \gamma )

By replacing the above result in

0.43=cos(α)+sin(α)sin(α)0.61cos(γ)0.43 = \cos( \alpha )+ \frac{\sin( \alpha )}{\frac{\sin( \alpha )}{0.61} } \cos( \gamma )

that simplifies to

cos(γ)=0.43cos(α)0.61(11)\cos( \gamma ) = \frac{0.43 - \cos(\alpha)}{0.61} \qquad(11)

The identity rule gives

sin2(γ)+cos2(γ)=1sin^2( \gamma ) + cos^2( \gamma ) = 1

Replacing Eq.(10) and Eq.(11) into the identity rule results in

(sin(α)0.61)2+(0.430.61cos(α)0.61)2=1\Big( \frac{\sin(\alpha)}{0.61} \Big) ^2 + \Big( \frac{0.43}{0.61} - \frac{cos(\alpha)}{0.61} \Big)^2 = 1

Expanding results in

(sin(α)0.61)2+(0.430.61)22(0.430.61)(cos(α)0.61)+(cos(α)0.61)2=1\Big( \frac{\sin(\alpha)}{0.61} \Big) ^2 + \Big( \frac{0.43}{0.61} \Big)^2 -2 \Big( \frac{0.43}{0.61} \Big) \Big( \frac{\cos(\alpha)}{0.61} \Big) + \Big( \frac{\cos(\alpha)}{0.61} \Big)^2 = 1

Placing the denominator to the right side simplifies this to

sin2(α)+0.4322(0.43)cos(α)+cos2(α)=0.612\sin^2(\alpha)+0.43^2-2 ( 0.43)\cos( \alpha ) + cos^2( \alpha ) = 0.61^2

which reduces further to

1+0.4322(0.43)cos(α)=0.6121 + 0.43^2 - 2 ( 0.43 ) cos( \alpha ) = 0.61^2

resulting in an expression for cos(α) as

cos(α)=1+0.4320.6122(0.43)cos( \alpha ) = \frac{ 1 + 0.43^2 - 0.61^2 }{ 2(0.43) }

Note that, computing the arcsine of angle γ gives 32.39 degrees which is in the first quadrant. Since we know that the angle γ needs to be obtuse, we are interested in the value in the second quadrant of 147.61 degrees which has the same value of sine. The third angle β then becomes

β=180αγ=18019.07147.61=13.32º\beta = 180 - \alpha - \gamma = 180 - 19.07 - 147.61 = 13.32 º

Therefore, in order for the satellites to fly at the target normalized distance to each other (as per values shown in Figure 12), the angular sensor in satellite A needs to measure 19.07 degrees (between satellite B and C), the one in satellite B needs to measure 147.61 degrees (between satellite A and C) and satellite C needs to read 13.32 degrees (between satellite A and B) .

2.4.2 Antenna Array Beamforming and Steering

Possibility. Imagine an antenna array composed of sensing elements uniformly spaced in a line. Beamforming is a process by which an interference pattern between the radiated or received signals of all the antenna elements allows the array to acquire directivity in reception/emission, by forming a high gain lobe (in which it is most sensitive to transmissions) located at the center of the array, and perpendicular to the array’s axis [18]. This process involves the use of sine functions and trigonometry, and hence we will expand further into its workings.

When two antenna elements are spaced by distance dd (Figure 13), and angled by α\alpha to the incoming transmitted wave front (each successive antenna elements will record the same wave with a delay), where this lag is trigonometrically governed by a right-angled triangle inherently governed by the Pythagoras theorem. This forms the basis for the processing required to calculate the array’s gain sensitivity to direction or beamforming. If X(t)X(t) is the signal received by the antenna element 0 (Figure 13), then antenna element 1 receives the same signal (as element 0) with a time delay τ\tau, expressed mathematically as

X1(t)X(tτ)X_1(t) \cong X(t-\tau)

The gain of an array is typically computed from the summation of all signals, and is used to construct the array radiation pattern, quantifying its directivity.

Figure 13. Time delay in reception of EM wave in a linear antenna array governed by right triangles

For these two particular elements, the time delay τ\tau occurs due to the waves angle of arrival α\alpha — i.e. the extra distance BCBC that the EM waves need to cover due to the inclination %alpha of the antenna array — thus forming with the distance d=BAd=BA in between elements a right triangle ABC\triangle ABC, and is defined as


where cc is the speed of light in the medium air (close to that in vacuum), and dd is the distance between successive elements. Beam steering is a process that provides the ability to change the direction of an incoming or outgoing beam on a uniform linear antenna array, and it is achieved by introducing an arbitrary time delay Δt\Delta t between each subsequent pair of antenna elements. This alters the signal received/emitted by element 1 into

X1(t)X(tτΔt)(12)X_1(t) \cong X(t-\tau-\Delta t) \quad (12)

In reality, this time delay Δt\Delta t can be interpreted as a distortion in the trigonometric relation between the distances composing the right triangle ABC\triangle ABC. Geometrically, the time delay Δt\Delta t is translated into an added distance, transforming BCBC (Figure 13) into BCBC' (Figure 14). This consequentially transforms the right triangle ABC\triangle ABC (Figure 13) into the scalene triangle ABC\triangle ABC' (Figure 14).

Figure 14. Time delay in reception of EM wave in a linear antenna array governed by scalene triangles

The new time delay τ\tau' that accounts for the artificial modification is


This formula indicates that τ(α,γ)\tau'(\alpha,\gamma) accounts for the two delay components: (1) delay between successive elements due to array inclination and element spacing — the original τ\tau [this is governed by angle %alpha and forms the right triangle ABC\triangle ABC], and (2) the fictitious delay introduced mathematically resulting in the artificial elongation/reduction of the wave travel distance (that compounds with the array elements) [the alteration of angle γ\gamma morphs the right triangle into a scalene triangle; the difference of the base of the two defines this delay]. Thus, the extended sine function enables the signal mathematical expression to change to

X1(t)X[tτ(α,γ)]withτ(α,γ)=δc[sin(α)sin(γ)](13)X_1(t) \cong X\big[ t-\tau'(\alpha,\gamma)\big] \qquad \textrm{with} \qquad \tau'(\alpha,\gamma)=\frac{\delta}{c} \Big[ \frac{\sin(\alpha)}{\sin(\gamma)} \Big] \quad (13)

where sin(α)\sin(\alpha) governs the directivity of the antenna array [via the control of angle α\alpha], and sin(γ)\sin(\gamma) governs the direction of the emitted/received signal [via the angle %gamma]. Hence, both the influence of τ\tau and Δt\Delta t can be modeled within one expression, and controlled via two angles α\alpha and γ\gamma. Since the fundamental operating principle of beamforming is the same between 2D and 3D, by legacy the application of the proposed modification (of the mathematical modelling using the extended sine function) could be extended to three dimensional beamforming.

2.4.3 Enhancing Data Aerial Transmission

Possibility. Binary digital modulation techniques (used in modern wireless data transmission) are simple in concept, but are not efficient in terms of their spectral density. Augmentation of spectral efficiency (i.e., boost transmission bit rate without affecting bandwidth requirements) is commonly achieved via the adoption of quaternary signaling schemes, like quadrature phase-shift keying (QPSK) [19]. In QPSK, the number of bits that are combined are 2 so this makes M=4. Quaternary signaling schemes embed information in carrier phase modifications, while at the same time keeping the carrier amplitude and frequency the same.

Digital signals Q and I are used to modulate a carrier wave by altering its phase in four possible ways (i.e., 45, 135, 225 and 315 degrees), each representing a symbol (Figure 15a). The in-phase signal I is along the x-axis in Figure 15a, and is represented in magenta in Figure 15b. The quadrature signal Q is along the y-axis in Figure 15a, and is represented in blue in Figure 15b. In an orthogonal system of axis, the x-axis is at 90 degrees to the y-axis, which means that the in-phase signal I is phase shifted by 90 degrees from the quadrature signal Q. The novelty introduced by the extended sine and cosine functions is in the ability to change the angle between the system’s axis (other than γ=90\gamma=90 deg), providing the potential to increase the keys available by having two (or more) interchangeable system of axis in operation. This is discussed in more detail as follows.

Figure 15. Quadrature Phase Shift Keying Modulation [QPSK]

The digital QPSK modulated signal is given by the equation

S(t)=AC{[sin(ϕ)sin(α)]Q+[cos(ϕ)cos(α)]I}(14)S(t) = A_C \Big\{ [ \sin( \phi ) \sin( \alpha) ]_Q + [ \cos ( \phi ) \cos( \alpha ) ]_I \Big\} \qquad(14)

where angle ϕ\phi controls the normalized amplitude of the carrier signals Q and I — such that when they add up, the carrier is perceived to undergo a phase shift (i.e. ϕ=45,135,225\phi=45, 135, 225 or 315315 deg) — and angle α=2πfct\alpha= 2 \pi f_c t governs the oscillation of the carrier and modulated waves with time. The amplitude of the signal is given as Ac=2E/TA_c=\sqrt{2E/T} , where E=P×TE=P \times T (i.e., power times time interval) is the energy content in a bit duration. In other words, changing ϕ\phi offsets in Figure 15b the dashed line horizontally back and forth. The coefficients that enable the carrier to shift by the above desired phases are given in Table 1.

Table 1. Coefficients of modulated signal for γ=90\gamma=90 deg

ϕ\phi (deg) - binary

mQ(t)=sin(ϕ)m_Q(t)=\sin( \phi )

mI(t)=cos(ϕ)m_I(t)=\cos ( \phi )

45 - (0)00



135 - (0)01



225 - (0)10



315 - (0)11



Until now, the phase distance between the carrier wave Q and I was fixed to 90 degrees, as sine and cosines are governed by the dynamics of a right-angled triangle. With the extended sine and cosines functions, this phase distance can be modified to an arbitrary value given by the angle γ. Since the system axes are now no longer static (frozen in orthogonal mode), this approach is hereby termed the Dynamic-Axis Quadrature Phase Shift Keying Modulation or DA-QPSK Modulation. Replacing the sine and cosine by their extended functions in Eq.(2) and Eq.(3), allows the x-axis to be at virtually any angle γ\gamma to the y-axis, which means that the in-phase signal I is phase shifted by angle γ\gamma from the quadrature signal Q.

Digital signals Q and I are used to modulate a carrier wave by altering its phase in four possible ways (i.e., 45, 135, 225 and 315 degrees), each representing a symbol (Figure 15a).

Figure 16a shows that by altering the angle between the two axes (in this case, to γ=120\gamma=120 degrees) allows the creation of additional four phases (i.e., 30, 120, 210 and 300 degrees) or symbols different from Figure 15a. Note that these phase angles are associated with a system of axis (i.e., γ=120\gamma=120 degrees) different from the orthogonal (i.e., γ=90\gamma=90 degrees), and thus a 120 degrees has different meaning in either system of axes. For the case where γ=120\gamma=120 degrees, the in-phase signal I has a 120 degree phase shift from the quadrature signal Q (Figure 16b).

Figure 16. Dynamic-Axis (γ=120°) Quadrature Phase Shift Keying Modulation [DC-QPSK]

The amplitude modulations of the quadrature signal Q and in-phase signal I produce the phase-shifting of the wave produced by their sum (dashed red in Figure 16b for ϕ=30\phi=30 deg implying symbol 100) — of the same frequency and amplitude as the carriers — is given as follows

S(t,γ,ϕ)=[sin(ϕ,γ)sin(α,γ)]Q+[cos(ϕ,γ)cos(α,γ)]I(15)S(t,\gamma,\phi)=[sin^*(\phi,\gamma)\sin^*(\alpha,\gamma)]_Q + [cos^*(\phi,\gamma)\cos^*(\alpha,\gamma)]_I \qquad(15)

where ϕ\phi is the phase shift of the modulated signal with respect to Q, and controls via the extended sine and cosine coefficients the amplitudes of both Q and I such that the their sum is always constant. Note that the generalized Eq.(15) encompasses the particular cases of Figure 15 where γ=90\gamma=90 deg and Figure 16 where γ=120\gamma=120 deg. In both cases it produces a wave, same as the carrier, except it has a phase offset. Equation (15) can be written in a more compact manner as

S(t,γ,ϕ)=[mQ(t)sin(α,γ)]Q+[mI(t)cos(α,γ)]I(16)S(t,\gamma,\phi)=[m_Q^*(t)\sin^*(\alpha,\gamma)]_Q + [m_I^*(t)\cos^*(\alpha,\gamma)]_I \qquad(16)

where the extended mQ(t)m_Q^*(t) and mI(t)m_I^*(t) functions are determined by expanding using Eq.(2) and Eq.(3) as

mQ(t)=sin(ϕ,γ)=sin(ϕ)sin(γ);mI(t)=cos(ϕ,γ)=sin(ϕ+γ)sin(γ)m_Q^*(t)=sin^*(\phi,\gamma)=\frac{\sin(\phi)}{\sin(\gamma)} \quad ; \quad m_I^*(t)=cos^*(\phi,\gamma)=\frac{\sin(\phi+\gamma)}{\sin(\gamma)}

For the particular case of an orthogonal axis system (γ=90\gamma=90 deg), the above functions reduce back to the original versions mQ(t,90)=sin(ϕ)m_Q^*(t,90)=\sin(\phi) and mI(t,90)=cos(ϕ)m_I^*(t,90)=\cos(\phi). As an example of its application, let us conceive a change in the angle formed between the y-axis for Q and x-axis for I to the particular case of γ=120 degrees. In such a case, the extended function mQ(t)m_Q^*(t) becomes

mQ(t)=sin(ϕ,120)=sin(ϕ)sin(120)=23sin(ϕ)(17)m_Q^*(t)=\sin^*(\phi,120)=\frac{\sin(\phi)}{\sin(120)}=\frac{2}{\sqrt{3}}\sin(\phi) \qquad (17)

and the extended function mI(t)m_I^*(t) becomes

mI(t)=cos(ϕ,120)=sin(ϕ+120)sin(120)=23{sin(ϕ)(12)+cos(ϕ)32}m_I^*(t)=\cos^*(\phi,120)=\frac{\sin(\phi+120)}{\sin(120)}=\frac{2}{\sqrt{3}} \Big\{ \sin(\phi)\big(-\frac{1}{2}\big)+\cos(\phi)\frac{\sqrt{3}}{2} \Big\}

which re-arranged further simplifies to

mI(t)=13sin(ϕ)+cos(ϕ)(18)m_I^*(t)=-\frac{1}{\sqrt{3}} \sin(\phi)+\cos(\phi) \qquad (18)

For the particular case of γ=120\gamma=120 deg, the signal expression in Eq.(15) transforms to

S(t,120,ϕ)=[{23sin(ϕ)}sin(α,120)]Q+[{13sin(ϕ)+cos(ϕ)}cos(α,120)]I(19)S(t,120,\phi)=\Big[\Big\{\frac{2}{\sqrt{3}}\sin(\phi)\Big\}\sin^*(\alpha,120)\Big]_Q + \Big[\Big\{-\frac{1}{\sqrt{3}} \sin(\phi)+\cos(\phi)\Big\}\cos^*(\alpha,120)\Big]_I \qquad(19)

Consider the case of the symbol 100 (in the first quadrant, where ϕ=30\phi=30 deg). Then the function mQ(t)m_Q^*(t) [given by Eq.(17)] becomes the fractional number


Similarly, the function mI(t)m_I^*(t) [given by Eq.(18)] becomes also a fractional number

mI(t)=13sin(30)+cos(30)=13(12)+32=33m_I^*(t)=-\frac{1}{\sqrt{3}} \sin(30)+\cos(30)=-\frac{1}{\sqrt{3}} \Big(\frac{1}{2}\Big)+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}

Thus, the coefficients that enable the carrier to shift by the above desired phases (i.e., 30, 120, 210 and 300 degrees) can be computed using the same approach as above, resulting in Table 2.

Table 2. Coefficients of modulated signal for γ=120\gamma=120 deg

ϕ\phi (deg) - binary



30 - (1)00



120 - (1)01



210 - (1)10



300 - (1)11



Figure 17 shows an example of the conceptual block diagram of the transmitter that could produce the Dynamic-Axis QPSK modulation. The key difference between DA-QPSK and QPSK modulation is the ability to alter the phase between the Q and I carrier waves beyond the customary 90 degrees, a choice that is controlled by the binary status of the first of the 3 Bits via an additional switch. Hence, in DA-QPSK, the number of bits that are combined are 3 so this makes M=8 (an improvement from QPSK where M=4).

Figure 17. Transmitter block diagram for Dynamic-Axis QPSK modulation

The reception and interpretation of the signal implies, for each phase change determine what is the magnitude. If it is an angle multiple of 90/2=45 degrees, then the first digit is 0. Likewise, if it is a multiple of 120/2=60 degrees, then the first digit is 1. So, the first digit is determined by the resulting factorization of the phase. Second, the other two digits are determine by the magnitude of the phase, and is located in the axes plus circle diagram in the same manner as for QPSK modulation. The only difference here, is that if the first bit is 1, then the 120 degree system of axis is used, and the location in the diagram will change accordingly to Eq.(17).

3.5 Vibration Theory

Possibility. Sine and cosine are functions that known to govern the oscillation of the most fundamental vibratory system — i.e., the pendulum and the mass-spring-damper system (Figure 18) [20].

Figure 18. Traditional mass-spring system

In such a system, the instantaneous conversion of kinetic energy KEKE into potential energy PEPE without losses means that the total energy TETE in the system remains the same. The governing equation (from an energy balance perspective) is

KE+PE=TE12mx˙2+12kx2=constant(20)KE+PE=TE \equiv \frac{1}{2}m\dot{x}^2+\frac{1}{2}kx^2=constant \qquad (20)

which, noting that there are two squared terms, it is equivalently represented by the Pythagoras theorem as an area balance

x2+y2=z2(21)x^2+y^2=z^2 \qquad(21)

where x=m/2x˙x=\sqrt{m/2}\dot{x}, y=k/2xy=\sqrt{k/2}x and zz is the square root of the total energy. Later, this constant will disappear with the differentiation. The Pythagoras Theorem is satisfied by the sine function y=sin(α)y=sin(\alpha) and cosine function x=cos(α)x=cos(\alpha), being expressed non-dimensional as

sin(α)2+cos(α)2=1(22)\sin(\alpha)^2+\cos(\alpha)^2=1 \qquad(22)

Diferentiating Eq.(20) provides the (unforced and undamped) governing equation in its mostly known formation

mx¨+kx=0(23)m\ddot{x}+kx=0 \qquad(23)

which has a solution of the type

x(t)=Acos(ωt)x(t)=A\cos(\omega t)

where A is an arbitrary constant to be defined for particular initial conditions of the system (e.g., holding the spring at a given height, and then releasing it).

Now, consider the possibility that there is a lag in converting kinetic into potential energy. The mass is the same, and the stiffness is the same. Only that kinetic energy does not convert into potential energy immediately, as there is a degree of freedom acting as a buffer in the system. It means that when the mass is at the energetic extremes, all the the energy in the system is converted into kinetic energy half-way in between the oscillations, and all the energy is transformed into potential at maximum displacement. However, in between these extremes part of the energy is stored in a buffer (to which the kinetic and potential energy of the mass-spring system is coupled).

If the Pythagoras theorem governs the energy balance in a conventional mass-spring system, then one that presents a lag is governed by the more general version of such an equation — the Law of Cosines. Prior research has shown that there are other extended version of the Pythagoras theorem using triangles [21] and hexagons [22] that differ from the former by the presence of a coupling element. Figure 19 shows in yellow the coupling area in both extended versions that make them different from the original Pythagoras theorem.

Figure 19. Pythagoras theorem side by side with extended version (left) using triangles [21] and (right) using hexagons [22]

In the case of a modified spring-mass system, the coupling is modeled by replacing the mass block by a spring with distributed mass along its length (Figure 20). This mass distribution has to do the coupling of inertia and stiffness that will be explained better later. The direction of displacement of this “buffer” spring is at an angle γ to the displacement direction of the main mass-spring system. The angle γ defines how much coupling exists between the buffer and the main system. For instance, if γ=90\gamma=90 deg then there is no coupling, and the “buffer” spring reduces to a rigid mass block.

Figure 20. Energy-coupled mass-spring system

Accounting for this coupling results in the modified energy governing equation as

KE+Coupling+PE=TE(24)KE+Coupling+PE=TE \qquad(24)

which can be more readily modeled and understood by replacing the particular case of the Pythagoras theorem by the general case of the Law of Cosines (that reduces to the Pythagoras theorem for γ=90\gamma=90 deg)

x22xycos(γ)+y2=z2(25)x^2 - 2 xy\cos( \gamma ) + y^2 =z^2 \qquad(25)

Note that the coupling term here is 2xycos(γ)-2xy\cos(\gamma), which contains a component of mass represented by variable xx and a component of stiffness represented by variable yy. The angle γ\gamma defines the amount of coupling between inertia (mass) and stiffness effects. Re-applying the connections defined for the case of the Pythagoras theorem (a particular case) where x=m/2x˙x=\sqrt{m/2}\dot{x} and y=k/2xy=\sqrt{k/2}x, the Law of Cosines would translate into

12mx˙2mkx˙xcos(γ)+12kx2=constant(26)\frac{1}{2} m{ \dot{x} }^2 - \sqrt {mk} { \dot{x} } x \cos( \gamma ) + \frac{1}{2} k{x}^2 =constant \qquad(26)

whose differentiation leads to the more generalized governing equation [more than Eq.(23)]. Note the appearance of the new coupling term mkx˙xcos(γ)- \sqrt {mk} { \dot{x} } x \cos( \gamma ) by comparing Eq.(26) to Eq.(23). The mathematical differentiation process is out of the scope for this paper, however, the solution to the governing equation is anticipated to be achieved by replacing the functions sine or cosine with their extended or gamma versions. Also, from a normalized perspective, the Law of Cosines is satisfied by the extended sine and cosine functions [x=cos(α)x=cos^*(\alpha), y=sin(α)y=sin^*(\alpha) and z=1z=1 are the normalized lengths of the sides of a scalene triangle], and expressed in non-dimensional form as

sin(α,γ)22sin(α,γ)cos(α,γ)cos(γ)+cos(α,γ)2=1(27)sin^*( \alpha,\gamma )^2 -2 sin^*( \alpha,\gamma ) cos^*( \alpha,\gamma ) cos( \gamma ) + cos^*( \alpha,\gamma )^2 =1 \qquad (27)

which was already proven to be true [19], and the extended sine and cosine functions have been shown to be given by Eq.(2) and Eq.(3), respectively. The angle γ\gamma quantifies the lag or coupling between the conversion of the two states of energy in the system. The mathematical details governing the energy-coupled mass-spring system, and its proof, have already been published [23]. Future addition of damping effects due to resistance, and the impact of this coupling on the analogous electrical RLC circuit, will be a subject for discussion in future articles.


[1] Curtis, H. D. (2010). Orbital Mechanics for Engineering Students 2nd Edition. Butterworth-Heinemann.

[2] Rawlins, J. C. (2000). Basic AC Circuits (2nd Edition). Newnes Publishing.

[3] Howard Mark, H. and Workman Jr. J. (2018). Chemometrics in Spectroscopy (2nd Edition). Academic Press.

[4] Parisher, R. A and Rhea, R. A. (2012). Pipe Drafting and Design (3rd Edition) Gulf Professional Publishing.

[5] Reys, B. J., Dingman, S., Nevels, N. & Teuscher, D. (2007). High School Mathematics: State-Level Curriculum Standards and Graduation Requirements. Center for the Study of Mathematics Curriculum.

[6] Canadian Ministry of Education (2020). The Ontario Curriculum, Grades 1–8: Mathematics.

[7] Australian Curriculum, Assessment and Reporting Authority. (n.d.). Australian curriculum: Mathematics F–10.

[8] Maor, E. (2007). The Pythagorean Theorem: A 4,000 Year History. Princeton University Press.

[9] Pickover, C. A. (2012). The Math Book: From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics. Sterling Milestones

[10] Teia, L. (2022). Extended Sine and Cosine Functions for Scalene Triangles. Journal of Mathematics Research.

[11] Teia, L. (2022). Extended Angle Sum and Difference Identity Rules for Scalene Triangles. Journal of Mathematics Research.

[12] Feng, G. T. (2013). Introduction to Geogebra – Version 4.4. 0249/Introduction_to_Introduction_to_GeoGebra

[13] Pipinato, A. (2021). Innovative Bridge Design Handbook: Construction, Rehabilitation, and Maintenance (2nd Edition). Elsevier.

[14] Topcon Corporation. (2016). DT-200 Series : Advanced Digital Theodolite.

[15] Haselbach, F., and Taylor, M., (2013). Axial Flow High Pressure Turbine Aerodynamic Design. Lecture Series—Von Karman Institute for Fluid Dynamics. Chap. 7

[16] Wuhan Huazhiyang Technology Corporation (2020). Security EOS Electro Optical Systems , Radar Tracking System For Vessel / Aircraft.

[17] Moorefield Jr., F. D. (2020) GPS Standard Positioning Service (SPS) Performance Standard. Office of the Department of Defense. U.S. Government.

[18] Liu, W. & Weiss, S. (2010). Wideband Beamforming - Concepts and Techniques. John Wiley & Sons Inc. New York

[19] Grami, A. (2016). Introduction to Digital Communications. Academic Press.

[20] Timoshenko, S. P., Young, D. H., & Weaver, W. (1974). Vibration Problems in Engineering (4th Edition) John Wiley & Sons Inc., New York

[21] Teia, L. (2021). Extended Pythagoras Theorem Using Triangles, and its Applications to Engineering. The Journal of Open Engineering. 

[22] Teia, L. (2022). Extended Pythagoras Theorem Using Hexagons. Journal of Mathematics Research.

[23] Teia, L. (2022). Gamma Derivatives of the Extended Sine and Cosine Functions for the Upgraded Mass-Spring Oscillatory System. Journal of Mathematics Research. (Reviewed and accepted. To be published soon)

Timothy Bond:


Luis Teia:

Corrected. Thanks

Timothy Bond:

Luis Teia:

There was a factor of 2 missing. Thanks, now it is corrected.

Timothy Bond:

check numbering

is this a new section?

Luis Teia:


Timothy Bond:

check numbering (following from previous)

Luis Teia:

All numbering was updated. Cheers

Timothy Bond:

check numbering

Luis Teia:

This was also updated. Thx

Timothy Bond:

use of ‘deg’ above instead of degree symbol?

Luis Teia:

Degree symbol inserted. Thanks

Timothy Bond:


I am not sure what this means, nor how this is supported by figure 1, and I can’t easily find a definition elsewhere.

Luis Teia:

The first two sentences were re-written to expand the meaning that was being conveyed. The point was that the relative position between satellites was assumed to have an optimum where they functioned best. This was further explained. Also, the figure number was corrected to the appropriate picture below. Cheers

Timothy Bond:

does not change with respect to

Luis Teia:

Corrected on new release. Cheers

Timothy Bond:


It may be helpful to include a diagram as part of your answer to this question

Luis Teia:

Diagram included to better clarify the solution to the problem.

Timothy Bond:

distance from engineers A to the projection at point O

Luis Teia:

Corrected. Cheers

Luis Teia:


Timothy Bond:

alpha comma gamma

Luis Teia:

Figure number updated. Thanks

Timothy Bond:

Check figure number

Timothy Bond:

Check figure number

Luis Teia:

Figure number updated

Timothy Bond:

“does not change relative to point”

Timothy Bond:

+ 1 more...
Timothy Bond:



Luis Teia:

Spelling modified as proposed.

Timothy Bond:


I think you need to indicate before here that the engineers have equal height (above the river?)


What if the engineers don’t have equal height?

Luis Teia:

That is a good challenge ! By adding an inclination to AB of say 10 degrees, the computation of the new height of C to the line AB can be achieved by re-applying the extended sine function to the triangle BOC (O being the base of the height), with the modification that the obtuse angle BOC is now 90 + 10 degrees. This results in the desired answer, which has been included in the new draft of the paper. Thanks for that !

Timothy Bond:

Check figure number

Luis Teia:

Updated. Cheers

Timothy Bond:


Anyone who knows the cosine rule knows the relative positions of the side lengths and the angle gamma, but it still frustrates when it isn’t explicitly indicated.

Up to you if you want to add a phrase in this paragraph to clarify where the angle is.

Luis Teia:

Yes, I see your point. To make it clear, a figure highlighting the location of the angles and sides x, y and z in a scalene triangle has been included. Cheers

Timothy Bond:


Luis Teia:


Timothy Bond:

Not sure if it’s required

desired perhaps?

Luis Teia:

Agreed and replaced

Timothy Bond:


“Dealing with” seems a little unclear - what are we actually doing with the scalene triangle?

I’m also not sure about the word ‘The” at the start - implies that there is only one way that these kinds of problems are solved. Maybe “A typical approach…”?

Ok, so “A typical approach to finding the side lengths and angles of a scalene triangle is to subdivide it into two right triangles and employ trigonometric operations.”

Luis Teia:

Indeed the words needed to be polished. I have replaced the sentence with your expression, for it is more framed to transmit the targeted intention. Thanks

Timothy Bond:


Missing angle notation?

Luis Teia:

Yes, here the program did not copy the angles from another draft holding the original draft. Angles added now. Thanks

Timothy Bond:


Something is missing in here

Luis Teia:

Values of angles 60 and 120 degrees added. Thanks for highlighting this.