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

Published onSep 13, 2022
Beyond 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 sine sin(α)\sin(\alpha) and cosine 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  or  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  and  ― or simply put, into  and  [where the asterisk sign * implies that they are different from the original sine and cosine].

The contemporary approach to dealing with a scalene triangle is to subdivide it into two right triangles and employ trigonometric operations to the two in order to find the lengths of the sides and angles of the scalene triangle. This resorting to orthogonality to solve non-orthogonal problems is not an organic approach, and a more direct trigonometric and mathematical path is required. If an 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 sine sin(α)\sin(\alpha) and cosine cos(α)\cos(\alpha) [valid only for right triangles] need to be replaced by a more generic expression herein termed the extended sine and cosine functions or sin(α,γ)\sin^*(\alpha,\gamma) and cos(α,γ)\cos^*(\alpha,\gamma).

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

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.

1.2 Extended Functions

It has been proved [10] that the expressions for the extended sine function sin(α,γ)\sin^*(\alpha,\gamma) and the extended 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 3], 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 3. An angle summation within a scalene triangle and associated projections

For the angle difference identity rule (where α=AB\alpha=A-B)[Figure 4], 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 3 and 4 are practically the same, except that the scalene triangle ABC\triangle ABC is upwards in Figure 3 (adding angle B to A), and it is downwards in Figure 4 (subtracting angle B from A).

Figure 4. An angle subtraction within a scalene triangle and associated projections

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 5. 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 1 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 850m. 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?

Figure 6. Finding distance from shores to a point on the inaccessible bridge

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

AC=850cos(38,118)=850[sin(38+118)sin(118)]=391.6mAC = 850 \cos^*(38,118) = 850 \Big[ \frac{\sin(38+118)}{\sin(118)} \Big] = 391.6m

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

BC=850sin(38,118)=850[sin(38)sin(118)]=592.7mBC = 850 \sin^*(38,118) = 850 \Big[ \frac{ \sin(38) }{ \sin(118) } \Big] = 592.7m

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=391.6sin(38)=241.1mCO=391.6 \sin(38) =241.1m

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 1 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 1 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 3,000 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 7. 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 1 (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? Sub-sequentially, what is the height of the aircraft at the different points in time?

Figure 8. 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 at which they were taken. While the three ground angles were measured with the theodolites, the third internal angle of each scalene triangle is 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 30 degrees to the vertical, what is the distance of engineer A to the projection at point O? If the vertical angular distance increases by another 10 deg 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), what is the new distance from engineer at point A to the shadow (use the angle sum identity rule)?

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 changing to point A means that, as the aircraft rotates in its climb, the distance AT2=224.2mAT_2=224.2m (in Figure 8) remains fixed. The extended angle sum identity rule [given by Eq.(7)] is used here to determine the added vertical angle as a result of the aircraft displacement. With the rotation, the original angle Aα2=32A\equiv\alpha_2=32 deg becomes A+B=32+10=42A+B=32+10=42 deg. The new horizontal projection of the shadow becomes therefore


The projected direction of the shadow is still γ=110\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]

That is, the shadow moved from 146.9m146.9m to a distance from engineer A of 110110 meters, a total of 36.9m36.9m. 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 of 10 deg, instead of 20 deg). Then, it is only needed to re-apply the above equations, updating with the new Sun incidence angle.

2.2.3 Satellite Constellation Trajectories

Problem. Three GPS satellites trailing each other on the same orbit operate best at an optimum normalized relative position to each other as shown in Figure 1. It is worth noting that in reality such a constellation of properly geometrically-spaced GPS satellites orbiting the Earth has typically 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 optimum orbital distance to each other?

Figure 9. Satellite constellation flying in optimum 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=13.32deg\beta = 180 - \alpha - \gamma = 180 - 19.07 - 147.61 = 13.32 deg

Therefore, in order for the satellites to fly in an optimum distance deployment, 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 10), 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 10), 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 10. Time delay in reception of EM wave in a linear antenna array governed by right triangles

For these two particular elements (Figure 10), 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 10) into BCBC' (Figure 11). This consequentially transforms the right triangle ABC\triangle ABC (Figure 10) into the scalene triangle ABC\triangle ABC' (Figure 11).

Figure 11. 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 12a). The in-phase signal I is along the x-axis in Figure 12a, and is represented in magenta in Figure 12b. The quadrature signal Q is along the y-axis in Figure 12a, and is represented in blue in Figure 12b. 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 12. 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 12b 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 12a).

Figure 13a shows that by altering the angle between the two axes (in this case, to γ=120\gamma=120 degrees) allows the creration of additional four phases (i.e., 30, 120, 210 and 300 degrees) or symbols different from Figure 12a. 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 13b).

Figure 13. 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 13b 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 12 where γ=90\gamma=90 deg and Figure 13 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 1 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 14. 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 [20].

Figure 15. 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=mx˙x=\sqrt{m}\dot{x} and y=kxy=\sqrt{k}x. 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 16 shows in yellow the coupling area in both extended versions that make them different from the original Pythagoras theorem.

Figure 16. Pythagoras theorem side by side with extended version (left) using triangles [21] and (right) using hexagons [22]

In the case of amodified spring-mass system, the coupling is modeled by replacing the mass block by a spring with distributed mass along its length (Figure 17). 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 17. 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 x 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=mx˙x=\sqrt{m}\dot{x} and y=kxy=\sqrt{k}x, the Law of Cosines would translate into

mx˙22mkx˙xcos(γ)+kx2=constant(26)m{ \dot{x} }^2 - 2 \sqrt {mk} { \dot{x} } x \cos( \gamma ) + k{x}^2 =constant \qquad(26)

whose differentiation leads to the more generalized governing equation [more than Eq.(23)]. Note the coupling term 2mkx˙xcos(γ)- 2 \sqrt {mk} { \dot{x} } x \cos( \gamma ) by comparing Eq.(26) to Eq.(23). The differentiation is out of the scope of this paper, however, the solution to the governing equation is anticipated to be a replacement of the sine or cosine functions by the extended 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.


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