MTH 122: Trigonometry and Analytical Geometry

Department of Statistics School of Applied Sciences Federal Polytechnic Orogun Delta State, Nigeria

Lecture 1

Trigonometry

• Trigonometry is the branch of mathematics that deals with how both the angles and sides of triangles relate.
• This is done via the use of basic ratios, Pythagoras theorem and unit circle, etc.
• Trigonometry is applied in several disciplines which include:
  • Enginering
  • Architecture
  • Physics
  • Astronomy
  • Navigation and surveying
  • Computer graphics
  • Statistics and many more.

Triangle

• Triangle is a polygon which has 3 sides and 3 angles.
• The angles are measured in degree while the sides are measured in metres.
• Triangles could also be formed within a circle as it relates to geometry. Figures above are examples of an equilateral and right-angled triangle respectively.

Basic Trigonometric Ratios

• Basic trigonometric ratios can be defined in two ways: (1) traditional and (2) modern

Traditional Definition

• The basic trigonometric ratios (BTR) can be defined in terms of the sides of a right-angled triangle.

Figure 1

• In triangle \(<ABC\), which is a right-angled triangle, \(<ABC = \theta\) and \(<CAB = 90^0\). We define the three basic trigonometric ratios as follows: \[sin\theta = \frac{opp}{hyp} = \frac{CA}{BC} = \frac{b}{a}\] \[cos\theta = \frac{adj}{hyp} = \frac{AB}{BC} = \frac{c}{a}\] \[tan\theta = \frac{opp}{adj} = \frac{CA}{AB} = \frac{b}{c}\]
• Obviously from above, \(tan\theta = \frac{sin\theta}{\cos\theta}\).

Modern Definition

• The modern definition is given in the light of the figure below:

Figure 2

• where \(P\) with coordinates \((x,y)\) is a variable point on unit circle, centre \(O\). \(OP\) makes angle of \(\theta\) with the positive values \(x-axis\). \(x\) can be considered as the projection of \(OP\) of unit length on \(x-axis\) and \(y\) can be considered as the vertical displacement of \(P\) from \(x-axis\). \(sin \theta\) and cos \(\theta\) are defined in terms of \(y\) and \(x\) respectively \[sin \theta = y\] \[cos \theta = x\] \[tan \theta = \frac{sin \theta}{cos \theta} = \frac{y}{x}\].

• In order to find the remaining ratios when one ratio is given, the traditional approach makes it easy.

Pythagoras Theorem

• This theorem states that the square of the longest side of the right-angled triangle is equal to the sum of the squares of the other two sides: \[ hyp^2 = opp^2 +adj^2\] Using Figure 1, \[a^2 = b^2 +c^2\]

Corollary 1

• Additional to the above three (3) basic trigonometric ratios, we also have their reciprocals as trigonometric ratios:
  • Secant of angle \(\theta\) = sec \(\theta\) = \(\frac{1}{cos \theta}\)
  • Cosecant of angle \(\theta\) = cosec \(\theta\) = \(\frac{1}{sin \theta}\)
  • Cotangent of angle \(\theta\) = cot \(\theta\) = \(\frac{1}{tan \theta}\)

Sine and Cosine Formulas

• Using the relationship infer by the sides and angles of a triangle on the basis of both basic trigonometric ratios and the Pythagoras theorem, we can obtain the sine and cosine formulas as follows:

Figure 3

• \[sin C = \frac{h}{b} \rightarrow h=bsin C\]

• \[sin B = \frac{h}{c} \rightarrow h=csin B\]

• Hence, \[\frac{b}{sin B} = \frac{c}{sin C}\]

• Generally, we have the sine rule: \[\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}\]

• The cosine formula can be derived as follows using Figure 3

  1. \(\triangle ABC \rightarrow cosB =\frac{x}{c} \rightarrow x = ccosB\) by BTR.
  2. \(\triangle ABC \rightarrow c^2 = x^2+h^2 \rightarrow h^2 = c^2 -x^2\) by Pythagoras
  3. \(\triangle BCA \rightarrow b^2 +h^2 +{(a-x)}^2 = h^2 +a^2 -2ax+x^2\)
  4. By substituting \(c^2 -x^2\) for \(h^2\); \(b^2 = (c^2 -x^2) +a^2 - 2ax+x^2 = c^2 +a^2 -2ax\)
  5. Finally, by subtituting \(ccosB\) for \(x\) from 1 into 4; \(b^2\) becomes:

\[b^2 = c^2+a^2-2accosB = a^2+c^2-2accosB\]

• Thus;the following are corresponding formulas with respect to angle A and C

\[a^2 = b^2+c^2-2abcosC\] \[c^2 =a^2+b^2-2abcosB\]

Class Activity 1a

1. Given that \(tan\theta = \frac{3}{4}\), find (i) sin \(\theta\) (ii) cos \(\theta\) (iii) What is the longest side of the triangle?
2. Using the appropriate formula, what is the value of \(|AB|\) if \(|BC|=3cm\), \(|AC|=3.6\) and \(\angle ABC = 48\)?

3. Using the result obtained in b, what is the value of \(\angle BCA\)? Hint: \(sin\phi = g \Rightarrow \phi = sin^{-1}{g}\).

Product to Sum of Trigonometric Ratios

Case 1: Sine Function \[sin(A+B) = sinAcosB + cosAsinB \tag{1}\] \[sin(A-B) = sinAcosB - cosAsinB \tag{2}\]

• Add equation 1 and 2, we have: \[sin(A+B)+sin(A-B) = 2sinAcosB \tag {1*}\]

• subtract equation 2 from 1, we have: \[sin(A+B)-sin(A-B) = 2cosAsinB \tag{2*}\]

Case 2: Cosine Function \[cos(A+B) = cosAcosB - sinAsinB \tag{3}\] \[cos(A-B) = cosAcosB + sinAsinB \tag{4}\]

• Add equation 3 and 4, we have: \[cos(A+B)+cos(A-B)=2cosAcosB \tag{3*}\]

• subtract equation 4 from 3, we have \[cos(A-B)-cos(A+B)=2sinAsinB \tag{4*}\]

Sum to Product of Trigonometric Ratios

• Let \(A+B = \theta\) and \(A-B = \phi\) \[\Rightarrow (A+B)+(A-B) = \theta +\phi\] \[2A = \theta +\phi\] \[A = \frac{\theta + \phi}{2}\]

• Also; \[\Rightarrow (A+B)-(A-B) = \theta -\phi\] \[2B = \theta -\phi\] \[B = \frac{\theta - \phi}{2}\]

• By substituting the results in all eqautions (*), we have:

1. \(sin\theta +sin\phi = 2sin\frac{\theta + \phi}{2}cos\frac{\theta - \phi}{2}\)
2. \(sin\theta - sin\phi = 2cos\frac{\theta + \phi}{2}sin\frac{\theta - \phi}{2}\)
3. \(cos\phi + cos\theta = 2cos\frac{\theta + \phi}{2}cos\frac{\theta - \phi}{2}\)
4. \(cos\phi - cos\theta = 2sin\frac{\theta + \phi}{2}sin\frac{\theta - \phi}{2}\)

Class Activity 1b

1. Find the expression of \(sin75^\circ + sin15^\circ\)
2. Express as a sum or difference \(2sin37.5^\circ cos7.5^\circ\).

Learner’s Activity (LA) 1

1. Show that \(sin^{2}{\theta} + cos^{2}{\theta} = 1\).
2. Investigate if \(tan^{2}{\theta}+1 =sec^{2}{\theta}\).
3. Given the following properties of \(\triangle ABC\): \(\hat{A}=40^\circ\), \(\hat{B} = 65^\circ\), and \(c=10cm\). Find
   1. lengths of all other sides of the triangle.
   2. the third angle of the triangle.
4. Prove that \(sin(2x-y)= 2sinxcosxcosy - 2cos^{2}{x}siny +siny\)
   
5. Without using trigonometric table, if \(sinx = \frac{3}{5}\), \(siny=\frac{5}{13}, 0^\circ < x <90^\circ\) and \(90^\circ < y < 180^\circ\), calculate:
   1. \(sin(2x-y)\)
   2. \(cos(x+2y)\)

LECTURE TWO

Angles of Elevation and Depression

• The angles between the line of sight and the horizontal for which an observer views an object either from the the top (having need to lower his or her line of sight) or from below (having need to raise his line of sight) are referred to as angles of depression and elevation respectively.

• The problems around these angles are usually of obtaining other angles formed formed by creating a triangle from the scenario or by obtaining corresponding distances associated with it.

• These problems are mainly solved using techniques explored in Lecture 1 - Pythagoras theorem, BTR, Sine and Cosine rules.

Illustration 1

• A man who is 183cm tall casts a 72cm long shadow on the horizontal ground. What is the angle of elevation of the sun to the nearest tenth of a degree?

Solution

\[tan \theta = \frac{Opp}{Adj} = \frac{183}{72} = 2.5417\] \[\theta = tan^{-1}(2.5417)\] \[\theta =68.5^\circ\]

The angle of elevation of the sun is approximately \(68.5^\circ\).

Bearing and Distance

• Bearing the clockwise angular distance covered between two points.

• Steps to Solving Bearing and Distances

  1. Taking reading in bearing starts from the North Pole in clockwise direction and ends also at the North Pole. i.e. North-North Pole reading.
  2. All angles formed while taking reading in bearing is equal to \(360^\circ\).
  3. All questions in bearing leads into the formation of a triangle.

Illustration 2

• A man drove 6km from a location X on a bearing of \(065^\circ\) and thereafter 13km on a bearing of \(136^\circ\). What is the distance and bearing of the driver from X?

Solution

• Using Cosine Rule, \[b^2 = a^2 + c^2 - 2acCosB\] \[y^2 = x^2 + z^2 -2xyCosY\]

• \(y^2 = 13^2 +6^2 - 2\times 13 \times 6 Cos109^\circ\)

• \(y=\sqrt{255.7886}\)

• \(y=15.9934 \approx 16km\)

• To find the bearing of the driver from X, we apply the Sine Rule;

\[\frac{x}{Sin X} = \frac{y}{Sin Y}\]

• \(Sin \theta = \frac{13\times Sin 109^\circ}{16} = 0.7682\)
• \(\theta = Sin^{-1}(0.7682)= 50.1955^\circ\)

Revised Trigonometry

Double Angle

\[Sin(2A) = Sin(A+A) = SinACosA + CosASinA = 2SinACosA\] \[Cos(2A) = Cos(A+A) = CosACOSA - SinASinA = Cos^{2}{A} - Sin^{2}{A} = 2Cos^{2}{A}-1\] \[Tan(2A) = \frac{Sin(2A)}{Cos(2A)}=\frac{2TanA}{1-Tan^{2}{A}}\]

Inverse Circular Functions

• Circular functions as used to describe trigonometric functions as they revolved around a unit circle \((360^\circ)\). The inverse circular functions are the inverse of the basic trigonometric functions: Sine, Cosine, and Tangent.

• The inverse circular functions are:

  1. \(Sec \theta = \frac{1}{Cos \theta}\)
  2. \(Cosec \theta = \frac{1}{Sin \theta}\)
  3. \(Cot \theta = \frac{1}{Tan \theta}\)

Learner’s Activity 2a

1. If \(Sin^{2}{\theta} + Cos^{2}{\theta} = 1\), What is \(Sin\theta\) in term of \(Cos\theta\)?
2. Find a suitable expression for \(Sin(3A)\).
3. Give the following in their lowest terms:
   1. \(\frac{2SinA}{Tan{A}}\)
   2. \(Cos^{2}{x}Cot^{2}{x}\)
   3. \(\frac{Sec^{2}{x}}{Cos{x}}\)

Learner’s Activity 2b

LECTURE THREE

Circular Measure

• Circular measure is the way angles are measured using the radian instead of degrees. this technique is commonly used in the field of mathematics, physics and engineering because it simplifies many formulas.

Definition of Radian

• One radian is the angle subtended at the center of a circle by an arc whose length is the equal to the radius of the circle.

Relationship Between Degrees and Radians

\[180^\circ = \pi \hspace{0.2cm} radians\]

• \(1\) radian \(= \frac{180^\circ}{\pi} \approx 57.3^\circ\)
  
• \(1^\circ = \frac{\pi}{180}\) radian
• Conversion: To convert degree to radian
  1. Radians = Degrees \(\times \frac{\pi}{180}\)
  2. Degrees = Radian \(\times \frac{180}{\pi}\)

Properties of the Circle Using Radian

1. Length of an Arc: An arc is a part of the circumference of a circle. The length of an arc is the distance along the curved line that makes up the arc. It can be obtained as: \[ LA = r\theta\] where:

   • LA = length of the arc
   • r = radius of the circle
   • \(\theta\) = central angle in radians.

   Illustration 4: If a circle has radius \(r = 10cm\) and the angle subtended at the center is \(\theta = \frac{\pi}{3}\) radians, the length of arc becomes: \[LA = 10 \times \frac{\pi}{3}= \frac{10\pi}{3}\approx 10.47cm\]

2. Area of Sector: The sector is a portion of the circle enclosed by two radii and the arc between them. It looks like a “slice of pie.” It is calculated as: \[AS = \frac{1}{2}r^{2}{\theta}\] where:

   • AS = Area of the sector
   • r = radius
   • \(\theta\) = angle subtended at the center in radian.

3. Length of Chord: A chord is a straight line joining any two points on the circumference of a circle. When two radii form a central angle \(\theta\) the chord connects the endpoints of the arc subtended by that angle. It is expressed as: \[\text{Length of Chord} = 2rSin\left(\frac{\theta}{2}\right)\]

4. Area of Segment: Area of segment is the region between the chord and the corresponding arc of a circle. It is found by subtracting the triangle formed by the two radii and the chord from the area of the sector. It is expressed as: \[\text{Area of Segment} = \frac{1}{2}r^{2}{(\theta -Sin\theta)}\]

Surface Area and Volumes of Simple Shapes

Cylinder

• A cylinder is a three-dimensional shape with two parallel circular bases and a curved surface connecting them.
• The total surface area of the cylinder is the sum of the area of the circular bases and the curved surface. It is expressed as: \[\text{Total Surface Area}= 2\pi r^{2}+2 \pi rh\] where:
  • \(r\) = radius of the base
  • \(h\) = height of the cylinder
• The volume of a cylinder is the area of the base time the height. It is expressed as: \[\text{Volume} = \pi r^{2}h\]

Cone

• A cone is a three-dimensional shape with a circular base and a pointed top (apex).
• The total surface area of a cone is sum of the base area and lateral area. It is expressed as: \[\text{TSA} = \pi r^{2} + \pi rl\] where:
  • \(l\) = slant height of the cone
• Since the cone is \(\frac{1}{3}\) of a cylinder with same base and height, its volume is expressed as: \[\text{Volume} = \frac{1}{3}\pi r^{2}h\]

Area and Volume of Irregular Shapes

Simpson’s Rule

• Simpson’s rule is a method for estimating the area under a curve or the volume of an irregular shape often when exact formulas can’t be applied.

• The area of an irregular shape using Simpson’s rule is expressed as: \[\text{Area} = \frac{h}{3}[y_0 + 4(y_1, y_3,..., y_{n-1} ) +2(y_2, y_4, ..., y_{n-2})+ y_n]\]

  where:

  • \(h\) = spacing between the ordinates (assumed equal)
  • \(y_0, y_1, y_2, ..,y_n\) = lengths of the vertical lines (ordinates)
  • \(n\) = even number of intervals (i.e. number of stripes = even)

Illustration 5: The shape of a land in Ugono-Orogun is shown below. To estimate the area of the land, a surveyor takes measurement at intervals of 50m, perpendicular to the straight portion with the results shown (the dimensions being in metres). Estimate the area of the land in hectares (\(1 ha = 10^{4}m^{2}\)).

Solution: Using Simpson’s rule with 6 intervals each of width 50m gives: \[Area \approx \frac{50}{3}[(140+4(160+190+130)+2(200+180)+0)\] \[=\frac{50}{3}(140+1920+760) = 47000\] Hence, in hectares, area \(\approx \frac{4700m^2}{10^{4}m^{2}/ha} \approx 4.70ha\)

• However, the volume of an irregular shape could also be obatined using the Simpson’s formulas: \[\text{Volume} = \frac{L}{3}[A_0 + 4(A_1, A_3,..., y_{n-1} ) +2(A_2, A_4, ..., A_{n-2})+ A_n]\] where:
  • \(L\) = distance between the first and last section (length)
  • \(A_0, A_1, A_2, ..,A_n\) = Areas of cross-sections at regular intervals
  • \(n\) = even number of intervals.

Illustration 6: The circumference of a 12m log of timber of varying circular cross-section is measured at intervals of 2m along its length and the results are:

Distance from one end (m)	0	2	4	6	8	10	12
Circumference (m)	2.80	3.35	3.94	4.32	5.16	5.82	6.36

Estimate the volume of the timber in cubic metres.

Solution: If circumference \(c =2\pi r\) then:

• radius, \(r = \frac{c}{2\pi}\)

• Cross-section area = \(\pi r^2 = \pi\left(\frac{c}{2\pi}\right)^2 = \frac{c^2}{4\pi}\)

• Hence, the cross-sectional areas are: \(\frac{2.80^2}{4\pi} = 0.6239m^2, \frac{3.25^2}{4\pi}=0.8405m^2, 1.2353m^2, 1.4851m^2, 2.1188m^2, 2.6955m^2, 3.2189m^2\)

• The volume of the timber is therefore approximately: \[\frac{2}{3}[0.6239+ 4(0.8405+1.4851+2.6955)+2(1.2353+2.1188)+3.2189]\] \[=\frac{2}{3}(30.6354)=20.42m^3\]

Mid-Ordinate Rule

• The mid-ordinate rule is an approximation technique that is used to estimate the area under a curve by summing up the areas of rectangles whose height are the values of the function at the midpoints of each interval. It is expressed as: \[Area = h\times \sum_{i=1}^{n}y_i\] where
  • \(h\) = width of each interval (strip width)
  • \(y_i\) = mid-ordinate height (the value of the curve at the midpoint of each interval)
  • \(n\) = number of strips (intervals).
• Mid-ordinate rule should only used when midpoint values are known, for quick approximations and when curve is fairly smooth.
• The mid-ordinate rule does not give accurate results as Simpson’s rule because it does not account for curvature. It therefore assumes the curve is flat over each interval- thus, underestimate or overestimate curve regions.

Lecture Four

Introduction to Analytical Geometry

• In the study of analytical geometry, there are tow dimensional coordinate systems, namely:

  • Cartesian, and
  • Polar

• Let the position of a point in a plane be described such that: \(OX\) is fixed line, 0 is the origin, being fixed point on it. The position of point \(P\) is known if the angle \(POX\) and the distance \(OP\) are given. The \(\angle POX (\theta)\) is called the vectorial angle and is considered positive when measured in an anticlockwise sense. The distance \(OP (r)\) is called the radius vector.

• Thus, the duo \((\theta, r)\) are known as the Polar coordinates of \(P\). It is possible to make a transformation from polar coordinates to Cartesian coordinates or vis-a-vis.

• \(P(x,y)\) is the Cartesian coordinates. Then we can write: \[x = ON = OPCos\theta \hspace{0.2cm} \text{or} \hspace{0.2cm} rCos\theta =x\] \[y = PN = OPSin\theta \hspace{0.2cm} \text{or} \hspace{0.2cm} rSin\theta =y\]

• And \(tan\theta = \frac{y}{x}\) is the slope of the line \(OP\) or \(m = \frac{y}{x} \rightarrow tan\theta = m\) while \(r^2 = x^2+y^2\). Thus; \[r=\sqrt{x^2+y^2}\]

Distance between Two Points

• Let \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) be two given points. Then, the distance between the two points is expressed as: \[|PQ|=\sqrt{(x_2 - x_1)^2 + (y_2 -y_1)^2}\]
• Above is the formula for finding the distance between two points.

Illustration 4.1: Given that there exist 3 points on a plane labelled: \(P\), \(Q\) and \(R\) which have coordinates respectively as $(5,-6), (-3,0), (-1,2). Find all possible distances between two points on the plane.

Solution

\[|PQ| = \sqrt{(5-(-3))^2+(-6-0)^2} =\sqrt{100} = 10 units\] \[|PR| = \sqrt{(5-(-1))^2+(-6-2)^2} =\sqrt{100} = 10 units\] \[|QR| = \sqrt{(-1-(-3))^2+(2-0)^2} =\sqrt{8} = 2\sqrt{2} units\]

Mid-Point of Two Points

• Let \(T(x_1, y_1)\) and \(V(x_2,y_2)\) be any two points on a straight line. Then, the mid-point between them is given as: \[U\left(\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2}\right)\]

Illustration 4.2: Find the coordinates of the mid-point of a line segment \(DE\) with respective coordinates given as: \(D(-4,2)\) and \(E(3,6)\).

Solution Let the mid-point be denoted by \(\Upsilon(x,y)\). Then, \[\Upsilon(x,y) = \Upsilon\left(\frac{-4+3}{2}, \frac{2+6}{2}\right)=\Upsilon\left(-\frac{1}{2},2\right)\]

The Straight Line

• The equation of a straight line is the relation between \(x\) and \(y\) which is satisfied by the coordinates of each and every point on the line and by those of no other point.

• The intercept is defined as the point where the straight line intersects the \(y-axis\).

• The gradient of line passing through two points, say \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) is defined as the ratio of the increase in \(y\)-coordinate to the increase in the \(x\)-coordinate in going from one point to another on the line. It is expressed as: \[m = \frac{QR}{PR} = \frac{y_2-y_1}{x_2-x_1}\]

Illustration 4.3: Find the gradient of the line joining \((3,2)\) and \((7,10)\) and the angle of slope of the line.

Solution: Let \(m\) be the gradient of the line. Then, \[m = \frac{y_2-y_1}{x_2-x_1}=\frac{10-2}{7-3}=\frac{8}{4}=2\] Let \(\theta\) be the angle of slope of the line, thus; \[tan \theta = 2 \rightarrow \theta = tan^{-1}(2) = 63.43^\circ\]

Equations of a Straight Line

• Let \(A(x,y)\) be any variable point on the line \(PQ\). If the gradient is \(m\) and the intercept of the line on the \(y-axis\) is \(c\), then; \[\frac{y-c}{x}=m \longrightarrow y-c=mx\] \[\therefore y=mx+c \tag{4.1}\] An equation of the form (4.1) is called equation of a straight line given the gradient and the intercept.
• The equation of a straight line given the coordinates of a point \((x_1, y_1)\) with slope \(m\) is given as: \[y-y_1=m(x-x_1)\tag{4.2}\]