THEORY OF TRIGONOMETRY

1. INVESTIGATING TRIGONOMETRY

1.1 Six trigonometry ratios

If one angle of a triangle is 90 degrees and one of the other angles is known, the

third is thereby fixed, because the three angles of any triangle add up to 180 degrees.

The two acute angles therefore add up to 90 degrees: they are complementary angles.

The shape of a right triangle is completely determined, up to similarity, by the angles.

This means that once one of the other angles is known, the ratios of the various sides

are always the same regardless of the overall size of the triangle. The-ratios are defined

as follows:

Table 19-1.-Trigonometric ratios.

Figure 19-1.-Relationship of sides and angles in a right triangle. (A) Names of the sides;

(B) symbols used to designate the sides.

The hypotenuse is the side opposite to the 90 degree angle in a right triangle; it is the

longest side of the triangle, and one of the two sides adjacent to angle A. The adjacent

leg is the other side that is adjacent to angle A. The opposite side is the side that is

opposite to angle A.

The reciprocals of these functions are named the cosecant (csc or cosec),

secant (sec) and cotangent (cot), respectively. The inverse functions are called the

arcsine, arccosine, and arctangent, respectively. There are arithmetic relations

between these functions, which are known as trigonometric identities. The secant,

cosecant, and cotangent are calculated, when needed, by using their relationships to

the three principal ratios. These relationships are as follows:

Negative Angles

We will just state:

sin (−θ) = −sin θ  

cos (−θ) = cos θ  

tan (−θ) = −tan θ

1.2 SOLUTION OF TRIANGLE

Sine Rule

The sine rule states that:

The triangle does not have to have a right angle in it for this rule to work.

The sine rule states that:

                                                

This can also be interpreted as three equations:

sin B

b = 

sin C

c,  

sin A

a = 

sin C

c,   and  

sin A

a = 

sin B

B

Since the three versions differ only in the labeling of the triangle, it is enough to verify

one just one of them, so we'll just consider the version stated first.

An explanation of the law of sine is fairly easy to follow, but in some cases we'll have to

consider sine of obtuse angles.

Example

 

We might want to find out what the angle opposite the side of length 20 is.

We can use the sine rule to solve for this angle, as follows:

                                              Let a = the side of length 50

                                       Let A = the angle opposite ‘a’ (100°)

                                              Let b = the side of length 20

                                            Let B = the angle opposite ‘b’

                                                    

Inverting both sides gives:

                                                    

Multiplying both sides by 20 gives:

                                              

Ambigous case

By definition, the word ambiguous means open to two or more interpretations.

Such is the case for certain solutions when working with the Law of Sines.

  •  If you are given two angles and one side (ASA or AAS),

   the Law of Sines will nicely provide you with ONE solution

   for a missing side.

•  Unfortunately, the Law of Sines has a problem dealing with SSA.

   If you are given two sides and one angle (where you

   must find an angle), the Law of Sines could possibly provide you

   with one or more solutions, or even no solution.

Before we investigate this situation, there are a few facts we need to remember.

       Facts we need to remember:

1.  In a triangle, the sum of the interior angles is

180º.

2.  No triangles can have two obtuse angles.

3.  The sine function has a range of

.

4.  If the = positive decimal < 1, the can

lie in the first quadrant (acute <) or in the second

quadrant (obtuse <).

Let's look at some cases.  In each example, decide whether the given information points

to the existence of one triangle, two triangles or no triangles.

Example 1:  In , a = 20, c = 16, and m<A = 30º.  How many distinct triangles can

be drawn given these measurements?

Use the Law of Sines:

                  

C = sin-1 (0.4) = 24º (to the nearest degree) - in Quadrant I.

Sine is also positive in Quadrant II.  If we use the reference angle 24º in Quadrant II,

 the angle C is 156º.

But, with m<A = 30º and m<C = 156º the sum of the angles would exceed 180º.

Not possible!!!!

Therefore, m<C = 24º, m<A = 30º, and m<B = 126º and only ONE triangle is possible.

Example 2:  In , a = 10, b = 16, and m<A = 30º.  How many distinct triangles can

be drawn given these measurements?

Use the Law of Sines:

                  

B = sin-1(.8) = 53.13010 = 53º.

Angles could be 30º, 53º, and 97º :  sum 180º

The angle from Quadrant II could create angles 30º, 127º, and 23º :  sum 180º

TWO triangles possible.

 

Cosine rule

c2 = a2 + b2 – 2ab cos C

There are two other versions of the law of cosines,

a2 = b2 + c2 – 2bc cos A

and

b2 = a2 + c2 – 2ac cos B.

Since the three versions differ only in the labeling of the triangle, it is enough to verify

one just one of them. We'll consider the version stated first.

Example :

 

So for this triangle…

You are given angles A and C and side c,

so use

B=180-(60+45)=75 and now

The cosine rule is also used when the three sides are given.

Using these methods, any triangle can be ‘solved’. That is all of the angles and sides of

a triangle can be worked out, if enough information is given to start with.

.

Area of Triangle

It is well known that to work out the

area of a rectangle you multiply the

length by the width.

Almost as well known is the formula

for the area of a triangle.

That is, the area of a triangle is half of

the area that the rectangle that the

triangle fits into.

But this formula requires the length of the base and the height of the triangle. There is

another formula working out the area of a triangle uses the lengths of two sides and the

angle between them.

As CB = b, and letting the area of

triangle ABC represented by T, we

have

T=12×base ×height

12×b×h ……… equation 1

From ∆ BNA ,

sin B=¿ hc¿

h=c sinB

substitute h=c sinB into equation 1,

T=12×b×c sin B=1

2bc sin B

By similar procedure above, we can also show that 𝑇= 1

2bc sin A and T=1

2ab sinC

So for this triangle:-

Of course, the way that the sides and

angles are labelled will not affect the

formula, as long as two sides and the

angle between them is used.

There is another formula, which is used when the lengths of the triangle's three sides

are known. It is Herons formula

             In mathematics, Hero’s formula is the formula used to find the area of triangle in

geometry. The hero’s formula is defined by the mathematician named Heron of

Alexandria. The Pythagoras theorem is used to prove the hero’s formula which is found

in his book, Metrica. Hero’s formula is also known as Heron area formula. Let as

assume the sides of triangle as a, b and c.

The semi perimeter of the hero’s formula is given as,

s=a+b+c2

The hero’s formula is given as,                                                                      

             Area= √s (s−a ) ( s−b ) (s−c )

Given that area ∆=12ab sinC ……………(1)

Trigonometry identity sin2C+cos2C=1 sinC=√(1¿−cos2C)¿………….(2)

Cosine rule : c2 = a2 + b2 – 2ab cos C

cosC=a ²+b ²+c ²2ab

…………….(3)

Substitute (3) into (2) into (1)

12ab¿

√ ¿¿

14

√[(2ab−a¿¿2¿−b2+c2)(2ab+a2+b2−c2)]¿¿

14√ [c2−(a−b )2] [(a+b )2−c2]

14 √ [ (c−a+b ) (c+a−b ) ] [ (a+b−c ) (a+b+c ) ]

s=a+b+c2

,2 s=a+b+c

14

√ (2 s−a−b−a+b ) (2 s−a−b−a+b ) (a+b−2 s+a+b ) (2 s )

14

√16 (s−a ) ( s−b ) ( s−c ) ( s)

√164

√ (s−a ) ( s−b ) (s−c ) ( s)

√s (s−a ) ( s−b ) (s−c )

Example :

       Find the area of a triangle using hero’s formula where every side is 8m long?

            Given:

                        a = b = c = 8

            Solution:

The semi perimeter of the hero’s formula is given as,

                                    S = (a + b + c)/2 or perimeter/2.

                                    S = (8+8+8)/2

                                    S = (24)/2

                                    S =12

                        The hero’s formula is given as,    

                                    Area=√ (s(s-a) (s-b) (s-c)),

                                    Area= √ (12(12-8) (12-8) (12-8))

                                    Area= √ (12(4) (4) (4))

                                    Area = √ (768)

                                    Area = 27.71

            Therefore, the area of triangle is 27.71

1.3 PROBLEMS IN THREE DIMENSIONS

Normal to a Plane

In 3D graphics, an imaginary line that is perpendicular to the surface of a

polygon. It may be computed at the vertex of a triangle, in which case it is the

average of all the vertices of adjoining triangles. Or, it may be computed for

each pixel in the triangle as in Phong shading. Surface normals are used to

derive the reflectivity of a light source shining onto an object.

A straight line is normal to a plane if the line is perpendicular to all straight line

drawn on the plane.

diagram 1

In diagram 1, the straight line VO, drawn from a point V to meet the plane at

O, is a normal to the plane ABCD, as VO is perpendicular to the lines AO and

OE and other lines lying on the plane.

1.4 Angle between a Line and a Plane

The angle between a line AB and a plane π can be determined by making an

orthogonal projection from B to the plane π. Diagram 2 shows this orthogonal

projection from B meets the plane π at a point C. note that angle ACB=90⁰

Diagram 2

Diagram 3

Angle between two Planes

Two non-parallel planes π1 and π2, will meet at a common line. In diagram 4,

this common line is PQ. To find the angle between these two planes, take a

point, say A, on this common line. From this point A, draw two perpendicular

lines, AB and AC, to PQ with one of these lines lying on the plane π1and the

other line lying on the plane π2. Diagram 3 shows these two lines AB and AC,

with <BAP = <CAP =90⁰. Hence, from finding the angle between two planes,

we reduce this to find the angle between two lines, that is angle between the

line AB(representing plane π1) and the line AC(representing the plane π2).

The angle between two planes is the angle BAC.

Diagram 3

Problem in three dimensions

Question 1

Sally and Kate stand some distance away from a building (B) which is 10m

high. Sally is on a bearing of 030° from the building. From where she is

standing, the angle of elevation of the top of the building is 35°. Kate is on a

bearing of 090° from the building. From where she is standing, the angle of

elevation of the top of the building is 50°. This information is shown in the

diagram.

What is the distance between Sally and Kate?

Answer

Using trigonometry on the right-angled triangles OBS

so = 14.3 m (3 s.f.)

In the same way we can find OK = 8.39m (3 s.f.)

Applying the cosine rule to triangle KOS, we get:

KS2 = 8.392 + 14.32 - ( 2 × 8.39 × 14.3 × cos60°)

KS2 = 154.534

KS = 12.4m (3 s.f.)

1.4 ANGLE OF ELEVATION AND ANGLE OF DEPRESSION

Angle of Elevation

The word “elevation” means “rise” or “move up”.

Angle of elevation is the angle between the horizontal and the line of sight to an object

above the horizontal.

Example :

The angle of elevation of an object as seen by an observer is the angle between the

horizontal and the line from the object to the observer's eye (the line of sight).

Angle of Depression

The angle below horizontal that an observer must look to see an object that is lower

than the observer. (this assumes the object is close enough to the observer so that the

horizontals for the observer and the object are effectively parallel; this would not be the

case for an astronaut in orbit around the earth observing an object on the ground).

 

If the object is below the level of the observer, then the angle between the horizontal

and the observer's line of sight is called the angle of depression.

Note: The angle of depression is congruent to the angle of elevation

The picture below illustrates an example of an angle of depression and an angle of

elevation

Example of calculation :

The angle of elevation of the top of a pole measures 48° from a point on the ground 18

ft away from its base. Find the height of the flagpole.

 Solution:

 Step 1: Let’s first visualize the situation.

 Step 2: Let ‘x’ be the height of the flagpole. 

  Step 4: x = 18 × tan 48° = 18 × 1.11061… = 19.99102… 20 Step 5: So, the flagpole is about 20 ft high.

2. EXTENDING TRIGONOMETRY

2.1 TRIGONOMETRY FORMULA

Basic identities

The picture shows a point on the unit circle. Since we know that any point on the unit

circle can be described by cos θ, sin θ. It is possible to draw the triangle that describes

this point. As you can see from the picture, the length of one side is cos θ and the

length of the other side is sin θ and, by definition, the radius of the unit circle is 1.

From these facts, the primary Pythagorean identity can be shown. sin2θ + cos2θ = 1.

This identity is just an application of the Pythagorean Theorem to the unit circle.

Formula for the Pythagorean Identities.

I. sin² θ + cos² θ = 1

II. tan² θ + 1 = sec² θ

III. 1 + cot² θ = csc² θ

Pictured below is the formula for the aptly named Pythagorean identity:

sin²θ + cos²θ = 1

Compound angle formula

The following are important trigonometric relationships:

sin(A + B) = sinAcosB + cosAsinB

cos(A + B) = cosAcosB - sinAsinB

tan(A + B) =   tanA + tanB

                  1 - tanAtanB

To find sin(A - B), cos(A - B) and tan(A - B), just change the + signs in the above

identities to - signs and vice-versa:

sin(A - B) = sinAcosB - cosAsinB

cos(A - B) = cosAcosB + sinAsinB

tan(A - B) =   tanA - tanB

                1 + tanAtanB

Compound Angle Proof

The compound angle formula is: [4.1]

We construct a triangle, ABC, with CX being perpendicular to AB, and of length h. The line CS divides the angle C into two angles α and β.

We recall that the area of triangle ABC is:

[4.2]

Also, the area of triangle ACX is:

[4.3]

And the area of triangle XBC is:

[4.4]

We also note that h=b·cos α=a·cos β  [4.5]

And for the set up, we note that the area of ABC is equal to the sum of the areas of triangles AXC and XBC:

[4.6]There are choices in substituting for h from Equation 4.5, and choosing the appropriate one to give 1/2ab throughout gives:

[4.7]Choosing an inappropriate value leads to more algebra, but the same result! A similar method is used to prove Pythagoras' Theorem

On division throughout by 1/2·a·b gives us the compound angle formula:

Double-angle formulae

sin(A + B) = sin A cos B + cos A sin B

Replacing B by A in the above formula becomes:

sin(2A) = sin A cos A + cos A sin A

so: sin 2A = 2 sin A cos A

similarly:

cos 2A = cos2 A - sin2 A

Replacing cos2A by 1 - sin2A (see Pythagorean identities) in the above formula gives:

cos 2A = 1 - 2sin2A

Replacing sin2 A by 1 - cos2 A gives:

cos2 A = 2 cos2 A - 1

It can also be shown that:

tan 2A =     2 tan A  

        1 - tan2A 

Example. Check the identities

Answer. We will check the first one. the second one is left to the reader as an exercise. We have

Hence

which implies

Half-angle formulae

We start with the formula for the cosine of a double angle

cos 2θ = 1− 2sin2 θ

Now, if we let

then 2θ = α and our formula becomes:

cos α = 1 − 2sin2 (α/2)

We now solve for

(That is, we get sin(α/2) on the left of the equation and everything else on the right):

2sin2(α/2) = 1 − cos α

sin2(α/2) = (1 − cos α)/2

Solving gives us the following sine of a half-angle identity:

The sign of depends on the quadrant in which α/2 lies.

If α/2 is in the first or second quadrants, the formula uses the positive case:

If α/2 is in the third or fourth quadrants, the formula uses the negative case:

Using a similar process, with the same substitution of (so 2θ = α) we subsitute

into the identity

cos 2θ = 2cos2 θ − 1 (cosine of a double angle)

We obtain