Measurement of Triangles

Knowledge of trigonometric identities helps simplify complex trigonometric expressions into more manageable forms and prove mathematical results. In physics and engineering, trigonometric identities help in studying wave motion, mechanics, analysing electrical circuits and mechanical systems. In architecture and design, the concept aids in designing structures with curved surfaces. This section introduces you to the concept of trigonometric identities, deriving and applying the sine and cosine rule and solving trigonometric equations.

KEY IDEAS

• Multiple angles formulas express trigonometric functions of multiples of an angle (e.g., 2θ , 3 θ) in terms of the trigonometric functions of the original angle.

• Trigonometric equations involve trigonometric functions and require solving for the angles or other variables. These equations often leverage trigonometric identities and properties for solutions.

• Trigonometric Identities are fundamental relationships between trigonometric functions that are true for all values of the variables involved, within their domain. These include Pythagorean identities, Reciprocal Identities and sum and difference identities.

In year one, we learnt about the fundamental identities in trigonometry:

sinθ = opposite________ hypotenuse, cosθ = adjacent________ hypotenuse, tanθ = opposite_______ adjacent , cscθ = hypotenuse________ opposite , secθ = hypotenuse________ adjacent and cotθ = adjacent_______ opposite From this we can derive the reciprocal identities.

Activity 9.1: Reciprocal Identities

In small groups or pairs, work through the following activity.

1. Choose a number (e.g., 2)

2. Write down the reciprocal of the number (e.g. , 1/2)

3. Now, write down the reciprocal of the sine identity as 1___ sinθ.

4. Express sinθ as opposite________ hypotenuse within the reciprocal to get 1________ opposite________ hypotenuse .

5. Note that 1________ opposite________ hypotenuse is the same as hypotenuse________ opposite .

6. What relationship do you see between the result of 1___ sinθ and the cosec (csc) identity?

7. Hopefully you see that cscθ = 1___ sinθ.

8. Now, repeat the steps from 3 to 6 using the identities cosθ , tanθ , secθ , cscθ and cotθ.

9. Hopefully, you realise that:

10. cosθ = 1_____ (sec θ); tanθ = 1____ cotθ , secθ = 1____ cosθ , cotθ = 1____ tanθ , sinθ = 1____ cscθ.

Now that we have established the reciprocal identities through Activity 9.1, let us explore the Quotient Identities.

Activity 9.2: Quotient Identities (in small groups or individually) In small groups or pairs, work through the following activity.

1. Write out sinθ____ cosθ as opposite________ hypotenuse_______ adjacent________ hypotenuse .

2. Simplify the resulting expression:

opposite________ hypotenuse_______ adjacent________ hypotenuse = opposite________ hypotenuse ÷ adjacent________ hypotenuse = opposite________ hypotenuse × hypotenuse________ adjacent = opposite_______ adjacent

3. Which trigonometric identity can be expressed as opposite_______ adjacent?

4. Identify the relationship between sinθ____ cosθ and tanθ .

5. Hopefully, you were able to conclude that sinθ____ cosθ = tanθ .

6. Now go through steps 1 to 3 using cosθ____ sinθ and identify which trigonometric identity is equal to cosθ____ sinθ .

7. Hopefully, you came to the conclusion that cosθ____ sinθ = cotθ and that cotθ and tanθ are reciprocal identities as well.

Pythagorean Identities

Using knowledge of Pythagoras theorem and algebra, we can deduce three Pythagorean Identities. Let us go through this activity to deduce the three identities.

Activity 9.3: Proving Pythagorean Identities

In small groups or pairs, work through the following activity.

Identity 1

Figure 9.1: Right-angled Triangle for Activity 9.3

1. Draw a right-angle triangle.

2. Choose variables to represent the lengths of the hypotenuse, opposite and adjacent sides to the angle (β).

r = hypotenuse, x = adjacent and w = opposite.

3. Write out the Pythagoras theorem using identified lengths.

r²= x²+ y²4. Divide both sides by r²and simplify.

r²_ r²= x²_ r²+ y²_ r²1 = (x_ r)²+ (ʸ_ r)²5. Rewrite (x_ r)²and (ʸ_ r)²using the sides they represent.

( adjacent________ hypotenuse) 2 = cos²β and ( opposite________ hypotenuse) 2 = sin²β

6. Substitute cos²β and sin²β into the simplified equation in Step 4 1 = cos²β + sin²β Identity 2

1. From Step 3 in Identity 1, divide both sides by x²and simplify.

r²_ x²= x²_ x²+ y²_ x²(r_ x)²= 1 + (ʸ_ x)²2. Rewrite (r_ x)²and (ʸ_ x)²using the sides they represent.

(hypotenuse________ adjacent ) 2 = sec²β and (opposite_______ adjacent) 2 = tan²β

3. Substitute sec²β and tan²β into the simplified equation in Step 1 of Identity 2.

sec²β = 1 + tan²β Identity 3

1. From Step 3 in Identity 1, divide both sides by y²and simplify.

r²_ y²= x²_ y²+ y²_ y²(ʳ_ y)²= (ˣ_ y)²+ 1

2. Rewrite (ʳ_ y)²and (ˣ_ y)²using the sides they represent.

(hypotenuse________ opposite ) 2 = csc²β and (adjacent_______ opposite) 2 = cot²β

3. Substitute csc²β and cot²β into the simplified equation in Step 1 of Identity 3.

csc²β = cot²β + 1 Hopefully, you were able to follow through all the steps in the activity to arrive at the Pythagorean identities, which are:

1. cos²β + sin²β = 1

2. sec²β = 1 + tan²β

3. csc²β = cot²β + 1 Compound angles identities There are other trigonometric identities that involve two variables. These identities are commonly referred to as compound identities or the sum and difference formula. The compound angle formulas are important because they help describe rotations. If an object rotates by two angles A and B, the formulas combine these rotations into a single equivalent angle. This concept is vital in designing 3D models and simulating movements in animation or robotics. Compound angle identities are used in modelling sound waves, light waves and electromagnetic waves.

The concept of compound angles is based on a unit circle. A unit circle (circle with radius of 1 unit) can be drawn such that the starting side of the angle, (β− α) lies on the positive x −a x i s and hence intersects the circle at (1, 0) while the ending side is rotated such that instead of intersecting the circle at (cosβ, sinβ), it intersects at (cos(β− α) sin(β− α)). This ensures the length of the chord between the sides of angle, (β− α) is maintained, as shown in fi gure 9.2.

Activity 9.4: Research on compound angle identities (cos β, sin β) (cos α, sin α) α β y x (cos(β – α), sin(β – α)) (β –α) (1, 0) x y

Figure 9.2: Graphical illustration of compound angles In small groups or pairs, undertake the following research.

1. Carry out research on the compound angle identities outlined from a. – f.

a. cos(α − β) = cos(α)cos(β) + sin(α)sin(β)

b. cos(α + β) = cos(α)cos(β) − sin(α)sin(β)

c. sin(α + β) = sin(α)cos(β) + cos(α)sin(β)

d. sin(α − β) = sin(α)cos(β) − cos(α)sin(β)

e. tan(α + β) = tan(α) + tan(β)_____________ 1 − tan(α)tan(β)

f. tan(α − β) = tan(α) − tan(β)_____________ 1 + tan(α)tan(β)

2. Consult textbooks and online resources to work out the algebraic proofs.

3. Present your findings to your classmates.

Application of compound angle identities In year one we learnt about special angles (30⁰, 45⁰, 60⁰, 90⁰). We will now apply this knowledge on special angles and the compound angle identities to solve problems.

Example 9.1

Determine the exact value of sin( 120⁰) without using calculators.

Solution

Step 1: Split the given angle into two special angles sin( 120⁰) = sin( 60⁰+ 60⁰)

Step 2: Express the split angle based on sin(α + β) compound angle identity sin(60⁰+ 60⁰) = sin(60⁰)cos(60⁰) + cos(60⁰)sin(60⁰)

Step 3: Substitute the values for the trigonometric identity angles and simplify = (√3/2 )(1/2) + (1/2)(√3/2 ) = √3/4 + √3/4 = 2 √3/4 = √3/2 Therefore, sin( 120⁰) = √3/2 .

Example 9.2

Determine the exact value of cos( 135⁰) without using calculators.

Solution

Step 1: Split the given angle into two special angles cos( 135⁰) = cos( 90⁰+ 45⁰)

Step 2: Express the split angle based on cos(α + β) compound angle identity cos( 90⁰+ 45⁰) = cos(90⁰)cos(45⁰) − sin(90⁰)sin(45⁰)

Step 3: Substitute the values for the trigonometric identity angles and simplify = (0)(√2/2 ) − (1)(√2/2 ) = −√2/2 Therefore, cos( 135⁰) = −√2/2 .

Example 9.3

Find the value of tan( 120⁰) without using calculators.

Solution

Step 1: Split the given angle into two special angles tan( 120⁰) = tan( 60⁰+ 60⁰)

Step 2: Express the split angle based on tan(α + β) compound angle identity tan( 60⁰+ 60⁰) = an( 60⁰) + tan( 60⁰)________________ 1 − tan( 60⁰)tan( 60⁰)

Step 3: Substitute the values for the trigonometric identity angles and simplify = √3 + √3/1 − (√3) (√3) = 2 √3____ − 2 Therefore, tan( 120⁰) = −√3.

Multiple-angle identities Multiple–angle identity corresponds to multiples of an angle, such as 2α, 3α, or nα, where n is a positive integer. These identities express trigonometric functions of a multiple of an angle in terms of the trigonometric functions of the base angle, α. Multiple angles can be double and triple angles. However, here we only look at double angles.

Double angle identities Double angle identities focus on angles that can be expressed as a sum of two of the same angles (i.e., γ = α + α ). In such cases, compound angle identities can be used to find the solution given that α = β . Based on our knowledge of multiple angles, if we substitute α with β given that α = β , we will realise that the three identities below hold:

1. cos(α + α) = cos(α)cos(α) − sin(α)sin(α) cos2α = cos²α− sin²α And using the Pythagorean identities this means that:

cos2α = 1 − 2 sin²α and cos2α =2cos²α − 1

2. sin(α + α) = sin(α)cos(α) + cos(α)sin(α) sin2α =2sin(α)cos(α)

3. tan(α + α) = tan(α) + tan(α)____________ 1 − tan(α)tan(α) tan2α = 2tan(α)_________ 1 − 2tan(α) If an angle is α , half of the angle results in α__

2. This leads us into the concept of half-angles.

Let us go through the Activity 9.5 to prove the half-angle identities.

Activity 9.5: Research on half-angle Identities (work in groups) In small groups, undertake the following research.

1. Conduct research to prove that:

a. sin(^(α)__ 2 ) = ±√________ 1 − cos(α)_________ 2

b. cos(^(α)__ 2 ) = ±√________ 1 + cos(α)_________ 2

c. tan(^(α)__ 2 ) = ±√________ 1 − cos(α)_________ 1 + cos(α)

2. Include graphical illustrations to support your algebraic proofs.

3. Present a report on your findings to your classmates and teacher.

Verifying Identities

An identity is an equation that is true for all values of the variables within its domain. It is a fundamental relationship that holds universally and does not depend on specific variable values. We can apply the trigonometric identities to verify whether an equation is an identity or not.

For example, using graphs to illustrate identities, sketch the graph of y = tan(x) − cot(x)___________ sin(x) cos(x) and y = sec²(x)− csc²(x).

Figure 9.3: Graphical illustration of y = tan(x)-cot(x))__________ (sin(x) cos(x) Comparing Figures 9.3 and 9.4, it is evident that, for any value of x , the results of the two equations will be the same. Thus, tan(x) − cot(x)___________ sin(x) cos(x) and sec²(x)− csc²(x ) are identities.

Figure 9.4: Graphical illustration of y = sec²(x)− csc²(x) Mathematically, we can represent them as tan(x) − cot(x)___________ sin(x) cos(x) ≡ sec²(x)− csc²(x).

Example 9.4

Prove that tan(x) − cot(x)___________ sin(x) cos(x) ≡ sec²(x)− csc²(x), algebraically.

Solution

Step 1: Rewrite Left-hand side expression in terms of trigonometric identities:

tan(x) − cot(x)___________ sin(x) cos(x) = sin(x)_____ cos(x) − cos (x ) _ sin (x ) _________________ sin(x) cos(x)

Step 2: Simplify the numerator of the fraction:

sin²(x) − cos²(x)____________ cos(x)sin(x)___________ sin(x) cos(x)

Step 3: Simplify the double fraction:

sin²(x) − cos²(x)____________ cos(x)sin(x) × 1 __ in(x) cos(x) sin²(x) − cos²(x)____________ cos²(x) sin²(x)

Step 4: Split the fraction in two terms and simplify:

sin²(x)___________ cos²(x) sin²(x) − cos²(x)___________ cos²(x) sin²(x) 1_ cos²(x) − 1_____ sin²(x)

Step 5: Rewrite 1______ cos²(x) and 1_____ sin²(x) in terms of secant (sec) and cosec (csc) trigonometric identities.

1______ cos²(x) = sec²(x) and 1_____ sin²(x) = csc²(x) Therefore, 1______ cos²(x) − 1_____ sin²(x) results in sec²(x) − csc²(x).

Step 6: Compare the righthand side (RHS) of the original equation with the lefthand side (LHS) result and draw a conclusion.

Since the LHS resulted in sec²(x) −csc²(x) which is same as the original RHS expression, tan(x) − cot(x)___________ sin(x) cos(x) ≡ sec²(x)− csc²(x).

Activity 9.6: Algebraic Verification of Identities

In small groups, undertake the following research.

1. Research on the algebraic proof of tan(x) + cot(x) ≡ csc(x)_____ cos(x) .

2. Provide a graphical illustration (using graphing tool or manually) to support your algebraic proof.

3. Make a presentation to your classmates and teacher.

When we talked about vectors in Section 5, we looked at the Sine and Cosine rule that helped in finding angles and lengths of lines of oblique (non-right-angle) triangles. While we used measurements to understand the relationships, we will use algebraic proofs to establish the Sine and Cosine rules in this section. The cases in which we can apply a sine or cosine rule are:

1. When two angles and any side length are known.

2. When two side lengths and an angle opposite one of the known side lengths are known.

3. When three side lengths are known.

4. When two side lengths and the angle included between the two sides are known.

Sine Rule

From Section 5 on vectors, we found out that r____ sinR = m____ sinM = t____ sinT given the triangles in Figure 9.5 below.

Figure 9.5: Oblique triangles supporting exploration of Sine rule The Sine rule is applicable when two angles and any side length is known.

Let us go through the process of deriving the Sine rule using the steps outlined below.

Deriving the sine rule This derivation is based on the triangles in Figure 9.5.

1. Using < MTR, sin(T) = h/r from trigonometric identities.

2. Making h the subject, we get h = r sin(T)

3. Also, using < TRM, sin(R) = h/t

4. Making h the subject, we get h = t sin(R)

5. This reveals that h = r sin(T) = t sin(R)

6. Rewrite rsinT = t sin(R) by dividing both sides by sin(R) sin(T) :

r sin(T)_________ sin(R)sin(T) = sin(R)_________ sin(R)sin(T) Simplifying the equation results in r_____ sin(R) = t_____ sin(T)

Activity 9.7: Verifying the Sine Rule

In small groups, undertake the following proof.

Follow the steps used for deriving the sine rule and verify if m_____ sin(M) = r_____ sin(R).

Application of the sine rule Now that we are familiar with the concept of the sine rule, let us see how we can apply them using the worked examples below.

Example 9.5

A ship is moving from Point A to Point B, then to Point C. At A, the angle between the paths to B and C is 45⁰. At B, the angle between the paths to A and C is 60⁰. If the distance between A and B is 8km, find the distance between B and C.

Solution

Step 1: Sketch the movement of the ship:

Figure 9.7: Graphical representation of the ship’s movement

Step 2: Since we have two angles, we can calculate the third angle in the triangle:

< ACB = 180⁰− 45⁰− 60⁰= 75⁰Figure 9.6: Moving Ship

Step 3: Applying the sine rule:

|AB|_ sin 75⁰= ||BC|_ sin 45⁰Step 4: Substitute the value for |AB| and simplify the equation 8_ sin 75⁰= ||BC|_ sin45⁰|BC| = 8 (sin45 0) _ sin 75 0 |BC| = 5.86 km (to 3sf) Therefore, the distance between Point B and C is 5.86 km .

Example 9.6

A drone needs to deliver medicine to a hospital to help save a patient’s life. At the hospital (H), the angle between the drone’s position (D) and the supply depot (S) is 50⁰. From the drone’s current position, the angle to the hospital and a nearby supply depot is 70⁰. The distance between the drone and the supply depot is 12 km.

a. Find the distance between the:

i. drone and the hospital.

ii. hospital and the supply depot.

b. If the drone has to pick up the medicine from the supply depot and deliver it to the hospital, how many kilometres will it travel?

Solution

a. i. Step 1: Sketch the movement of the drone

Figure 9.9: Drone movement

Figure 9.8: Drone delivering Medicine

Step 2: Since we have two angles, we can calculate the third angle in the triangle.

< HSD = 180⁰− 50⁰− 70⁰= 60⁰Step 3: Apply the sine rule:

|HD|_ sin60⁰= ||DS|_ sin50⁰Step 4: Substitute the value for |DS| and simplify the equation |HD|_ sin60⁰= 12_____ sin50⁰|HD| = 12 (sin60 0) _ sin50 0 |HD| = 13.57km Therefore, the distance between the hospital and drone’s current position is 13.6km (to 3sf).

ii. Step 1: Apply the sine rule:

|HS|_ sin 70⁰= ||DS|_ sin50⁰Step 2: Substitute the value for |DS| and simplify the equation |HS|_ sin70⁰= 12_____ sin50⁰|HS| = 12 (sin70 0) _ sin50 0 |HS| = 14.72km Therefore, the distance between the hospital and supply depot is 14.7km (to 3sf).

b. Distance travelled = |DS| + |HS| = 12 + 14.7 = 26.7km (to 3sf) The drone needs to travel 26.7km in order to get to the hospital with the medicine.

Example 9.7

Joel moves from the school entrance walking at 5 km/h on a bearing of 050⁰.

Edith leaves the same point (school entrance) as Joel at the same time and cycles on a bearing of 125⁰travelling at a constant speed. Find Edith’s average cycling speed if she and Joel are 60km apart after 6 hours.

Solution

Step 1: Use the speed distance formula to determine the kilometres walked by Joel Speed = distance______ time 5 = distance/6 Distance = 6 × 5 = 30km

Figure 9.10: Graphical illustration of Joel and Edith’s movement

Step 2: Sketch the movement of Joel and Edith

Step 3: Calculate the distance covered by Edith using the sine rule Let θ be angle formed at Edith’s final position 60_ sin 75⁰= 30___ sinθ sinθ = 30sin 75⁰________ 60 θ = sin⁻¹(30sin75⁰_ 60 ) θ = 28.88⁰Now, 180⁰− 75⁰− 28.88⁰= 76.12⁰(sum of interior angles in a triangle) Let distance covered by Edith be represented by d.

60_ sin 75⁰= d_______ sin76.12° d = 60sin 76.12°_________ sin 75⁰ d = 60.30km

Step 4: Use the distance – speed relationship to determine Edith’s cycling speed.

Edith’s Speed = 60.30/6 Edith’s Speed = 10.05km / h Edith’s average cycling speed is 10.05km/h Cosine Rule In section 5, on vectors, we learnt about the cosine rule and proved it using measurements. We also discovered that when three sides are known or when two sides and an angle included between the two sides are known, the cosine rule can assist in finding the unknown angles and lengths. Let us see how algebra can assist us in proving the cosine rule. Suppose ∆ DEF in Figure 9.11 has the coordinates of vertex F as (x, y), then sin(E) = y__and cos(E) = x__. Hence, y = dsin(E) and x = dcos(E) as depicted in Figure 9.11

Figure 9.11: Oblique triangle supporting Cosine Rule

Activity 9.8: Algebraic proof of the cosine rule In small groups, work through the following activity.

1. Draw a triangle such as the one in Figure 9.11

2. Define the coordinates of the vertices as E(0, 0), D(f, 0) and F(dcos(E), dsin(E).

3. Write the distances in terms of the coordinates ( |ED| = f, |EF| = √_______________ dcos(E)²+ dsin(E)², |^(DF)| = √_______________________ (dcos(E) − f)²+ (dsin(E) − 0)²).

4. Calculate |DF|²and simplify using

5. |^(DF)|²= (dcos(E) − f)²+ d²sin²(E) as a guide.

6. You should have |DF|²= d²+ f²− 2dfcos(E)

7. Substitute |DF| with e to get e²= d²+ f²− 2dfcos(E)

8. Share your results with a classmate

9. You have successfully proved the cosine rule.

10. Investigate what happens to the cosine rule when dealing with a right- angle triangle.

Generalisation(s)

1. e²= d²+ f²− 2dfcos(E)

2. d²= f²+ e²− 2fecos(D)

3. f²= d²+ e²− 2decos(F)

Example 9.8

Given that ∆ GMN has |GM| = 12m, |MN| = 17.8m and <GMN = 75⁰, determine the value(s) of:

a. <MGN

b. <GNM to the nearest whole number.

Solution

Step 1: Sketch the triangle:

Figure 9.12: Graphical representation of ∆ GMN

a. Step 2: Represent <MGN by θ

Step 3: Since two sides and the included angle is known, apply the cosine rule to find |GN|.

|GN|²= |GM|²+ |MN|²− 2|ᴳᴹ||MN|cos(75) |GN| = √__________________________ 12²+ 17.8²− 2(12)(17.8)cos(75) |GN| = 18.72m

Step 4: Substitute |GN| = 18.72 into the cosine rule to find θ |MN|²= |GM|²+ |GN|²− 2|GM||GN|cos(θ) 17.8²= 12²+18.72²− 2(12)(18.72)cos(θ)

Step 5: Make cos(θ) the subject of the equation 449.28cos(θ) = 12²+ 18.72²− 17.8²cos(θ) = 177.598/449.28

Step 6: Take co s⁻¹of 177.598/449.28 θ = cos⁻¹(177.598_ 449.28 ) θ = 66.72⁰Therefore <MGN = 67⁰to the nearest whole number.

b. Step 1: Represent <GNM by β

Step 2: Since two sides and the included angle is known, apply the cosine rule.

|GM|²= |GN|²+ |MN|²− 2|ᴳᴺ||MN|cos( β) 12²= 18.72²+17.8²− 2(18.72)(17.8)cos( β)

Step 3: Make cos( β) the subject of the equation 666.432cos( β) = 18.72²+ 17.8²− 12²cos(β) = 523.2784/666.432

Step 4: Take co s⁻¹of 523.2784/666.432 β = cos⁻¹(523.2784_ 666.432 ) β = 38.26⁰Therefore < GNM = 38⁰to the nearest whole number.

Just as we are able to solve algebraic equations, we are also able to solve trigonometric equations. In the same way as algebraic equations, trigonometric equations can come in the form of linear, quadratic, cubic form, etc. Here we will focus on linear and quadratic trigonometric equations.

Linear Equations

A linear equation is a mathematical statement that shows a relationship between variables where each term is either a constant or a product of a constant and a single variable and the equation represents a straight line when graphed on a coordinate plane.

Example 9.9

Find the value of x given that 2cosx − 1 = 0

Solution

Step 1: Write the equation 2cosx − 1 = 0

Step 2: Group like terms 2cosx = 1

Step 3: Divide both sides by 2 and simplify 2cosx_ 2 = 1/2 cosx = 1/2

Step 4: Take cos⁻¹of both sides x = cos⁻¹(1_ 2) x = 60⁰Therefore, x = 60⁰.

Example 9.10

Find the value of w given that 2sinw − √3 = 0

Solution

Step 1: Write the equation 2sinw − √3 = 0

Step 2: Group like terms 2sinw = √3

Step 3: Divide both sides by 2 and simplify 2sinw_ 2 = √_ 3/2 sinw = √_ 3/2

Step 4: Take sin⁻¹of both sides w = sin⁻¹(√3_ 2 ) w = 60⁰Therefore, w = 60⁰.

Example 9.11

Find the exact solution for tan2v = √3

Solution

Step 1: Write the equation tan2v = √3

Step 2: Take tan⁻¹of both sides and divide by 2 v = tan⁻¹(√3)_ 2 v = 30⁰Therefore, when v = 30⁰, tan2v = √3.

Example 9.12

Find the exact solution for 4 cos3t + 2 = 0

Solution

Step 1: Write the equation 4 cos3t + 2 = 0

Step 2: Group like terms 4 cos3t = − 2

Step 3: Divide both sides by 4 and simplify 4 cos3t_ 4 = −2_ 4 cos3t = −1_ 2

Step 4: Take cos⁻¹of both sides and divide by 3 3t = cos⁻¹(− 1_ 2 ) t = 120⁰_ 3 t = 40⁰Therefore, when t = 30⁰, 4 cos3t + 2 = 0.

Quadratic Equations

A quadratic equation is a mathematical equation where the highest power of the variable is 2. It represents a parabolic curve when graphed on a coordinate plane.

It is usually in the form ax²+ bx + c where a, b, c are constants.

Let us go through some examples to see how this works in trigonometric equations.

Example 9.13

Find the exact solution(s) for 2sin²m − sinm = 0 .

Solution

Step 1: Write the equation 2sin²m − sinm = 0

Step 2: Factorise sinm sinm(2sinm − 1) = 0

Step 3:

sinm = 0 , or (2sinm − 1) = 0 sinm = 0, or sinm = 1 __

Step 4: take sin⁻¹of both sides and simplify m = sin⁻¹(0), or m = sin⁻¹(1/2) m = 0⁰, or m = 30⁰Therefore, when m = 0⁰and m = 30⁰, 2sin²m − sinm = 0.

Example 9.14

Find the exact solution(s) for sin²y = 2 − 2 cosy .

Solution

Step 1: Write the equation sin²y = 2 − 2 cosy sin²y + 2 cosy = 2

Step 2: Replace sin²y with 1 − cos²y (trigonometric identities) and group like a quadratic equation:

1 − cos²y + 2 cosy = 2 cos²y − 2 cosy + 1 = 0

Step 3: Factorise:

(cosy − 1)(cosy − 1) = 0 cosy = 1

Step 4: Take cos⁻¹of both sides and simplify:

y = cos⁻¹(1) y = 0⁰Therefore, when y = 0⁰, sin²y = 2 − 2 cosy.

Example 9.15

Find the exact solution(s) for 2cos²x + cosx − 1 = 0

Solution

Step 1: Write the equation:

2cos²x + cosx − 1 = 0

Step 2: Factorise:

(cosx + 1)(2 cosx − 1) = 0

Step 3: Solve the equations:

cosx = - 1, 2 cosx − 1 = 0 cosx = - 1, cosx = 1/2

Step 5: take cos⁻¹of both sides and simplify:

x = 180⁰or x = 60⁰Therefore, when x = 180⁰and x = 60⁰, 2cos²x + cosx − 1 = 0.

1. Given that sinρ = 5/8, find cscρ .

2. If sinβ = 12/13, and cosβ = 5/13, find the value of cotβ.

3. Swedru School of Business’ rectangular football field’s diagonal forms an angle θ with one of its shorter sides. If the field’s length is 40 metres and its width is 30 metres:

a. Calculate the value of sinθ + cosθ.

b. Prove that sin²θ + cos²θ = 1 using the dimensions of the field.

4. Determine the exact value of cos ( 120⁰) without using calculators.

5. Simplify (1 - cos²φ ) sec²φ

6. Eliminate δ from the equations v = 9cosδ and r = 7sinδ .

7. Find the value of tan( 150⁰) without using calculators.

8. Find the value of cos( 45⁰) without using calculators.

9. Find the value of 2 sin 30⁰cos 30⁰without using calculators.

10. Simplify tan 45⁰− tan30⁰______________ 1 + tan45⁰tan 30⁰without using calculators.

11. Find the angles between 0⁰and 180⁰which satisfy the equations

a. sinv = 0.45 .

b. cos k = − 0.63 .

c. tanβ = 2.15.

12. Prove that cot σ + tan σ ≡ cosec σ sec σ

13. Show that:

a. sin⁴φ − cos⁴φ ≡ 1 − 2 cos²φ

b. sec⁴β − sec²β ≡ tan²β + tan⁴β

c. tan²σ − sin²σ ≡ sin⁴σ sec²σ

14. Find values of α in the interval 0⁰≤ α ≤ 180⁰for which tan 2 α − tan α = 2.

15. In a triangle MNP, it is known that < N MP = 73⁰, < MNP = 49⁰and |NP|= 12.2m. Find the value(s) of:

a. < N PM

b. |MN|

c. |MP|

16. Two buses (A and B) set out from a bus terminal at the same time. Bus A moves North while bus B moves on a bearing of 057⁰. When bus B had travelled 10km, the two buses were 15km apart. How far was bus A from the bus terminal?

17. Find the value of y in the equation 2sin(y) − 1 = 0.

18. Determine the value of v given that sin(v) − 2sin(v)cos(v) = 0.

19. A vertical tower JK of height 40m is observed from two points G and H in the same horizontal plane as K, the foot of the tower. The points K, G and H lie a straight-line KG = GH. Given the angle of elevation of J from H is 60⁰, calculate:

a. The distance of G from the foot of the tower.

b. The angle of elevation of J from G.