STUDY KNOWHOW PROGRAM STUDY AND LEARNING CENTRE. Trigonometry

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STUDY KNOWHOW PROGRAM

STUDY AND LEARNING CENTRE

Trigonometry

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Trigonometry Booklet Page 2 of 24 Nov 2017

Contents

Pythagoras’ Theorem ... 3

Right Triangle Trigonometry ... 5

Sine Rule ... 9

Cosine Rule ... 13

Angular measurement - Radians ... 15

Circular Functions ... 17

Trigonometric Equations ... 20

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PYTHAGORAS’ THEOREM

Right-angled triangles

A right-angled triangle has an angle of 90⁰_{. In a right-angled triangle the side opposite the right angle}

is called the hypotenuse. It is also the longest side.

Pythagoras’ Theorem Pythagoras’ theorem states:

In a right angled triangle the square of the length of the hypotenuse (h) is equal to the sum of the squares of the other two sides.

Pythagoras’ Theorem can be used to find a side length of a right angled triangle when you know the other two side lengths.

Examples:

Finding the hypotenuse

Find the length of the hypotenuse (h) in the following triangle.

Finding a shorter side

Find the value of x in the triangle below.

h

2

= a

2

+ b

2

4.2

2

= 2.7

2

+ x

2

4.2

2

- 2.7

2

= x

2 10.35 =

x

2 x = 3.22

h

2

= a

2

+ b

2

hypotenuse

h

b

a

Right angle

(90o₎

Use Pythagoras theorem

h

2

= a

2

+ b

2

with a = 6cm. and b = 8

cm:

4.2 2.7 x h 6 8

h

2

= a

2

+ b

2

h

2

= 6

2

+ 8

2

h

2

= 100

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Pythagorean triples

In some right-angled triangles all three sides have integer values. These three values form a

Pythagorean triple.

Some examples are triangles with sides: (3,4,5), (5,12,13), (7,24 25) and (8,15,17) Check! Multiples of these, such as (6,8,10) and (9,12,15) are also Pythagorean triples.

Exercises

Find the missing sides in the following. (1) (2) (3) (4) Answers: (1) 15 mm. (2) 19.13 (3) 7.84 cm. (4) 12.12 cm. 14 cm

a

6.2 cm 4.8 cm 20.2 6.5 h 12mm 9mm 7cm

c

a

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RIGHT TRIANGLE TRIGONOMETRY

Trigonometry is a branch of mathematics involving the study of triangles, and has applications in fields such as engineering, surveying, navigation, optics, and electronics. The ability to use and manipulate trigonometric functions is necessary in other branches of mathematics, including calculus, vectors and complex numbers.

Right-angled Triangles

In a right-angled triangle the three sides are given special names.

The side opposite the right angle is called the hypotenuse (h) – this is always the longest side of the triangle.

The other two sides are named in relation to another known angle (or an unknown angle under consideration).

Trigonometric Ratios

In a right-angled triangle the following ratios are defined for a given angle θ

sine θ =

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑙𝑒𝑛𝑔𝑡ℎ

cosine θ =

𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑙𝑒𝑛𝑔𝑡ℎ

tangent θ =

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ

These ratios are abbreviated to sin θ, cos θ, and tan θ respectively.

A useful memory aid is SOH CAH TOA: Sin=Opp/Hyp Cos=

A

dj/Hyp Tan=Opp/Adj

These ratios can be used to find unknown sides and angles in right-angled triangles. Examples

Evaluating ratios

In the right-angled triangle below evaluate sin θ, cos θ, and tan θ.

sin θ = o h = 4 5 = 0.8 cosθ = 𝑎 h = 3 5 = 0.6 tan θ = o a = 4 3 = 1.33

If this angle is known or under consideration



this side is called the opposite side because it is opposite the angle

This side is called the adjacent side

because it is adjacent to or near theangle

h 4 5 3  θ θ θ

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Finding angles.

Find the value of the angle  in the triangle below

In this triangle we know two sides and need to find the angle .

The known sides are the opposite side and the hypotenuse.

The ratio that relates the opposite side and the hypotenuse is the sine ratio.

sin θ = _hypotenuseopposite

sin θ = 13.4

19.7 sin θ = 0.6082

θ = sin-1_(0.6082)_{[Using sin}-1_{function on the calculator]}

θ = 42.9⁰

Finding side lengths

Find the value of the indicated unknown side length in each of the following right-angled triangles.

(a) tan θ = opposite adjacent tan 27⁰₌ b 42

b = 42 × tan27⁰ b = 21.4cm (b)

cos θ =

adjacent hypotenuse

cos 35⁰₌7 x x = 7 × cos 35⁰ _{x = 8.55}_{See Exercise 2} Method

1.

Determine which ratio to use.

2.

Write the relevant equation.

3.

Substitute values from given information.

4.

Solve the equation for the unknown

In this problem we know an angle, and the adjacent side.

The side to be determined is the opposite side. The ratio that relates these two sides is the tangent ratio.

In this problem we know an angle, and the adjacent side. The unknown side is the hypotenuse.

The ratio that relates these two sides is the cosine ratio.

13.4 19.7  b c b x 35o 7 27o 42 cm

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Special angles and exact values

There are some special angles for which the trigonometric functions have exact values rather than decimal approximations.

Applying the rules for sine, cosine and tangent to the triangles below, exact values for the sine, cosine

and tangent of the angles 30⁰_{, 45}⁰_{and 60}⁰_{can be found.}

Exercises Exercise 1

Using the right-angled triangle below find:

(a) sin θ, (b) tan θ, (c) cos α, (d) tan α (Hint: Use Pythagoras theorem to find the hypotenuse)

Exercise 2

Find the value of the indicated unknown (side length or angle) in each of the following diagrams.

(a) (b)

o o o

1

sin 45 =

, cos 45 =

,

tan 45 =

2

o

3

o

1

o

sin 60 =

, cos 60 =

, tan 60 = 3

2

o

1

o

3

o

1

sin 30 =

, cos 30 =

, tan 30 =

2

3

47o _{4.71 mm}

a

62o 14cm

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27o

(c) (d)

(e) (f)

(g) In a right-angled triangle sin ∅ = 0.55 and the hypotenuse is 21 mm. Find the length of each of the other two sides.

Answers Exercise 1 (a) 12/13 = 0.9231 (b) 12/5 = 2.4 (c) 12/13 = 0.9231 (d) 5/12 = 0.4167 Exercise 2 (a) = 6.6 cm (b) a = 3.4 mm (c) z = 7.8,  = 37.7 (d) θ =18.8, x = 19.1 (e) a = 44.4 (f) b = 47.1 (g) 11.6mm. and 17.5mm. z



b 500 a 34



20.2 6.5 6.2 cm 4.8 cm 42 x

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Trigonometry Booklet Page 9 of 24 Nov 2017 C A b a B c p sin P= r sin R

SINE RULE

The sine rule can be used to find angles and sides in any triangle (not just a right-angled triangle) when given:

(i) One side and any two angles

OR (ii) Two sides and an angle opposite one of the given sides.

In the triangle ABC below:

angles A,B,C, are the angles at the vertices A,B,C respectively a,b,c are the side lengths opposite the angles A,B,C respectively.

The sine rule states: or or

Examples

1. In triangle PQR find:

a) side length p b) side length q

a) Side length p:

Use the sine rule in the form

The relevant part of the formula is

b) Side length q:

Angle Q is found using the fact that the sum of the three interior angles of a triangle add to 180°.

 Q = 180°  (70° + 30°) = 80° a sin A= b sin B= c sin C sin A a = sin B b = sin C c p sin P= q sin Q= r sin R p sin 70°= 15 sin 30° p =15 x sin 70° sin 30° p = 28.2 cm. P Q R p q 70° _30°

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Trigonometry Booklet Page 10 of 24 Nov 2017 a sin A= b sin B sin A a = sin B b = sin C c sin B b = sin C c sin126° 20 = sin C 12 C = sin−1_{( 0.485)} sin C = 12 𝑥 sin 126° 20 a sin A= b sin B= c sin C

From the sine rule p

sin P

=

q

sin Q

2. In triangle ABC find:

a) angle C

b) angle A

c) side length a.

a) Angle C:

C = 29° b) Angle A: A = 180°  (126° + 29°) → A = 25° c) Side length a:

a 28.2 sin 70°= q sin 80° q =28.2 x sin 80° sin 70° q = 29.6 cm. A B C 126° b = 20 m a sin 25°= 20 sin 126° a =20 x sin 25° sin 126° sin C = 0.485 a = 10.4 m.

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Trigonometry Booklet Page 11 of 24 Nov 2017 sin A a = sin C c sin 30° 8 = sin C 12 sin C =12 x sin 30° 8 sin C = 0.75 C = sin−10.75

3. Given a triangle ABC with angle A = 30°, adjacent side = 15 cm and opposite side = 8 cm as shown below, find angle C.

In this case there are two possible solutions. This is called the ambiguous case of the sine rule.

C = 48.6° This solution gives the triangle ABC1

Another solution to 3 is ( ) = 131.4° = (180° 48.6°). This solution gives triangle

ABC2

Note: In any non-right-angled triangle, where two sides and the non-included angle are given check for the ambiguous case if:

 the angle is acute and

 the length of the side adjacent to the angle is greater than the length of the side opposite the angle.

Exercises

Exercise 1

For the following triangles find the unknown sides.

a) b) c) r = ? t = ? q = ? c = ? b = ? 15° 27 130° R S 30° T B r = ? C A 10 30° C1 B A C2 65° C = sin−1( 0.75) Q 38 85° 45° R P

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Trigonometry Booklet Page 12 of 24 Nov 2017 35° 21 C 16 B A Exercise 2

For the following triangles find all unknown angles and sides.

a) b) c) Answers. Exercise 1 a) b = 18.1, c =19.9. b) q = 53.5, r = 41.2. c) t = 9.1, r = 20.2. Exercise 2 a) L = 35.4° , N = 64.6°, n = 7.8. b) Ambiguous case: C = 131.2°, B = 13.8°, b = 6.7 or C = 48.8, B = 96.2, b = 27.7 c) B = 41.7°, A = 73.3°, a = 6.8 a = ? n= ? B 5 80° 8.5 L M N 6.4 4.7 65° C A b = ?

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COSINE RULE

The Cosine Rule can be used to solve non-right triangles

The angles α, β and γ are respectively opposite the sides a, b, and c.

N.B. The side on the left hand side of the equation is opposite the angle listed at the end of the Equation:

a2_{= b}2_{+ c}2 _{– 2bc cos α}

Use the Cosine Rule when you are given

 two sides and the angle between them

 three sides

Examples

1. Find the value of a in this triangle

2. Find the size of angle β in this triangle:

b2_{= a}2_{+ c}2 _{– 2ac cos β} 112 _{= 5}2_{+ 7}2 _{– 2×5×7×cos β} 121 = 25 + 49– 70 cos β 47 = – 70 cos β cos β = −47 70 β = 132⁰_11’ b =12 830 c = 15 a ° a = 5 b =11 c = 7 Cosine Rule a2_{= b}2_{+ c}2 _{– 2bc cos α} b2_{= a}2_{+ c}2 _{– 2ac cos β} c2_{= a}2_{+ b}2 _{– 2ab cos Υ} a2_{= b}2_{+ c}2 _{– 2bc cos α} a2_{= 12}2_{+ 15}2 _{– 2×12×15 cos 83}o a2_{= 144 + 225}_{– 360 × cos 83}o a2_{= 369 – 43.87} a2_{= 325.13} a = 18.03

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Exercise

1. Use the sine OR cosine rule to find the pronumeral shown:

a) b) c) d) 2. Find θ a) b) Answers 1 a) 28.7 b) 4.38 c) 8.33 d) 1.05 2 a) 34..₂o _{b) 42}.₉o 7.21 4.93

β

0 9.99 15.6 9.2

α

0 10.1 0.93 1.25 780 a 6.6 2.3 1610 b 23.1 19.6 840 a 13.6 8.2 350 c

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ANGULAR MEASUREMENT -

RADIANS

Definition of a radian

Though angles have commonly been measured in degrees they may also be measured in units known

as radians. One radian is the angle created by bending the radius length around the arc of a circle.

Converting between radians and degrees

Because the circumference of a circle is given by the formula C = 2πr, we know 2π radians (2πc_{) is a}

complete rotation and the same as 360 degrees.

Similarly half a rotation or 180 degrees = radians (180⁰_{= π}c_).

Angles that represent fractional parts of a circle can be expressed in terms of π:

Angle in Degrees Angle in Radians

90 π 2 45 π 4 60 π 3 30 π 6 270 _{3 × 90 = 3 ×}π 2 = 3π 2

For other angles rearranging πc_{= 180}⁰_gives:

Examples 1. Convert 60⁰_{to radians.} 1𝑂 = 𝜋 𝐶 180 60𝑂_{= 60 ×}𝜋𝐶 180 2. Convert 240⁰_{to radians.} 1𝑂 = 𝜋 𝐶 180

240𝑂_{= 240 ×}𝜋𝐶 180 1c₌180 𝑂 𝜋 and 1 𝑂₌𝜋𝐶 180

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Trigonometry Booklet Page 16 of 24 Nov 2017 60𝑂 = 𝜋 𝐶 3 ≅ 3.142 2 = 1.05 C ₂₄₀𝑂₌4𝜋 𝐶 3 3. Convert 𝜋 4 radians to degrees 1c₌180° 𝜋 𝜋 𝐶 4 = π 4 × 18045° 𝜋

𝜋𝐶 4 = 45 ⁰ 4. ‘Convert 6.5c_{to degrees} 1c₌ 180 ° 𝜋 6.5C_{= 6.5 ×}180° 𝜋

6.5C_{= 372.4}⁰

Note: The symbol for radian, c , is often omitted.

Exercise

1. Convert the following degrees to radians

a) 30⁰ _{b) 270}⁰ _{c) 20}⁰

d) 450⁰ _{e) 135}⁰ _{f) 57.3}⁰

2. Convert the following radians to degrees

a) π 2

d)3.5π

b) 5π 4

e)

π

c)

11π 6 f) 1 radian Answers 1. a) π 6 d) 5π 2 b) 3π 2 e) 3π 4 c) π 9 f) 1 radian 2. a) 90⁰ _{b) 225}⁰ _{c) 330}⁰ d) 630⁰ _{e) 180}⁰ _{f) 57.3}⁰

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CIRCULAR FUNCTIONS

The trigonometric ratios that have been defined in right-angled triangles can be extended to angles

greater than 90⁰_{by considering angles as rotations}

within a unit circle. The centre of the unit circle is at the point (0,0) and it has a radius of one unit:

Angles are considered as rotations from the positive x-axis.

Angles greater than 180⁰_{and negative angles can also be defined in terms of the unit circle.}

Anticlockwise rotations are considered positive and clockwise rotations are negative:

An angle of 135o_or3𝜋

4

c An angle of -75⁰ An angle of 210⁰ or – 150⁰

If P(x,y) is any point on the circle, and ∅ is the angle POQ in the triangle POQ as shown:

1 ∅

x

y



P(x, y)

x = cos(∅) y = sin (∅)

Q

O

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Then from the trigonometric properties of a right triangle

cos(∅) = OQ OP = x 1 = x, sin(∅) = PQ OP = y 1 = y, tan(∅) = PQ OQ = y x Examples

Cos, sin and tan values can also be calculated using a scientific calculator.

[Hint: make sure your calculator is in degrees or radian mode accordingly].

Exact values

Using the table of exact values and the symmetry of the unit circle it is also possible to find the exact

values for multiples of 30⁰_{, 45}⁰_{, 60}⁰_.

∅ ₃₀⁰₌𝜋 6 45 ⁰₌𝜋 4 60 ⁰₌𝜋 3 Sin ∅ 1 2 1 √2 = √2 2 √3 2 Cos ∅ √3 2 1 √2 = √2 2 1 2 Tan ∅ 1 √3 1 √3 Example Evaluate sin 330⁰

Plotting 330⁰_{on a unit circle shows that sin 330}⁰

is closely related to sin 30⁰_{. The y-coordinates}

differ only by sign because the distances from the x-axis are the same.

∴ sin 30⁰₌1

2 (from the table) and sin 330

⁰_{= −}1 2 . P(0.45, 0.89)



cos(∅) = x, sin (∅) = y, tan(∅) =

y

x 63o (i) sin 0 = 0 (ii) cos 180 = -1 (iii) sin (-90) = -1 (iv) sin 63o_{= 0.89} (v) tan(-180) = 0 −1 = 0 (vi) tan 63o₌0.89 0.45 = 1.98 (vii) cos (-297o_{) = 0.45}

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1) Evaluate cos 150⁰

3) Evaluate tan 4𝜋

3

Plotting 150⁰_{on a unit circle shows that cos 150}⁰

is closely related to cos 30⁰_{. The x-coordinates}

differ only by sign because the distances from the y-axis are the same.

∴ cos 30⁰₌√3

2 (from the table) and cos 150

⁰_{= −}√3 2 .

Plotting 4𝜋

3 on a unit circle shows that tan

4𝜋 3 is

closely related to tan 𝜋

3. Tan values of

diametrically opposite angles are the same.

∴ tan 𝜋

3 = √3 (from the table) and tan

4𝜋

3 = √3 .

Note: By drawing a sketch diagram it is always possible to use the symmetry of the unit circle and the values in the table to find the exact value of any multiple of 30⁰_{, 45}⁰_{, 60}⁰_{. The only difference will be a}

change of sign and the following diagram may assist in determining whether a negative sign will be required.

Rotating anticlockwise from the first quadrant:

All trigonometric functions are +ve in the first quadrant Sin is +ve for angles in the second quadrant

Tan is +ve for angles in the third quadrant Cos is +ve for angles in the fourth quadrant

It can be helpful to use a mnemonic such as

All Stations To Camberwell to remember these properties of the trigonometric functions

Exercise

1) What are the coordinates for points on the unit circle that make the following angles with the positive x-axis?

2) Find the exact value for:

Answers

a) 30⁰ b) 125⁰ c) -60⁰

d) 270⁰ e) -180⁰ f) 720⁰

a) sin 330𝑜 b) cos 210𝑜 c) sin(−30𝑜)

d)cos 90𝑜 e) tan 300𝑜 f) cos 180𝑜

g) sin(−120𝑜) h) cos 315𝑜

1. a) (0.87,0.5) b) (-0.56,0.82) c) (0.5,- 0.87) d) (0,-1) e) (– 1,0) f) (1,0)

2. a) -0.5 b) √3

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TRIGONOMETRIC EQUATIONS

Often the value of the trigonometric function is given and the corresponding angle(s), within a given domain, are required.

Examples

1) Given sin ∅ = 0.3 find all values of ∅ in the domain 0 ≤ ∅ ≤ 360⁰

The diagram shows that there are two angles in the given domain (one complete positive rotation) that will have a sin value of 0.3.

Using the calculator to find the first value:

sin ∅ = 0.3

→ ∅ = sin-1 _(0.3)

→ ∅ = 17.46⁰

Using symmetry the angle in the 2nd_{. quadrant}

is 180 – 17.46 = 162.54⁰

∴ The solutions are ∅ = 17.46⁰_{, 162.54}⁰

2) Solve 𝑐𝑜𝑠 ∝ = 0.5 over the domain -2𝜋 ≤ ∝ ≤ 2𝜋

The diagram shows that there will be four angles in the given domain (one complete positive rotation and one complete negative rotation) that will have a cos value of 0.5.

Using the table of exact values to find the first value:

cos ∝ = 0.5 → ∝= cos-1 _(0.5) → ∝ = 𝜋

3

Use symmetry to find the other angles.

∴ The solutions are ∝ = −5𝜋

3, − 𝜋 3, 𝜋 3, 5𝜋 3

3) Solve tan 2x = -3 over the domain -90 ≤ 𝑥 ≤ 180⁰

First, the domain must be adjusted to suit the angle 2x in the equation:

-90 ≤ x ≤ 180⁰ _{⟹ -180 ≤ 2x ≤ 360}⁰

Using the calculator to find the first value:

tan 2𝑥 = -3

→ 2𝑥 = tan-1 _(-3)

→ 2𝑥 = -71.57⁰

Tan values of diametrically opposite angles are the same and the diagram shows how to use symmetry to find other angles in the

domain 180 ≤ 2x ≤ 360⁰_:

2x = -71.57⁰_{, 108.43}⁰_{, 288.43}⁰

Finally we divide by 2 to find solutions for x:

x = -35.78⁰_{, 54.22}⁰_{, 144.22}⁰

Exercises

a) If sin ∅ = 0.25 find ∅ for 0⁰_{≤ ∅ ≤}₁₈₀⁰_{b) If tan ∅ = 0.8 find ∅ for 0}⁰_{≤ ∅ ≤}₃₆₀⁰

c) If cos ∅ = 0.4 find ∅ for -180⁰_{≤ ∅ ≤}₃₆₀⁰_{d) If cos ∅ = -0.4 find ∅ for -180}⁰_{≤ ∅ ≤}₃₆₀⁰

e) If tan ∅ = -1.5 find ∅ for 0⁰_{≤ ∅ ≤}₃₆₀⁰_{f) If cos ∅ = -0.3 find ∅ for 0}⁰_{≤ ∅ ≤}₃₆₀⁰

Answers

a) 14.5⁰_{, 165.5}⁰ _{b) 38.7}⁰_{, 218.7}⁰ _{c) 66.4}⁰_{, -66.4}⁰

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GRAPHS OF SINE AND COSINE

FUNCTIONS

The functions y = sin 𝑥 and y = cos 𝑥 have a domain of R and a range of [-1, 1]. The graphs of both

functions have an amplitude of 1 and a period of 2π radians (repeats every 2π units).

y = cos 𝑥

[Remember 𝜋≈3.142 so 2 𝜋≈ 6.284]

Change of amplitude and period

The graphs of both y = 𝑎 sin 𝑛𝑥 and y = 𝑎 cos 𝑛𝑥 have amplitude |𝑎| and period 2𝜋

𝑛

Examples

1) y = 3 sin 𝑥 2) y = 3cos 2𝑥

Vertical translation

The graphs of y = 𝑎 sin 𝑛𝑥 + k, y = 𝑎 cos 𝑛𝑥 + k are the graphs of y = 𝑎 sin 𝑛𝑥, y = 𝑎 cos 𝑛𝑥 translated up k units for k > 0 and down k units for k < 0.

x -6 -4 -2 2 4 6 y -3 -2 -1 1 2 3 4 x -8 -6 -4 -2 2 4 6 y -1.5 -1 -0.5 0.5 1 1.5 x -6 -4 -2 2 4 6 y -4 -2 2 4 y = 3 sin 𝑥 has: amplitude 3 (|𝑎| = 3) period 2π (n = 1) y = 3 cos 2 𝑥 has: amplitude 3 (|𝑎| = 3) period 2π 2 = π (n = 2) y = sin 𝑥

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Horizontal translation

Replacing x with (x ∅ ) shifts the graphs of y = sin 𝑥 and y = cos 𝑥 horizontally ∅ units to the right.

Replacing x with (x+∅ ) shifts the graphs of y = sin 𝑥 and y = cos 𝑥 horizontally ∅ units to the left.

Examples

1) y = 𝑠𝑖𝑛( 𝑥 − 𝜋

2) 2) y = 𝑐𝑜𝑠(𝑥 + 𝜋)

3) y = 3 𝑠𝑖𝑛( 4𝑥 − 𝜋)

First change y = 3 𝑠𝑖𝑛( 4𝑥 − 𝜋) to the form y = 3 𝑠𝑖𝑛 4( 𝑥 − 𝜋

4) so that the horizontal translation of

the graph is clear.

Reflection

Changing the sign of 𝑎 in the equations y = 𝑎 sin 𝑛𝑥 or y = 𝑎 cos 𝑛𝑥 results in a reflection about the x – axis. x -6 -4 -2 2 4 6 8 y -1.5 -1 -0.5 0.5 1 1.5 x -6 -4 -2 2 4 6 8 y -1.5 -1 -0.5 0.5 1 1.5 x -2 -1 1 2 3 y -3 -2 -1 1 2 3 4 y = 3 𝑠𝑖𝑛 4( 𝑥 − 𝜋 4) The period is 2𝜋₄ = 𝜋₂ The amplitude is 3.

There is a horizontal shift right of 𝜋

4.

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Trigonometry Booklet Page 23 of 24 Nov 2017 Example y = −3𝑐𝑜𝑠 2𝑥 Exercises

1.Sketch the graph of the following functions for one complete cycle stating the amplitude and period:

(a) y = 2cos 𝑥 (b) y = 2sin 3𝑥 (c) y = 1₂sin 2𝑥 (d) y = 3cos𝑥₂ (e) y = -2sin 3𝑥

2. Sketch the graph of the following functions for one complete cycle stating the amplitude and period.

(a) y = 2sin(𝑥 − 𝜋) (b) 3cos(x + 𝜋

2)

3. Sketch the graph of the following functions for one complete cycle stating the amplitude and period.

(a) y = 2sin(3𝑥 − 𝜋) (b) 3cos(4x − 2𝜋 ) (c) y = 2sin(2x + 𝜋

3)

Answers

1(a) 1(b) 1(c)

Amplitude = 2. Period = 2 Amplitude = 2. Period = 2/3 Amplitude = 0.5. Period = 

1(d) 1(e)

Amplitude = 3. Period = 4 Amplitude = 2. Period = 2/3

x -3 -2 -1 1 2 3 4 y -3 -2 -1 1 2 3 4 y = −3cos 2𝑥 x 1 2 3 4 5 6 y -2 -1 1 2 x 0.5 1 1.5 2 y -2 -1 1 2 x 0.5 1 1.5 2 2.5 3 y -0.5 0.5 x 2 4 6 8 10 12 y -3 -2 -1 1 2 3 x 0.5 1 1.5 2 y -2 -1 1 2

The graph of y = −3cos 2𝑥 is a reflection of y = 3cos 2𝑥 (dotted) in the x-axis

The period is 2𝜋₂ = 𝜋

(24)

2(a) 2(b)

Amplitude = 2. Period = 2 Amplitude = 1. Period = 2

3(a) 3(b) 3(c)

Amplitude = 2. Period = 2/3 Amplitude = 3. Period = /2 Amplitude = 2. Period = 

x 1 2 3 4 5 6 y -2 -1 1 2 x 1 2 3 4 5 y -1.5 -1 -0.5 0.5 1 1.5 x 0.5 1 1.5 2 y -2 -1 1 2 x 0.5 1 1.5 y -3 -2 -1 1 2 3 x 0.5 1 1.5 2 y -2 -1 1 2