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# Name - Penditamuda's Blog | Berkongsi Sesama Cahayasas.edu.my 6 5.2.2 Define cotangent, secant and cosecant of any angle in a Cartesian plane. sin 1 cos tan 2 1 1 r sin y 11 cos 11

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TRIGONOMETRIC FUNCTIONS

Name

........................................................................................

TRIGONOMETRIC FUNCTIONS

5.1 Positive Angle and Negative Angle

Positive Angle Negative Angle

Represent each of the following angles in a Cartesian plane and state the quadrant of the angle.

Example

60

1(a) 70 (b) 150

Example

215

2(a) 195 (b) 345

Example

395

3(a) 415 (b) 480

Example

5

4

4(a) 3

4 (b)

5

3

Example

45

5(a) 130 (b)

1

3

60

y

x O

y

x O

y

x O

II

I

III

IV

y

x

90 2

0

360 (2) 180 ()

270 3

2

215

y

x O

y

x O

y

x O

A positive angle is measured

in an anticlockwise direction

from the positive x-axis.

45

y

x O

y

x O

y

x O

y

x O

y

x O

5

4

y

x O

y

x O

y

x O

60

y

x O

Anticlockwise

direction

45

y

x O

Clockwise

direction

395

y

x O

35

360

A negative angle is measured

in a clockwise direction

from the positive x-axis.

5.2 Six Trigonometric Functions of any Angle (1)

5.2.1 Define sine, cosine and tangent of any angle in a Cartesian plane

1

sin

cos

tan

Conclusion :

2

r2 = 3

2 + 4

2

r = 2 23 4

r = 5

Conclusion :

Pythagoras’ Theorem :

2 2 2 2 2

2 2 2 2 2

2 2 2 2 2

c a b c a b

a c b a c b

b c a b c a

3. Find the length of OA and the values of sine, cosine and tangent of .

OP =

=

sin =

5

cos =

12

tan =

5

OP =

=

sin =

6

=

cos =

8

=

tan =

6

=

x

r y

Oppositesin

Hypotenuse

Hypotenuse

Oppositetan

3

r 4

b

c a

P (12, 5)

5

12 O x

y

P (8, 6)

x

y

O 8

6

Opposite to

Hypotenuse

OP =

=

sin =

3

cos =

4

tan =

3

=

OP =

=

sin =

12

cos =

5

tan =

12

(e) Conclusion:

Sin is positive for in quadrant ……. and …….

Cos is positive for in quadrant ……. and …….

Tan is positive for in quadrant ……. and …….

Sin is negative for in quadrant ……. and …….

Cos is negative for in quadrant ……. and …….

Tan is negative for in quadrant ……. and …….

4. Find the corresponding reference angle of .

(a)

Reference angle = 55

(b)

Reference angle = 110

= 70

(c)

Reference angle = 215

= 35

(d)

Reference angle = 300

= 60

y

x

90

0

360 180

270

Sin

Cos

Tan

Sin

Cos

Tan

Sin

Cos

Tan

+

Sin

Cos

Tan

P (4, 3)

x

y

4 O

3

P (5, 12)

x

y

5

O 12

55

y

x

Fill in with or + sign.

110

y

x 180 360

215

y

x 180 360

300

y

x 180 360

(e) Conclusion:

Reference angle (RA) is the acute angle formed between the rotating ray of the angle and the

______________________________

sin = sin (180 ) sin = sin ( 180) sin = sin (360 )

cos = cos (180 ) cos = cos ( 180) cos = cos (360 )

tan = tan (180 ) tan = tan ( 180) tan = tan (360 )

5. Given that cos 51 = 0.6293, find the trigonometric ratios of cos 231 without using a calculator or

mathematical tables.

Reference angle of 231 = 231

=

cos 231 =

=

6. Given that sin 70 = 0.9397, find the trigonometric ratios of sin 610 without using a calculator or

mathematical tables.

Reference angle of 610 = 610

=

sin 610 =

=

7. Given that tan 25 = 0.4663, find the trigonometric ratios of tan 335 without using a calculator or

mathematical tables.

Reference angle of 335 = 335

=

tan 335 =

=

R.A = R.A =

R.A = R.A =

y

x

y

x O 180

y

x O

180

y

x O

360

y

x

y

x

y

x

5.2.2 Define cotangent, secant and cosecant of any angle in a Cartesian plane.

1

sin

cos

tan

2

1 1 r

sin y

1 1

cos

1 1

tan

3. Definition of cotangent , secant and cosecant .

1cosec

sin1

seccos

1cot

tan

4. Since sin

tancos

, then

cot

5.

sin sin 90

cos cos 90

tan tan 90

y x

r r

x y

r r

y x

x y

6.

Complementary angles:

sin = cos (90 )

cos = sin (90 )

tan = cot (90 )

cosec = sec (90 )

sec = cosec (90 )

cot = tan (90 )

7. Given that sin 48 = 0.7431, cos 48 = 0.6991 and tan 48 = 1.1106, evaluate the value of cos 42.

cos 42 =

= =

8. Given that sin 67 = 0.9205, cos 67 = 0.3907 and tan 67 = 2.3559, evaluate the value of cot 23.

cot 23 =

= =

9. Given that sin 37 = 0.6018, cos 37 = 0.7986 and tan 37 = 0.7536, evaluate the value of sec 53.

sec 53 =

= =

x

r y

x

r y

90

x

r y

48

90 48

67

90 67

37

90 37

5.2.3 Find values of six trigonometric functions of any angle

1. Complete the table below.

30 45 60

sin 1

2

cos

1

2

tan

1

2. Use the values of trigonometric ratio for the special angles, 30, 45 and 60, to find the value of the

trigonometric functions below

Example: Evaluate sin 210 a. Evaluate tan 300

Draw diagram to determine positive or negative

sin

Draw diagram to determine positive or negative

Find reference angle

Reference angle of 210 = 210 180

= 30

Find reference angle

Solve

sin 210 = sin 30

= 1

2

Solve

b. Evaluate cos 150 c. Evaluate sec 135

Draw diagram to determine positive or negative

Draw diagram to determine positive or negative

Find reference angle

Find reference angle

Solve

Solve

60 60

60 2

2

2

60

30 2

1

2 22 1

3

1

1

2 21 1

2

45

45

1 1

1

1

y

x 180 360

cos ( ) = cos

sin ( ) = sin

tan ( ) = tan

y

x O

5.2.4 Solve trigonometric equations

A. Steps to solve trigonometric equation

1. Determine the range of the angle.

2. Find the reference angle using tables or calculator.

3. Determine the quadrant where the angle of the trigonometric function is placed.

4. Determine the values of angles in the respective quadrants.

1. Solve the following equation for 0 360.

Example: sin = 0.6428

a. cos = 0.3420

Range :

0 360

0 360

Reference angle :

= sin1

0.6428

= 40

Actual angles

= 40 , = 180 40

= 40 , 140

b. tan = 1.192

c. cos = 0.7660

Range :

Reference angle :

Actual angles

y

x

y

x

y

x 180 360

S A

T C

y

x 180 360

S A

T C

y

x

y

x

y

x 180 360

40

S A

T C

y

x 180 360

40

S A

T C

d. sin = 0.9397 e. tan = 0.3640

Range :

Reference angle :

Actual angles

f. cot = 1.4826 g. cosec = 2.2027

Range :

Reference angle :

Actual angles

2. Solve the following equation for 0 360.

example : sec 2 = 2 a. 2 sin 2 = 1.6248

Range : 0 360

0 2 720

Reference angle :

12

21

22

2 60

cos

cos

Actual angles

2 = 60, 360 60, 60 + 360, (36060) + 360

= 60, 300, 420, 660

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x 180 360,720

S A

T C

60

60

b. cos 3 = 0.9781 c. tan

2

= 2.05

Range

Reference angle :

Actual angles

d. sin ( + 10) = 0.7660

e. cos ( + 40) = 0.7071

f. tan ( + 15) = 1

g. cos ( 20) = 0.5

h. tan (2 10) = 2.082

i. sin (2 30) = 0.5

j. sin = cos 20

k. cos = sin 55

Example : 2 sin x cos x = cos x

2 sin x cos x cos x = 0

cos x ( 2 sin x 1) = 0

cos x = 0 , 2 sin x 1 = 0

sin x = 1

2

x = 30

x = 90 , 270

x = 30, 150

x = 30, 90, 150, 270

m. 2sin x cos x = sin x

n. 2 cos 2 + 3 cos = 1

o. 2 sin2 + 5 sin = 3

p. tan2 = tan

q. 3 sin = 2 + cosec

y

x 360

y

x 180 360

S A

T C

3. Given that px sin and 00 < x < 90

0.

Express each of the following trigonometric ratios in terms of p.

(a) sec x =

(b) cosec x =

(c) tan x =

(d) cot x =

(e) sin ( 900- x) =

(f) cos (900- x) =

(g) sec (900- x) =

(h) cosec (900 – x) =

(i) tan ( 90o - x) =

(j) cot ( 90o – x ) =

(k) sin(-x) =

(l) cos (-x) =

x

4. Given that 17

8sin x and 270

0< x < 360

0.

Without using tables or calculator, find the values

of.

5. Given that 17

8-cos x and 180

0< x < 270

0.

Without using tables or calculator, find the values

of

(a) cos x =

(a) sin x =

(b) tan x =

(b) tan x =

(c) cosec x =

(c) cosec x =

(d) sec x =

(d) sec x =

(e) cos (900 – x) =

(e) sec (900 – x) =

(f) sin ( 900 – x ) =

(f) cot ( 900 – x ) =

(g) sin (-x) =

(g) sin (-x) =

(h) tan (-x) =

(h) cos (-x) =

x

x x

5.4 Basic Identities

5.4.1 Prove Trigonometric Identities using Basic Identities

Three basic trigonometric identities :

sin 2 + cos

2 = 1

1 + tan 2 = sec

2

1 + cot 2 = cosec

2

Formula of compound angle :

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

cos (A B) = cos A cos B Ŧ sin A sin B

tan (A B) = tan tan

1 tan tan

A B

A B

Formula of double angle :

sin 2A = 2 sin A cos A

cos 2A = cos2 A − sin

2 A

= 2 cos2 A − 1

= 1 − 2sin2 A

tan 2A = A

A2tan1

tan2

Formula of half angle :

sin A = 2 sin 2

A cos

2

A

cos A = cos2

2

A − sin

2

2

A

= 2 kos2

2

A − 1

= 1 − 2sin2

2

A

tan 2A = 2

22

12

Atan

Atan

1. Prove the following identities

Example: cot + tan = cosec sec

2 2

1

cos sincot tan

sin cos

cos sin

sin cos

sin cos

cosec sec

a. tan2 (1 sin

2 ) = sin

2

b. 2

11

sincos

cos

c. sin2 + cot

2 = cosec

2 cos

2

cos2 = 1 – sin

2

sin2 = 1 – cos

2

d. 2 1

1

sinsec tan

sin

e.

12

1

sin x cos xsec x

cos x sin x

2. Solve the following equations for 0 x 360.

a. 3 sin x + 2 = cosec x

b. 2 cot2 x 5 cot x + 2 = 0 c. cos

2 x 3 sin

2 x + 3 = 0

d. cot2 x= 1 + cosec x e. 2 tan

2 x = 4 + sec x

5.2.1

5. cos 51= 0.6293

5.2.4

2a. 0 2 720 , 54.33

= 27.17, 62.83, 207.17, 242.83

3(a) 21

1

p

5.(a) 17

15

6. sin 70= 0.9397 b. 0 3 1080 , 12.01

= 56, 64, 176, 184, 296, 304 (b)

p

1 (b)

15

8

7. tan 25= 0.4663 c. 0

2

180 , 64

= 232

(c) 21 p

p

(c)

17

15

5.2.2

7. sin 48 = 0.7431

d. = 40, 120 (d)

p

p 21 (d)

17

8

8. tan 67 = 2.3559 e. = 5, 275 (e)

21 p

(e) 17

15

9. cosec 37 = 1.6617 f. = 30, 210 (f) p

(f) 8

15

5.2.3

2a. tan 300 = 3

g. = 80, 320 (g) p

1 (g) 15

17

b. cos 150 = 3

2

h. = 62.83, 152.83, 242.83, 332.83 (h)

21

1

p (h)

17

8

c. sec 135 = 1

2

i. = 30, 90, 210, 270 (i)

p

p 21

5.2.4

= 70, 290

j. = 70, 110 (j)

21 p

p

b. 0 360 , 50 ,

= 30, 330

k. = 145, 215 (k) -p

c. 0 360 , 40 ,

= 140, 220

m. = 60, 180, 300 (l) 21 p

d. 0 360 , 70

= 250, 290

n. = 120, 180, 240 4.(a)

17

15

e. 0 360 , 20.01

= 159.99, 339.99

o. = 30, 150 (b)

15

8

f. 0 360 , 34

= 146, 326

p. = 0, 45, 225 (c) 17

8

g. 0 360 , 27

= 207, 333

q. = 90, 199.47, 350.53 (d)

15

17

(e)

17

8

(f)

17

15

(g)

17

8

(h)

8

15