SPM Trigonometry F5

8/11/2019 SPM Trigonometry F5

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SPM Additional Mathematics- by KBJIM- [email protected]

Trigoniometry Ypositive

A: The General Angle

O x1. Angles measured along Ox in a counter-clockwise direction are positive,

and those measured in a clockwise direction are negative. negative

2. There are four quadrants to consider :

1st

quadrant(ALL trigonometrical ratios:- sin , cos and tan are positive.90

2nd

quadrant(ONLY sin is positive, cos and tan are negative. 2ndQuadrant 1stQuadrant

3rd

quadrant (ONLY tan is positive, cos and sin are negative. Sin All180 0

4

th

quadrant(ONLY cos is positive, tan and sin are negative. Tan CosYou can use mnemonics to remember it :- like All Science Teachers Crazy 270

3rd

Quadrant 4th

Quadrant

3. Lets take an angle where 0 90. (we normally express as an acute angle )

Let s say = 30 and from your calculator, sin 30 = 0.5, cos 30 = 0.866, tan 30 = 0.577All the values are positive because = 30 is in the 1stquadrant which is forAll orAcute

Now take =150, from your calculator, sin 150 = 0.5 (positive), 90(2ndQuadrant) cos 150 = 0.866 (negative)

tan 150 = 0.577 (negative)150

Now take = 210, from your calculator, sin 210 = 0.5 ( negative), 30(3rdQuadrant) cos 210 = 0.866 (negative) 180 0

tan 210 = 0.577 (positive)210

330

Now take = 330, from your calculator, sin 330 = 0.5 (negative), 270(4thQuadrant) cos 330 = 0.866 (positive)

tan 330 = 0.577 (negative)

4. For 0 x 360, there will always be two angles which have the same trigonometrical ratios.

Notice that (use your calculator):( sin 30 = sin 150, cos 30 = cos 330, tan 30 = tan 210 ; si n 210 = sin 330, cos 150 = cos 210, tan 150 = tan 330 )

Lets take x=150,150is in the 2nd

quadrant. (Sine is positive )180 360 +

Use this rule150 30

sin 150= sin (180 30) = sin 30= 0.5cos 150= cos (180 30) = cos 30= 0.866tan 150 = tan (180 30) = tan 30= 0.577 210 330

180 + 360 **Remember : is an acute angle
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Lets take x=210,210is in the 3rd

quadrant. (Tangent is pos itive )180 360 +

Use this rule150 30

sin 210= sin (180 + 30) = sin 30= 0.5cos 210= cos (180 + 30) = cos 30= 0.866tan 210 = tan (180 + 30) = tan 30= 0.577 210 330

180 + 360 **Remember : is an acute angle

Lets take x= 330,330is in the 4th

quadrant. ( cosine is positive )180 360 +

Use this rule150 30

sin 330= sin (360 30) = sin 30= 0.5cos 330= cos (360 30) = cos 30= 0.866tan 330 = tan (360 30) = tan 30= 0.577 210 330

180 + 360

**Remember : is an acute angle

Rules:1. determine which quadrant the angle belongs to.2. use the rule (highl ighted in yellow) to get the acute angle.3. put in the appropriate sign for the required trigonometrical ratio.

Example T.1:Express the following trigonometrical ratios in terms of acute angles.

(i) sin 170 (ii) tan 200 (iii ) cos 300 (iv) sin (50) (v) cos (45) (vi) sin 325 (vii ) cos 120

(vii i) tan (140) (ix) sec 140 (x) cosec 130 (xi) cot 260 (xii) sec (25) (xiii ) cot (60)

Solution:

(i) sin 170 = sin (18010) = sin 10 (ii) tan 200 = tan (180 + 20) = tan 20

170 is in the 2nd

quadrant, sine is positi ve. 200 is in the 3rd

quadrant, tangent is positi ve.We are using sin = sin (180 ) , is an acute angle. We are using tan = tan (180 + ) , is an acute angle.

(iii) cos 300 = cos (360 30) = cos 30 (iv) sin(50) = sin 50

300 is in the 4th

quadrant, cosine is positi ve ( 50) is in the 4th

quadrant, sine is negativeWe are using cos = cos (360 ) , is an acute angle. We are using sin ( ) = sin , is an acute angle.

Remember..negative angles are drawn clockwi se.

(v) cos ( 45) = cos 45 (vi) sin 325 = sin (360 35) = sin 35

( 45) is in th e 4th

quadrant, cosine is positi ve. 325 is in the 4th

quadrant, sine is negativeRemember..negative angles are drawn clockwi se. We are using sin = sin (360 ) , is an acute angle .

(vii) cos 120 = cos 60 (viii) tan(140) = tan 220 = tan(180 + 40)= tan 40

120 is in the 2nd

quadrant, cosine is negative.We are using cos = cos (180 ) , is an acute angle. ( 140) is in the 3rd quadrant. Tangent is positive.

140 drawn clockwise = 220 drawn anti-clockwise.


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(ix) sec 140 = 1 = 1 = sec 40 (sec = 1 ) , sec = secantcos 140 cos 40 cos

140 is in the 2nd

quadrant, cosine and secant are negative

(x) cosec 130 = 1 = 1 = cosec 50 (cosec = 1 ) , cosec = cosecantsin 130 sin 50 sin

130 is in the 2nd

quadrant, sine and cosec are p ositive.

(xi) cot 260 = 1 = 1 = cot 80 (cot = 1 ) , cot = cotangenttan 260 tan 80 tan

260 is in the 3rd

quadrant, tangent and cotangent are positive

(xii) sec ( 25) = 1 = 1 = sec 25 .cos ( 25) cos 25

( 25) is i n 4th

quadrant, and cosine and secant are positi ve

(xiii) cot ( 60) = 1 = cot 60 tan 60

( 60) is angle drawn clockw ise. It is in 4th

quadrant.( 60) is same as 300 drawn anticlockwise..tangent and cotangent are negative in the 4

thquadrant.

B. Graph of sin , cos, tan

sin tan 1 notice that sin graph starts at origin

and peaks at 90270

0 90180 3600 90 180

1

cos 1 notice that cos graph starts at 1

and is zero at 90 and 270180 tan 0 and tan180 = 0

0 90 270 360 tan 90 is infinity ()

1


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C. Trigonometrical ratios you need to know

0 30 45 60 90 180

sin 0 1 2 3 1 0

2 2 2

cos 1 3 2 1 0 12 2 2

tan 0 3 1 3 03

Example T.2: Write down the values of the following.

(i) sin 540 (ii) tan 135 (iii ) cos 270 (iv) sin (120) (v) cos (30) (vi) sin 405 (vii ) cos 210

(vii i) tan (120) (ix) sin 150 (x) tan 210 (xi) tan ( 30) (xii) sin (150) (xii i) cos ( 135)

Solution:

(i) sin 540 = sin (360+180) = sin 180= 0 (ii) tan 135 = tan (180 45) = tan 45 = 1

(iii) cos 270 = cos (180+ 90) = cos 90= 0 (iv) sin (120) = sin (240) = sin 60 = 32

(v) cos (30) = cos (30) = 3. (vi) sin 405 = sin (360+ 45) = sin 45 = 22 2

(vii) cos 210 = cos (180+ 30) = cos 30 = 3. (viii) tan (120) = tan (240) = tan 60 = 32

(ix) sin 150 = sin (180 30) = sin 30= 1 (x) tan 210 = tan 30 = 32 3

(xi) tan (30) = tan 30 = 3 (xii) sin ( 150) = sin (210) = sin 30 = 13 2

(xiii) cos ( 135) = cos 225 = cos 45 = 22

Example T.3: Find the values of from 0 to 360, inclusive, which satisfy the following equations.

(i) cos = (ii) tan =1 (iii) cosec =2 (iv ) sin = 0.866 (v) cos =0.6 (vi) tan =3

(vii) sin =cos150 (viii) cos =tan 135 (ix) sec =tan 120 (x) cos =sin 150 (xi) sin2 =

(xii) t an (70 ) =1 (xiii) sin ( 30) =2 (xiv )) cos (40 ) =0.5 (xv) cos ( + 60) =0.52


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Solution:

(i) cos = (1. Use your calculator to get the acute angle for cos = . Ignore the minus sign2. your calculator gives you = 60

= 120, 240 3. Now the minus sign i n front of the indicates that the required angles are in the 2ndand3

rdquadrant. Remember: there are always two trigonometrical ratios that have the same value.

4. 2nd

quadrant angle is (180 60) and 3rd

quadrant angle is (180 + 60). Use the rule highligh ted

in yellow in the notes on page 2 ).

(ii) tan = 1 (1. Use your calculator to get the acute angle for tan = 1.2. your calculator gives you = 45

= 45, 225 3. Now the plus sign in front of the 1 indicates that the required angles are in the 1st and3

rdquadrant. Remember: there are always two trigonometrical ratios that have the same value.

4. 1st quadrant angle is 45 and 3

rdquadrant angle is (180 + 45). Use the rule highligh ted


(iii) cosec = 2 It means that 1 = 2sin

sin = (1. Use your calculator to get the acute angle for sin = .2. your calculator gives you = 30

= 30, 150 3. The plus in front of indi cates that the required angles are in the 1st and2nd quadrant for sine .

4. 2nd

quadrant angle is (180 30) and 1st quadrant angle is 30. Use the rule highl ighted


(iv) sin = 0.866 (1. Use your calculator to get the acute angle for sin = 0.866. Ignore the minus sign2. your calculator gives you = 60

= 240, 300 3. Now the minus sign in front of the 0.866 indi cates that the required angles are in the3rd and 4

th quadrant for sine .

4. 4th

quadrant angle is (360 60) and 3rd

quadrant angle is (180 + 60). Use the rule

highlightedin yellow in the notes on page 2 ).

(v) cos = 0.6 (vi) tan = 3 (vii) sin = cos 150 = 0.866

= 53, 307 = 120, 300 = 240, 300

(viii) cos = tan 135 = 1 (ix) sec = tan 120 = 3 (x) cos = sin 150 = 0.5

= 180 cos = 1 = 0.577 = 60, 3003

= 125.3, 234.8

(xi) sin2= 0..25 (xii) tan (70 ) = 1 (xiii) sin ( 30) = 22

sin = 0.5 70 = 45, 225 30 = 45, 135

= 30, 150, 210, 330 = 35 , 205 = 75, 165

(xvi) cos (40 ) = 0.5 (xv) cos (+ 60) = 0.5

40 = 60, 300 + 60 = 60, 300

= 340, 100 = 0 , 240


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Example T.4: Solve the following equations for values 0 360.

(i) sin 2= (ii) tan 2=1 (iii) cos 3=2 (iv) tan 2= 1 (v) sin(2 +30) =0.8662

(vi) tan2 + tan = 0 (vii ) 2 sin 2+ sin =0 (viii) 2 cos2+ 3 cos +1=0 (ix) sin 2 = 1/2

Solution:

(i) sin 2= (ii) tan 2= 1

2 = 30 , 150, 390, 510 2 = 45 , 225, 405, 585

= 15 , 75, 195, 255 = 22.5, 112.5, 202.5, 292.5

2 means that you need to go two rotations.first rotation gives angles below 360 andsecond rotation gi ves angles from 360to 720

**3 means that you need to go three rotations.

(iii) cos 3=2 (iv) tan 2 = 12

2 = 135 , 315, 495, 6753 = 45 , 315, 405, 675, 765 , 1035

= 67.5, 157.5, 247.5, 337.5 = 15 , 105, 135, 225, 255 , 345

(vi) sin (2+ 30) =0.866 (vii) 2 sin2+ sin =0

2 +30 = 60 ,120, 420, 480 sin( 2 sin + 1 ) = 0

= 45, 75, 225, 255 sin = 0 or sin = 1/2

= 0, 210, 330

(viii) 2 cos2 + 3 cos + 1=0 (ix) sin 2 = 1/2

(2 cos +1) (cos+1) = 0 2 = 210, 330, 570, 690

cos = or cos = 1 = 105 ,165, 285, 345

= 120, 240 and 180

= 120, 180, 240


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D. Trigonometrical Identities you need to know

cos2 + sin2 = 1

1 + tan2 = sec2

cot2 + 1 = cosec2

Example T.5: Prove the following identities:

(i) tan + cot = 1 (ii) cosec + tan sec =cosec sec2 sin cos

(iii) cos4 sin4 = cos2 sin2 (iv) (sec2 1)(cosec2 1) = 1

(v) 2 tan = 2 sin cos (vi) 1 + 1 = 11 + tan 2 tan2 + 1 cot2 + 1

(vii) sec cosec = tan + cot tan cot sec +cosec

Solution:

(i) tan + cot = sin + cos = sin2 + cos2 = 1 (proved)cos sin sin cos sin cos

(ii) cosec + tan sec = 1 + sin x 1 = cos2 + sin2 = 1 = cosec sec2 sin cos cos sin cos 2 sin cos 2

(iii) cos4 sin4 = (cos2 + sin2)(cos2 sin2) = cos2 sin2

(iv) (sec2 1)(cosec2 1) = ( 1 1 )( 1 1) = (1 cos2)(1 sin2) = sin2 .cos2 = 1.cos2 sin2 cos2 sin2 cos2 sin2

(v) 2 tan = 2 sin /cos = 2 sin x cos 2 = 2 sin cos1 + tan2 sec2 cos

vi) 1 + 1 = 1 + 1 = cos2 + sin2 = 11 + tan2 cot2 + 1 sec2 cosec2

(vii) sec cosec = sec cosec x tan + cot = (sec cosec )(tan + cot )tan cot tan cot tan + cot tan2 cot2

= (sec cosec )(tan + cot ) = (sec cosec )(tan + cot )(sec2 1) (cosec2 1) sec2 cosec2

= (sec cosec )(tan + cot ) = tan + cot (proved)(sec cosec )(sec + cosec ) sec + cosec


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E. Trigonometrical Identities you need to know-( usually provided in exams)

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

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

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

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

tan(A + B) = tan A + tan B and tan(A B) = tan A tan B1 tan A tan B 1 + tan A tan B

Example T.6: If sin A =3/5 and sin B =5/13, where A and B are acute angles, find

(i) sin (A +B) (ii) cos (A +B) (iii) cos ( A B) (iv) tan (A B) (v) cot ( A B)

Solution:

(i) sin (A + B) = sin A cos B + cos A sin B 53 13 5

= 3/5 x 12/13 + 4/5 x 5/13A 4 B 12

= 36/65 + 20/65 = 56/65Draw the right-angled triangles and fill in all the sides .

sin A = 3/5, cos A = 4/5 and tan A = 3/4sin B = 5/13, cos B = 12/13 and tan B = 5/12

(ii) cos (A + B) = cos A cos B sin A sin B = 4/5 x 12/13 3/5 x 5/13 = 48/65 15/65 = 33/65

(iii) cos (A B) = cos A cos B + sin A sin B = 4/5 x 12/13 + 3/5 x 5/13 = 48/65 + 15/65 = 63/65

(iv) tan (A B) = tan A tan B = 3/4 5/12 = 1/3 = 1/3 x 48/63 = 16/331 + tan A tan B 1 + 3/4 x 5/12 1 + 15/48

(v) cot (A B) = 1 + tan A tan B = 1 + 3/4 x 5/12 = 1 + 15/48 = 63/48 x 3 = 63/16tan A tan B 3/4 5/12 1/3


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Example T.7: If sin A =4/5 and sin B =12/13, where A and B are obtuse angles, find

(i) sin (A +B) (ii) cos (A +B) (iii ) cos ( A B) (iv) tan ( A B) (v) cot ( A B)

Solution:

(i) sin (A + B) = sin A cos B + cos A sin B 54 13 12

= 4/5 x ( 5/13) + ( 3/5) x 12/13A 3 B 5

= 20/65 36/65 = 56/65Draw the right-angled triangles and fill in all the sides .

sin A = 4/5, cos A = 3/5 and tan A = 4/3sin B = 12/13, cos B = 5/13 and tan B = 12/5(for obtuse angle, cos and tan are negative).

(ii) cos (A + B) = cos A cos B sin A sin B = (3/5) x (5/13) (4/5 x 12/13) = 15/65 48/65 = 33/65

(iii) cos (A B) = cos A cos B + sin A sin B = ( 3/5) x (5/13) + (4/5 x12/13) = 15/65 + 48/65 = 63/65

(iv) tan (A B) = tan A tan B = 4/3 12/5 = 56/15 = 56/15 x 15/63 = 8/91 + tan A tan B 1 + 4/3 x 12/5 1 + 48/15

(v) cot (A B) = 1 + tan A tan B = 1 + 4/3 x 12/5 = 1 + 48/15 = 63/15 x 15/56 = 9/8tan A tan B 4/3 12/5 56/15

Example T.8: If tan (A +B) =3/4 and tan A =5/12, where A and B are acute angles, find the value of tan B

Solution:

tan (A + B) = tan A + tan B = 3/41 tan A tan B

tan A + tan B = 3/4 (1 tan A tan B) = 3/4 3/4(tan A tan B)

5/12 + tan B = 3/4 3/4 tanA.tan B

3/4 5/12 = tan B ( 1 + 3/4 tanA) = tan B ( 1 + 3/4 x 5/12)

13/6 tan B = 1/3

tan B = 1/3 x 6/13 = 2/13

Example T.9: If tan A =3/4 and tan B = 5/12, where A and B are acute angles, find the values of A + B

Solution:

tan (A + B) = tan A + tan B = 3/4 + 5/12 = 7/6 = 7/6 x 16/11 = 56/331 tan A tan B 1 3/4 x 5/12 1 15/48

A + B = tan-1( 56/33) = 59.5, 239.5


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Example T.10: Find the values of cos 45 cos 15 +sin 45 sin 15

Solution:

cos 45 cos 15+sin 45 sin 15 = cos (45 15) = cos 30 = 0.866

Example T.11: Find the values of x for 0 x 360(i) 3 cos ( x + 45 ) = sin ( x +45); (ii) tan ( x A) =1, where tan A=3

Solution:

(i) 3cos (x + 45) = sin (x + 45) (ii) tan ( x A)=1, where tan A = 3

tan (x + 45) = sin (x + 45) = 3 A = tan-1(3) = 60, 240cos (x + 45)

tan (x A) = tan (x - 60) = 1 and tan (x - 240) = 1x + 45 = 60, 240

x 60 = 45, 225 and x 240 = 45, 225x = 15, 195

x = 105, 285

F. The double angle formulae you need to know-( usually provided in exams)

cos 2 = cos 2 sin2 = 2 cos2 1 = 1 2 sin2

sin 2 = 2 sin cos

tan 2 = 2 tan

1 tan2

Example T.12: Solve the following equationsfor values of x from 0 to 360 inclusive.(i) 3 cos 2x + sin x = 1 (ii) tan x. tan 2x =1 (iii) sin 2x cos x +sin2 x =1 (iv) sin 2x =sin x(v) 3 tan x =tan 2x (vi) cos x +sin x =sec x

Solution:

(i) 3 cos 2x sin x = 2 (ii) tan x. tan 2x = 1

3( 1 2 sin2x) sin x = 2 tan x. 2 tan x = 11 tan2x

3 6 sin2

x sin x 2 = 01 tan2x 2 tan

2x = 06 sin2x+ sin x 1 = 0

3 tan2x= 1(3 sin x 1)( 2 sin x+ 1) = 0

tan x= 1/3sin x = 1/3 or sin x = 1/2

x= 30, 150, 210, 330x = 19.5, 160.5, 210, 330


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(iii) sin 2xcos x + sin2x = 1 (iv) sin 2x=sin x

2 sin x cosx.cos x = 1 sin2x 2 sin x cosx = sin x

2 sin xcos2

x = cos2x 2 sin x cosx sin x = 0

2 sin xcos2 x cos2x = 0 sin x(2 cos x 1) = 0

cos2x(2 sin x 1) = 0 sin x = 0 or cos x =

cos2x = 0 or sin x = for sin x = 0, x = 0, 360

for cos x = 0, x = 90, 270 for cos x = , x = 60, 300

for sin x= , x = 30, 150 therefore x = 0, 60, 300, 360

therefore x = 0, 30, 90, 150, 270

(v)3 tan x = tan 2x = 2tan x (vi) cos x + sin x = sec x1 tan2 x

cos x + sin x = 1 .3 tan x(1 tan

2x) = 2 tan x cos x

3 tan x 3 tan3

x = 2 tan x cos2x + sin x. cos x = 1

3 tan3 x tan x = 0 sin x. cos x = 1 cos2x

tan x(3 tan2x 1 ) = 0 sin x. cos x = sin2x

tan x = 0 or tan x = 1/3 sin x. cos x sin2x = 0

x = 0, 180, 360 when tan x= 0 sin x( cos x sin x) = 0

x = 30, 150, 210, 330 when tan x= 1/3 sin x = 0 or cos x = sin x

therefore x = 0 , 30, 150, 180, 210, 330, 360. For cos x = sin x, x = 45, 225

therefore x = 0 , 45, 225,180, 360.

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SPM Trigonometry F5