higher outcome 3 higher unit 2 trigonometry identities of the form sin(a+b) double angle formulae...

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Higher Outcome 3

Higher Unit 2Higher Unit 2

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Trigonometry identities of the form sin(A+B)Double Angle formulae

Trigonometric Equations

Exam Type Questions

Radians & Trig Basics

More Trigonometric Equations

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Higher Outcome 3

Trig Identities

The following relationships are always true for two angles A and B.

1a. sin(A + B) = sinAcosB + cosAsinB1b. sin(A - B) = sinAcosB - cosAsinB

2a. cos(A + B) = cosAcosB – sinAsinB2b. cos(A - B) = cosAcosB + sinAsinBQuite tricky to prove but some of following examples should show that they do work!!

Supplied on a

formula sheet !!

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Higher Outcome 3Examples 1

(1) Expand cos(U – V).

(use formula 2b )

cos(U – V) = cosUcosV + sinUsinV

(2) Simplify sinf°cosg° - cosf°sing°

(use formula 1b )

sinf°cosg° - cosf°sing° = sin(f – g)°

(3) Simplify cos8 θ sinθ + sin8 θ cos θ

(use formula 1a )

cos8 θ sin θ + sin8 θ cos θ =

sin(8 θ + θ)= sin9 θ

Trig Identities

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Higher Outcome 3

= cos30°

Example 2 By taking A = 60° and B = 30°,

prove the identity for cos(A – B).

NB: cos(A – B) = cosAcosB + sinAsinB

LHS = cos(60 – 30 )°

= 3/2

RHS = cos60°cos30° + sin60°sin30°= ( ½ X 3/2 ) + (3/2 X ½)= 3/4 + 3/4

= 3/2Hence LHS =

RHS !!

Trig Identities

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Higher Outcome 3Example 3

Prove that sin15° = ¼(6 - 2)

sin15° = sin(45 – 30)°

= (1/2 X 3/2 ) - (1/2 X ½)= (3/22 - 1/22)

= sin45°cos30° - cos45°sin30°

= (3 - 1) 22

X 2

2= (6 - 2) 4

= ¼(6 - 2)

Trig Identities

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41

404

3

Show that cos( - ) = 187/205

Triangle1

If missing side = x

x

Then x2 = 412 – 402 = 81So x = 9

sin = 9/41 and cos = 40/41

Triangle2

If missing side = y

y

Then y2 = 42 + 32 = 25

So y = 5

sin = 3/5 and cos = 4/5

Trig Identities NAB type Question

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Higher Outcome 3Trig Identities

cos( - ) = coscos + sinsin

= (40/41 X 4/5) + (9/41 X 3/5 )

= 160/205 + 27/205

= 187/205

Remember this is a NAB type Question

sin = 9/41 and cos = 40/41

sin = 3/5 and cos = 4/5

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Higher Outcome 3Example 5Solve sinxcos30 + cosxsin30 = -0.966

where 0o < x < 360o

By rule 1a sinxcos30 + cosxsin30 =

sin(x + 30)sin(x + 30) = -0.966 AS

T C

xo

180+xo

360-xo

180-xo

Quad 3 and Quad 4

sin-1 0.966 = 75

Quad 3: angle = 180o + 75o

x + 30o = 255o

x = 225o

Quad 4: angle = 360o – 75o

x + 30o = 285o

x = 255o

Trig IdentitiesNAB type Question

ALWAYS work out Quad 1

first

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Higher Outcome 3

Trig IdentitiesExample 6

Solve sin5 θ cos3 θ - cos5 θ sin3 θ = 3/2 where 0 < θ <

By rule 1b. sin5θ cos3θ - cos5θ sin3θ =sin(5θ - 3θ)

sin2θ = 3/2 AS

T C

θ

+ θ 2 - θ

- θQuad 1 and Quad

2sin-1 3/2 = /3

Quad 1: angle = /3 2 θ = /3

θ = /6

Quad 2: angle = - /3 2 θ = 2/3

θ = /3

= sin2θ

Repeats every

In this example repeats lie out

with limits

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Find the value of x that minimises the expression

cosxcos32 + sinxsin32

Using rule 2(b) we get cosxcos32 + sinxsin32 = cos(x – 32)

cos graph is roller-coaster

min value is -1 when angle = 180ie x – 32o = 180o

ie x = 212o

Trig Identities

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Simplify sin(θ - /3) + cos(θ + /6) + cos(/2 - θ)

sin(θ - /3) + cos(θ + /6) + cos(/2 - θ)

= sin θ cos/3 – cos θ sin/3

+ cos θ cos/6 – sin θ sin/6

+ cos/2 cos θ + sin/2 sin θ

= 1/2 sin θ – 3/2cos θ + 3/2 cos θ – 1/2sin θ + 0 x cos θ + 1 X sin θ

= sin θ

Trig IdentitiesPaper 1 type

questions

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Prove that

(sinA + cosB)2 + (cosA - sinB)2 = 2(1 + sin(A - B))

LHS = (sinA + cosB)2 + (cosA - sinB)2

= sin2A + 2sinAcosB + cos2B + cos2A – 2cosAsinB + sin2B= (sin2A + cos2A) + (sin2B + cos2B) + 2sinAcosB - 2cosAsinB = 1 + 1 + 2(sinAcosB - cosAsinB)= 2 + 2sin(A – B)

= 2(1 + sin(A – B))

= RHS

Trig IdentitiesPaper 1 type

questions

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Higher Outcome 3

sin 2 2sin cosA A A2 2

2

2

cos2 cos sin

2cos 1

1 2sin

A A A

A

A

cos2Two further formulae derived from the formulae.A

2 12

2 12

cos (1 cos2 )

sin (1 cos2 )

A A

A A

Double Angle Formulae

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Higher Outcome 3Mixed Examples:

4tan sin 2 cos2

3Given that is an acute angle and , calculate and .A A A A

sin 4

cos 3

A

A 2 2sin cos 1A A

Substitute form the tan (sin/cos)

equation2 234sin ( sin ) 1A A

16sin

25

4

5A +ve because A is

acute

Similarly:cos

3

5A 3-4-5

triangle !

2sin24

sin 25

cos2

AA A

2 2 9 16cos sin

7cos

252

25A A A

A is greater than 45 degrees – hence 2A is

greater than 90 degrees.

Double Angle formulae

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Higher Outcome 3

sin 75 .Find the exact value of o

sin(75 ) sin(45 30)o

30o1

1

245o

1

2 3sin(75 ) sin 45cos30 cos45sin30o

1 3 1 1 1 3

2 22 2 2 2

sin( )tan tan

cos cosProve that

sin( ) sin cos cos sin

cos cos cos cos

sin sin

cos cos

tan tan


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Higher Outcome 3

1L

M

N3

3

5cos .

5For the diagram opposite show that LMN

cos cos( )LMN

18 3 2Length of LM

3 210Length of MN 10

cos( ) cos cos sin sin

1 3 1 1

2 10 2 102 2 1

20 4 5

5

5

5


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Higher Outcome 3

4 4cos sin cos2 . Prove that,

2 2 ( )( )Using x y x y x y4 4 2 2 2 2cos sin (cos ) (sin )

2 2 2 2(cos sin )(cos sin ) 2 2cos sin 1

2 2cos sin 2 2cos2 cos sin

cos2


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Higher Outcome 3


Double angle formulae (like cos2A or sin2A) often occur in trig equations. We can solve these equations by substituting the expressions derived in the previous

sections.

cos2A = 2cos2A – 1 if cosA is also in the equation

Rules for solving equations

sin2A = 2sinAcosA when replacing sin2Aequation

cos2A = 1 – 2sin2A if sinA is also in the equation

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Higher Outcome 3

cos2 4sin 5 0 0 360 . Solve: for o o ox x x

cos2x and sin x,

so substitute 1-2sin2x2(1 2sin ) 4sin 5 0x x

26 4sin 2sin 0x x 26 4 2 0compare with z z

(6 2sin )(1 sin ) 0x x

sin 1 sin 3 or x x 0 sin 1 f or all real angles x

90 ox


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Higher Outcome 3

5cos2 cos 2 0 360 Solve: for o o ox x xcos 2x and cos x,

so substitute 2cos2 -125(2cos 1) cos 2 x x

210cos cos 3 0 x x

(5cos 3)(2cos 1) 0 x x

3 1cos cos

5 2 or x x

36

53.1

300 53.1 6.9

o

o

x

x

and 120

2

180 60

180 06 40

o

o

x

x

and


C

AS

T0o180

o

270o

90o

3

2

2

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Higher Outcome 3


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Higher Outcome 3

( ) sin ( ) sin

0 360 .

The diagram shows the graphs of and

for

o o

o

f x a bx g x c x

x

y

x

( )y f x

( )y g x

360o

-2

0

2

4

-4

x

y

Three problems concerning this graph follow.


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Higher Outcome 3

, . i) State the values of and a b cy

x

( )y f x

( )y g x

360o

x

y

( ) sin of x a bx

The max & min values of asinbx are 3 and -3 resp.

The max & min values of sinbx are 1 and -1 resp.

3a

f(x) goes through 2 complete cycles from 0 – 360o

2b

( ) sin og x c xThe max & min values of csinx are 2 and -2 resp.

2c


( ) sin og x c x

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Higher Outcome 3

( ) ( ) ii) Solve the equation algebraically.f x g x

From the previous problem we now have:( ) 3sin 2 ( ) 2sin and f x x g x x

Hence, the equation to solve is:

3sin 2 2sinx xExpand sin 2x3(2sin cos ) 2sinx x x

6sin cos 2sin 0x x x Divide both sides by 2

3sin cos sin 0x x x Spot the common factor in the terms?

sin (3cos 1) 0x x Is satisfied by all values of x for which:

1sin 0 cos

3 or x x


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Higher Outcome 3

0 360

iii) find the coordinates of the points of intersection

of the graphs for .ox

From the previous problem we have: 1

sin 0 cos3

and x x

sin 0x

0

180

360

o

o

o

x

x

x

Hence

1cos

3x

(3

70.5

289.560 70.5)

o

o

o

x

x


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Higher Outcome 3

Radian Measurements

Reminders

i) Radians 180radians o

Converting between degrees and radians:

120 121 0

0.8

o

2

3 radians

5 5.

6 6

180

150o

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Higher Outcome 3

ii) Exact Values

45o right-angled triangle:

1

1

2

45o

1cos45 sin 45

2o o

tan 45 1o

Equilateral triangle:

3sin 60

2o

1

2 2

1

60o

30o

3

1cos60

2o

1sin30

2o

3cos30

2o

tan 60 3o 1

tan303

o

Degree Measurements

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Higher Outcome 3

degrees

0o 30o 45o 60o 90o

radians

sin

cos

tan

6

0

3

4

3

2

0

1

0

1

23

2

1

2 13

2

1

2

1

2 01

3 1

Example: What is the exact value of sin 240o

?240 180 60 sin(180 ) sin 3

sin 2402

o

Radians / Degrees

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Higher Outcome 3

Sine Graph

Period = 360o

Amplitude = 1

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Higher Outcome 3

Cosine Graph

Period = 360o

Amplitude = 1

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Higher Outcome 3

Tan Graph

Period = 180o

Amplitude cannot be found for tan function

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Higher Outcome 3

Solving Trigonometric Equations

Example:

2cos3 1 0 (0 360 )Solve ox x

Step 1: Re-Arrange

Step 2: consider what solutions

are expected

C

AS

T0o180

o

270o

90o

3

2

2

2cos3 1 0 x

2cos3 1x

1cos3

2x

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Higher Outcome 3

1cos3

2x

cos 3x is positive so solutions in the first and fourth quadrants

0 360 Since has 2 solutionsox

0 3 1080 Then has 6 solutionsox

x 3 x 3


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Higher Outcome 3

300o

Step 3: Solve the equation

1cos3

2x 1 1

3 cos 602

ox

1st quad 4th quad cos wave repeats every 360o

x = 20o


100o 140o 220o 260o 340o

420o 660o 780o 1020o3x = 60o

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Higher Outcome 3

Graphical solution for


1cos3

2x

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Higher Outcome 3


Example:

Step 1: Re-ArrangeStep 2: consider what

solutions

are expected

C

AS

T0o180

o

270o

90o

3

2

2

1 2 sin 6 0 (0 180 ) Solve ot t

1sin 6

2t

sin 6t is negative so solutions in the third and fourth

quadrants 0 180 Since has 2 solutions ot

0 6 1080 Then has 12 solutions ot

x 6 x 6

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Higher Outcome 3

315o


1sin 6

2

t 1 1

6 sin2

t

3rd quad 4th quad sin wave repeats every 360o

x = 39.1o


52.5o 97.5o 112.5o

157.5o

172.5o

585o 675o 945o 1035o6t = 225o

1 1sin 45

2stalways 1 Quad fi rst

o

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Higher Outcome 3



1sin 6

2

t

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Higher Outcome 3


Example:


solutions are expected

C

AS

T0o180

o

270o

90o

3

2

2

2sin(2 ) 1 (0 2 )3

Solve x x

1sin(2 60 )

2ox

(2x – 60o ) = sin-1(1/2)

0 360Since has 2 solutions ox

0 2 720 Then has 4 solutions ox

x 2 x 2

The solution is to be in radians – but work in degrees and convert at

the end.

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Higher Outcome 3

210o


1st quad 2nd quad sin wave repeats every 360o

x = 45o


105o 225o 285o

450o 570o 2x = 90o

1sin(2 60 )

2 ox 1 1

2 60 sin2

ox 1 1sin 30

2st(1 quadrant)

o

4

7

12

5

4

19

12

2 60 30 150and o o ox

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Higher Outcome 3



1sin(2 60 )

2 ox

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Higher Outcome 3


Harder Example:


solutions are expected

C

AS

T0o180

o

270o

90o

3

2

2

2 solutions

1st and 3rd quads

2tan 3 (0 2 )Solve x x

tan 3x

The solution is to be in radians – but work in degrees and convert at the

end.

tan 3x

2 solutions

2nd and 4th quads

tan 3x

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Higher Outcome 3

120o


1st quad 2nd quad tan wave repeats every 180o


240o 300o x = 60o

3

2

3

4

3

5

3

1tan 3 (603

st in the 1 quadrant)o tan 3x tan 3x

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Higher Outcome 3



32 tan x

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Higher Outcome 3


Harder Example:

Step 1: Re-ArrangeStep 2: Consider what

solutions

are expected

C

AS

T0o180

o

270o

90o

3

2

2

23sin 4sin 1 0 (0 360 ) Solve ox x x

(3sin 1)(sin 1) 0x x

sin 1x1

sin3

x

One solutionTwo solutions

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Higher Outcome 3

160.5o


1stquad 2nd quad


90o

x = 19.5o

sin 1x1sin

3x


Overall solution x = 19.5o , 90o and 160.5o

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Higher Outcome 3



23sin 4sin 1 0 x x

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Higher Outcome 3


Harder Example:

Step 1: Re-ArrangeStep 2: Consider what

solutions

are expected

C

AS

T0o180

o

270o

90o

3

2

2


25sin 2 2cos (0 2 ) Solve x x x

25(1 cos ) 2 2cos x x

23 2cos 5cos 0 x x

(3 5cos )(1 cos ) 0 x x

3cos

5x cos 1x

2 2

2 2

2 2

sin cos 1

cos 1 sin

sin 1 cos

Remember this !

The solution is to be in radians – but work in degrees and convert at the

end.

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Higher Outcome 3

306.9o


1stquad 3rd quad


180o

x = 53.1o

cos 1x3cos

5x


Overall solution in radians x = 0.93 , π and 5.35

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Higher Outcome 3



2sin 2 2cos 5 x x

Higher Maths

Strategies

www.maths4scotland.co.uk

Click to start

Compound Angles

Maths4Scotland Higher

Compound Angles

The following questions are on

Non-calculator questions will be indicated

Click to continue

You will need a pencil, paper, ruler and rubber.


QuitQuit

This presentation is split into two parts

Using Compound angle formula for

Exact values

Solving equations

Choose by clicking on the appropriate button


Hint

Previous NextQuitQuit

A is the point (8, 4). The line OA is inclined at an angle p radians to the x-axis a) Find the exact values of: i) sin (2p) ii) cos (2p)

The line OB is inclined at an angle 2p radians to the x-axis. b) Write down the exact value of the gradient of OB.

Draw triangle Pythagoras80

Write down values for cos p and sin p8 4

cos sin80 80

p p

Expand sin (2p) sin 2 2sin cosp p p 4 8 64 42

80 580 80

Expand cos (2p) 2 2cos 2 cos sinp p p 2 28 4

80 80

64 16 3

80 5

Use m = tan (2p)sin 2

tan 2cos 2

pp

p 4 3 4

5 5 3

8

4p


Hint


In triangle ABC show that the exact value of

Use Pythagoras

Write down values for

sin a, cos a, sin b, cos b

1 1 1 3sin cos sin cos

2 2 10 10a a b b

Expand sin (a + b) sin( ) sin cos cos sina b a b a b

is2

sin( )5

a b

2 10AC CB

2 10

Substitute values1 3 1 1

2 10 2 10sin( )a b

Simplify3 1

20 20sin( )a b 4

20

4 4 2

4 5 2 5 5


Hint


Using triangle PQR, as shown, find the

exact value of cos 2x

Use Pythagoras


cos x and sin x

2 7cos sin

11 11x x

Expand cos 2x2 2cos 2 cos sinx x x

11PR

11

Substitute values 222 7

11 11cos 2x

Simplify4 7

cos 211 11

x 3

11


Hint


On the co-ordinate diagram shown, A is the point (6, 8) and

B is the point (12, -5). Angle AOC = p and angle COB = q

Find the exact value of sin (p + q).

Use Pythagoras


sin p, cos p, sin q, cos q

8 6 5 12

10 10 13 13sin , cos , sin , cosp p q q

Expand sin (p + q) sin ( ) sin cos cos sinp q p q p q

10 13OA OB

Substitute values

Simplify 126 63

130 65

6

8

512

10

13

Mark up triangles

8 12 6 5

10 13 10 13sin ( )p q

96 30

130 130sin ( )p q


Hint


Draw triangles Use Pythagoras

Expand sin 2A sin 2 2sin cosA A A

A and B are acute angles such that and .

Find the exact value of

a) b) c)

3

4tan A 5

12tan B

sin 2A cos 2A sin(2 )A B4

3A

12

5B

Hypotenuses are 5 and 13 respectively

5 13

Write down sin A, cos A, sin B, cos B 3 4 5 12

, , ,5 5 13 13

sin cos sin cosA A B B

3 4 24

5 5 25sin 2 2A

Expand cos 2A 2 2cos 2 cos sinA A A 2 2 16 9 74 3

25 25 255 5cos 2A

Expand sin (2A + B) sin 2 sin 2 cos cos 2 sinA B A B A B

Substitute 24 12 7 5 323sin 2

25 13 25 13 325A B


Hint


Draw triangle Use Pythagoras

Expand sin (x + 30) sin( 30) sin cos30 cos sin 30x x x

If x° is an acute angle such that

show that the exact value of

4

3tan x

4 3 3sin( 30) is

10x

3

4

x

Hypotenuse is 5

5

Write down sin x and cos x4 3

,5 5

sin cosx x

Substitute

Simplify

Table of exact values

4 3 3 1sin( 30)

5 2 5 2x

4 3 3sin( 30)

10 10x 4 3 3

10


Hint


Use Pythagoras

Expand cos (x + y) cos( ) cos cos sin sinx y x y x y

Write down

sin x, cos x, sin y, cos y.

3 4 24 5, , ,

5 5 7 7sin cos sin cosx x y y

Substitute

Simplify20 3 4 6

35

The diagram shows two right angled triangles

ABD and BCD with AB = 7 cm, BC = 4 cm and CD = 3 cm.

Angle DBC = x° and angle ABD is y°.

Show that the exact value of

20 6 6cos( )

35x y is

5, 24BD AD

24

5

4 5 3 24cos( )

5 7 5 7x y

20 3 24cos( )

35 35x y

20 6 6

35


Hint



2 5

3 32 2sin , cosx x

The framework of a child’s swing has dimensions

as shown in the diagram. Find the exact value of sin x°

Write down sin ½ x and cos ½ x

5h

Substitute

Simplify


3 3

4

xDraw in perpendicular

2

2

x

h5Use fact that sin x = sin ( ½ x + ½ x)

Expand sin ( ½ x + ½ x) 2 2 2 2 2 22 2sin sin cos sin cos 2sin cosx x x x x xx x

2 5

3 32 2sin 2x x

4 5sin

9x


Hint


Given that

find the exact value of


cos a and sin a

3 11cos sin

20 20a a

Expand sin 2a sin 2 2 sin cosa a a

20

Substitute values11 3

sin 2 220 20

a

Simplify

11tan , 0

3 2

3a

11sin 2

Draw triangle Use Pythagoras hypotenuse 20

6 11sin 2

20a

3 11

10


Hint


Find algebraically the exact value of

1 3cos 120 cos 60 cos 150 cos30

2 2

3 1sin 120 sin 60 sin 150 sin 30

2 2

Expand sin (+120) sin 120 sin cos120 cos sin120

Use table of exact values

1 3 3 1

2 2 2 2sin sin . cos . cos . sin . Combine and substitute

sin sin 120 cos( 150)


Expand cos (+150) cos 150 cos cos150 sin sin150

Simplify 1 3 3 1

2 2 2 2sin sin cos cos sin

0


Hint


If find the exact value of

a) b)


cos and sin

4 3cos sin

5 5

Expand sin 2 sin 2 2 sin cos


4cos , 0

5 2

5

4

3

Opposite side = 3

3 4 242

5 5 25

Expand sin 4 (4 = 2 + 2) sin 4 2 sin 2 cos 2

Expand cos 2 2 2cos 2 cos sin 16 9 7

25 25 25

Find sin 424 7

sin 4 225 25

336

625

sin 2 sin 4


Hint


Draw triangles Use Pythagoras

Expand sin (P + Q) sin sin cos cos sinP Q P Q P Q

For acute angles P and Q

Show that the exact value of

1213

P

53

Q

Adjacent sides are 5 and 4 respectively

5 4

Write down sin P, cos P, sin Q, cos Q 12 5 3 4

, , ,13 13 5 5

sin cos sin cosP P Q Q

Substitute

12 3and

13 5sin sinP Q

63

65sin ( )P Q

12 4 5 3sin

13 5 13 5P Q

Simplify 48 15sin

65 65P Q 63

65


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Solving Equations

Using Compound angle formula for

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Hint


Solve the equation for 0 ≤ x ≤ correct to 2 decimal places 3cos(2 ) 10cos( ) 1 0x x

Replace cos 2x with 2cos 2 2cos 1x x

Substitute 23 2cos 1 10cos 1 0x x

Simplify 26cos 10cos 4 0x x 23cos 5cos 2 0x x

Factorise 3cos 1 cos 2 0x x

Hence 1

3cos

cos 2

x

x

Discard

Find acute x 1.23acute radx

Determine quadrants

AS

CT

1.23 2 1.23or radsx

1.23

5.05

rads

rads

x

x


Hint

Previous NextQuitQuitTable of exact values

Solve simultaneously 2cos 2 3x

Rearrange 3

2cos 2x

0 0 2 2x x

Find acute 2x 62acute x

Determine quadrants

AS

CT

6 6

6 6 6 62 or radsx

5 7

12 12orx

The diagram shows the graph of a cosine function from 0 to .

a) State the equation of the graph.

b) The line with equation y = -3 intersects this graph

at points A and B. Find the co-ordinates of B.

Equation 2cos 2y x

Check range

7

12, 3isB B Deduce 2x

Functions f and g are defined on suitable domains by f(x) = sin (x) and g(x) = 2x

a) Find expressions for:

i) f(g(x)) ii) g(f(x))

b) Solve 2 f(g(x)) = g(f(x)) for 0 x 360°


Hint


2nd expression

Form equation 2sin 2 2sinx x

Rearrange

Determine

quadrants

AS

CT60 , 300x

1st expression ( ( )) (2 ) sin 2f g x f x x

Common factor

( ( )) (sin ) 2sing f x g x x

Replace sin 2x 2sin cos sinx x x

sin 2 sinx x

2sin cos sin 0x x x

sin 2cos 1 0x x

Hence1

or2

sin 0 2cos 1 0 cosx x x

Determine x

sin 0 0 , 360x x

1

2cos 60acutex x

0 , 60 , 300 , 360x

Functions are defined on a suitable set of real numbers

a) Find expressions for i) f(h(x)) ii) g(h(x))

b) i) Show that ii) Find a similar expression for g(h(x))

iii) Hence solve the equation


Hint


2nd expression

Simplify 1st expr.

Similarly for 2nd expr.

Determine

quadrants

AS

CT3,

4 4x

1st expression 4 4( ( )) sinf h x f x x

Use exact values

and4

( ) sin , ( ) cos ( )f x x g x x h x x

1 1( ( )) sin cos

2 2f h x x x

for( ( )) ( ( )) 1 0 2f h x g h x x

4 4( ( )) cosg h x g x x

4 4( ( )) sin cos cos sinf h x x x

1 1

2 2( ( )) sin cosf h x x x

4 4( ( )) cos cos sin sing h x x x

1 1

2 2( ( )) cos sing h x x x

Form Eqn. ( ( )) ( ( )) 1f h x g h x

2

2sin 1x Simplifies to

2 2 1

2 2 2 2sin x Rearrange:

acute x 4acute x

a) Solve the equation sin 2x - cos x = 0 in the interval 0 x 180°

b) The diagram shows parts of two trigonometric graphs,

y = sin 2x and y = cos x. Use your solutions in (a) to

write down the co-ordinates of the point P.


Hint


Determine quadrants

for sin x

AS

CT

30 , 150x

Common factor

Replace sin 2x 2sin cos cos 0x x x cos 2sin 1 0x x

Hence1

or2

cos 0 2sin 1 0 sinx x x

Determine x cos 0 90 , ( 270 )out of rangex x 1

2sin 30acutex x

30 , 90 , 150x

Solutions for where graphs cross

150x By inspection (P)

cos150y Find y value3

2y

Coords, P

3

2150 ,P


Hint


Solve the equation for 0 ≤ x ≤ 360°3cos(2 ) cos( ) 1x x


Substitute 23 2cos 1 cos 1x x

Simplify 26cos cos 2 0x x

Factorise 3cos 2 2cos 1 0x x

Hence2

3cos x

Find acute x 48acute x

Determine quadrants

AS

CT1

2cos x

60acute x


2

3cos x

AS

CT

1

2cos x

132

228

x

x

60

300

x

x

Solutions are: x= 60°, 132°, 228° and 300°

48acute x 60acute x


Hint


Solve the equation for 0 ≤ x ≤ 2 62sin 2 1x

Rearrange

Find acute x 62

6acute x

Determine quadrantsAS

CT


Solutions are:

6

1sin 2

2x

62

6x 6

52

6x

Note range 0 2 0 2 4x x

and for range 2 2 4x

6

132

6x 6

172

6x

7 3, , ,

6 2 6 2x

for range 0 2 2x


Hint


a) Write the equation cos 2 + 8 cos + 9 = 0 in terms of cos

and show that for cos it has equal roots.

b) Show that there are no real roots for

Rearrange

Divide by 2

Deduction

Factorise cos 2 cos 2 0

Replace cos 2 with 2cos 2 2cos 1

22cos 8cos 8 0

2cos 4cos 4 0

Equal roots for cos

Try to solve:

cos 2 0

cos 2

Hence there are no real solutions for

No solution

Solve algebraically, the equation sin 2x + sin x = 0, 0 x 360


Hint


Determine quadrants

for cos xAS

CT

120 , 240x

Common factor

Replace sin 2x 2sin cos sin 0x x x

sin 2cos 1 0x x

Hence1

or2

sin 0

2cos 1 0 cos

x

x x

Determine x sin 0 0 , 360x x

1

2cos 60acutex x

x = 0°, 120°, 240°, 360°

Find the exact solutions of 4sin2 x = 1, 0 x 2


Hint


Determine quadrants for sin x

AS

CT

Take square roots

Rearrange 2 1

4sin x

1

2sin x

Find acute x6

acute x

+ and – from the square root requires all 4 quadrants

5 7 11, , ,

6 6 6 6x


Hint


Solve the equation for 0 ≤ x ≤ 360°cos 2 cos 0x x


Substitute 22cos 1 cos 0x x

Simplify


Hence1

2cos x


Determine quadrants

AS

CTcos 1x

180x


1

2cos x

60

300

x

x

Solutions are: x= 60°, 180° and 300°

60acute x 22cos cos 1 0x x


Hint


Solve algebraically, the equation for 0 ≤ x ≤ 360°cos 2 5cos 2 0x x


Substitute 22cos 1 5cos 2 0x x

Simplify 22cos 5cos 3 0x x


Hence1

2cos x


Determine quadrants

cos 3x


AS

CT

1

2cos x

60

300

x

x

Solutions are: x= 60° and 300°

60acute x

Discard above


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30° 45° 60°

sin

cos

tan 1

6

4

3

1

2

1

23

2

3

2

1

21

21

3 3



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