Review Exercise 2 x = - 2 2 Þ sec x = - 2 Þ cos x = - 1 2 x = 3p 4, 5 p 4, 11p 4, 13p 4 14 a sinq cos q + cos q sinq = sin2 q + cos 2 q cos qsinq = 1 s inq cos q (us ing cos 2 q

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Review Exercise 2

1 Crosses y-axis when x = 0 at

sin3p

4=

1

2

Crosses x-axis when

sin x +3p

4

æ

èçö

ø÷= 0

x +3p

4= -p, 0, p, 2p

x = -7p

4, -

3p

4,p

4,5p

4

So coordinates are

0,1

2

æ

èçö

ø÷, -

7p

4,0

æ

èçö

ø÷, -

3p

4,0

æ

èçö

ø÷,

p

4,0

æ

èçö

ø÷,

5p

4,0

æ

èçö

ø÷

2 a

y = cos x -p

3

æ

èçö

ø÷ is y = cos x translated by the vector

π

3

0

b Crosses y-axis when

y = cos -p

3

æ

èçö

ø÷=

1

2

Crosses x-axis when

cos x -p

3

æ

èçö

ø÷= 0

x -p

3=

p

2,

3p

2

x =5p

6,11p

6

So coordinates are

0,1

2

æ

èçö

ø÷,

5p

6,0

æ

èçö

ø÷,

11p

6,0

æ

èçö

ø÷

2 c

cos x -p

3

æ

èçö

ø÷= -0.27, 0 £ x £ 2p

cos-1(-0.27) = 1.844 (3 d.p.)

Þ x -p

3» 1.844 and x -

p

3» 2p -1.844

Þ x = 2.89, 5.49 (2 d.p.)


3 a Let C be the midpoint of the line AB, then

AOC is a right-angled triangle and AC = 3cm , so

sinq

2=

3

5= 0.6 Þ

q

2= 0.6435…

q = 1.287 radians (3 d.p.)

b Use l = rq

So arc AB = 5´1.287 = 6.44cm (3 s.f.)

4 As ABC is equilateral, BC = AC = 8 cm

BP = AB – AP = 8 – 6 = 2 cm

QC = BP = 2 cm

ÐBAC =

p

3, PQ = 6 ´

p

3= 2p = 6.28cm (2 d.p.)

So perimeter = BC + BP + PQ + QC = 18.28 cm (2 d.p.)

Exact answer 12 + 2 p cm

5 a 12(r + 10)2q - 1

2r2q = 40

Þ 20rq +100q = 80

Þ rq + 5q = 4

Þ r =4

q- 5

b r =

4

q- 5 = 6q

Þ 4 - 5q = 6q 2

Þ 6q 2 + 5q - 4 = 0

Þ (3q + 4)(2q -1) = 0

Þq = -4

3 or

1

2

But cannot be negative, so q =

1

2, r = 3

So perimeter = 20 + rq + (10 + r)q = 20 +

3

2+

13

2= 28cm

6 a arc BD =10 ´ 0.6 = 6cm

b Area of triangle ABC =

1

2(13´10)sin0.6 = 65´ 0.567 = 36.7cm2 (1 d.p.)

Area of sector ABD =

1

2102 ´ 0.6 = 30cm2

Area of shaded area BCD = 36.7 - 30 = 6.7cm2 (1 d.p.)


7 a

So r =10

cos0.7= 13.07cm (2 d.p.)

Area of sector OAB =1

2r 2 ´1.4 = 119.7cm2 (1 d.p.)

b BC = AC = r tan0.7

So perimeter = 2r tan0.7 + r ´1.4

= (2 ´13.07 ´ 0.842) + (13.07 ´1.4) = 40.3cm


8 Split each half of the rectangle as shown.

EFB is a right-angled triangle, and by Pythagoras’ theorem EF =

3

2r

Let ÐEBF = q , so tanq = 3 Þq =p

3

So ÐFBC =p

2-

p

3=

p

6

Area S =1

2r 2 p

6=

p

12r 2

Area T =1

2´

3

2r ´

1

2r =

3

8r 2

Þ Area R =1

2r 2 - Area S - Area T =

1

2-

3

8-

p

12

æ

èç

ö

ø÷ r 2

Area of sector ACB =

1

2r2 p

2=

p

4r2

Area U = Area ABCD - Area sector ACB - 2R

= r 2 -p

4r 2 - 2

1

2-

3

8-

p

12

æ

èç

ö

ø÷ r 2

= r 2 3

4-

p

12

æ

èç

ö

ø÷

So area U = r 2 -p

4r 2 - 2R

= 1-p

4-1+

3

4+

p

6

æ

èç

ö

ø÷ r 2

= r 2 3

4-

p

12

æ

èç

ö

ø÷ =

r 2

123 3 - p( )

So shaded area = 2U =

r2

63 3 - p( )


9 a 3sin2 x + 7cos x + 3= 3(1- cos2x) + 7cos x + 3

= -3cos2 x + 7cos x + 6

= 3cos2x - 7cos x - 6

b 3cos2 x - 7cos x - 6 = 0

(3cos x + 2)(cos x - 3) = 0

cos x = -2

3 or 3

cos x cannot be 3

so cos x = -2

3

x = 2.30, 2p - 2.30 = 2,30, 3.98 (2 d.p.)

10 a For small values of q :

sin4q » 4q

cos4q »1- 12(4q )2 »1- 8q 2 ,

tan3q » 3q

sin4q - cos4q + tan3q » 4q - 1- 8q 2( ) + 3q

» 8q 2 + 7q -1

b -1

11 a y = 4- 2cosec x is y = cosec x stretched by a scale factor 2 in the y-direction,

then reflected in the x-axis and then translated by the vector 0

4

b The minima in the graph occur when cosec x = –1 and y = 6. The maxima occur when cosec x =1

and y = 2. So there are no solutions for 2 < k < 6.

12 a The graph is a translation of y = secq by a .

So a =

p

3

b As the curve passes through (0, 4)

4 = k sec

p

3Þ k = 4cos

p

3= 2


12 c

-2 2 = 2sec q -p

3

æ

èçö

ø÷

Þ cos q -p

3

æ

èçö

ø÷= -

1

2

Þq -p

3= -

5p

4, -

3p

4

Þq = -11p

12, -

5p

12

13 a

cos x

1- sin x+

1- sin x

cos xº

cos2 x + (1- sin x)2

cos x(1- sin x)

ºcos2 x +1- 2sin x + sin2 x

cos x(1- sin x)

º2 - 2sin x

cos x(1- sin x)

º2

cos x

º 2sec x

b By part a the equation becomes

2sec x = -2 2

Þ sec x = - 2

Þ cos x = -1

2

x =3p

4,5p

4,11p

4,13p

4

14 a

sinq

cosq+

cosq

sinq=

sin2q + cos2q

cosq sinq

=1

sinq cosq(using cos2q + sin2 q º 1)

=1

12sin2q

(using double-angle formula sin2q º 2sinq cosq )

= 2cosec2q


14b The graph of y = 2cosec2q is a stretch of the graph of y = cosecq by a scale factor of 12 in the

horizontal direction and then a stretch by a factor of 2 in the vertical direction.

c By part a the equation becomes

2cosec2q = 3

Þ cosec2q =3

2

Þ sin2q =2

3, in the interval 0 £ 2q £ 720°

Calculator value is 2q = 41.81° (2 d.p.)

Solutions are 2q = 41.81°,180° - 41.81°, 360° + 41.81°, 540° - 41.81°

So the solution set is: 20.9°, 69.1°, 200.9°, 249.1°

15 a Note the angle BDC =q

cosq =BC

10Þ BC = 10cosq

sinq =BC

BDÞ BD =

BC

sinq=

10cosq

sinq= 10cotq

b

10cotq =10

3

Þ cotq =1

3, q =

p

3

From the triangle BCD, cosq =DC

BD

Þ DC = BDcosq

So DC = 10cotq cosq

= 101

3

æ

èçö

ø÷1

2

æ

èçö

ø÷

=5

3


16 a sin2q + cos2q º1

Þsin2q

cos2q+

cos2q

cos2qº

1

cos2q(dividing by cos2q )

Þ tan2q +1º sec2q

b 2tan2q + secq =1

Þ 2sec2 q - 2 + secq = 1

Þ 2sec2 q + secq - 3 = 0

Þ (2secq + 3)(secq -1) = 0

Þ secq = -3

2, secq = 1

Þ cosq = -2

3, cosq = 1

Solutions are 131.8°, 360°-131.8°, 0°

So solution set is: 0.0°,131.8°, 228.2° (1 d.p.)

17 a

a =1

sin x=

112b

=2

b

b

4 - b2

a2 -1=

4 - b2

2

b

æ

èçö

ø÷

2

-1

=4 - b2

4

b2-1

=4 - b2

4 - b2

b2

= (4 - b2 ) ´b2

4 - b2

= b2

An alternative approach is to first substitute the trigonometric functions for a and b

4 - b2

a2 -1=

4 - 4sin2 x

cosec2 x -1

=4(1- sin2 x)

cot2 x

=4cos2 x

cot2 x

= 4sin2 x = b2

18 a y = arcsin x

Þ sin y = x

x = cos p2

- y( )Þ p

2- y = arccos x

b arcsin x + arccos x = y + p

2- y

= p

2

Using sinq = cos p

2-q( )


19 a arccos

1

x= p Þ cos p =

1

x

Use Pythagoras’ theorem to show that opposite side of the right-angle triangle with angle p is

x2 -1

So sin p =

x2 -1

xÞ p = arcsin

x2 -1

x

b If 20 1 then 1x x is negative and you cannot take the square root of a negative number.

20 a y = 2arccos x -

p

2 is y = arccos x stretched by a scale factor of 2 in the y-direction and then

translated by -

p

2 in the vertical direction

b 2arccos x -

p

2= 0

Þ arccos x =p

4

Þ x = cosp

4=

1

2

Coordinates are

1

2, 0

æ

èçö

ø÷

21

tan x +p

6

æ

èçö

ø÷=

1

6Þ

tan x + 3

3

1- 3

3tan x

=1

6 [using the addition formula for tan (A + B)]

6 tan x + 2 3 = 1-3

3tan x

18 + 3

3

æ

èç

ö

ø÷ tan x = 1- 2 3

tan x =3 1- 2 3( ) 18 - 3( )

18 + 3( ) 18 - 3( )

=72 -111 3

321


22 a sin(x + 30°) = 2sin(x + 60°)

So sin xcos30° + cos xsin30° = 2 sin xcos60° – cos xsin60°( ) (using the addition formulae for sin)

3

2sin x +

1

2cos x = 2

1

2sin x -

3

2cos x

æ

èç

ö

ø÷

3sin x + cos x = 2sin x - 2 3cos x (multiplying both sides by 2)

-2 + 3( )sin x = -1- 2 3( )cos x

So tan x =-1- 2 3

-2 + 3

=-1- 2 3( ) -2 - 3( )-2 + 3( ) -2 - 3( )

=2 + 6 + 4 3 + 3

4 - 3

= 8+ 5 3

b tan(x + 60°) =

tan x + tan60

1- tan x tan60

=8 + 5 3 + 3

1- 8 + 5 3( ) 3

=8 + 6 3

-14 - 8 3

=4 + 3 3( ) -7 + 4 3( )

-7 - 4 3( ) -7 + 4 3( )

=36 - 28 - 21 3 +16 3

49 - 48

= 8 - 5 3

23 a sin165° = sin(120°+ 45°)

= sin120°cos45°+ cos120°sin45°

=3

2´

1

2+

-1

2´

1

2

=3 -1

2 2

=6 - 2

4


23 b cosec165° =

1

sin165°

=4

6 - 2( )´

6 + 2( )6 + 2( )

=4 6 + 2( )

6 - 2

= 6 + 2

24 a cos A =

3

4 Using Pythagoras’ theorem and noting that sin A is negative as A is

in the fourth quadrant, this gives

sin A = -

7

4

Using the double-angle formula for sin gives

sin2A = 2sin Acos A = 2 -7

4

æ

èç

ö

ø÷

3

4

æ

èçö

ø÷= -

3 7

8

b cos2A = 2cos2 A-1=

1

8

Þ tan2A =sin2A

cos2A=

- 3 7

8( )18( )

= -3 7

25 a cos2x + sin x =1

Þ1- 2sin2 x + sin x = 1 (using double-angle formula for cos2x)

Þ 2sin2 x - sin x = 0

Þ sin x(2sin x -1) = 0

Þ sin x = 0, sin x =1

2

Solutions in the given interval are: –180°, 0°, 30°,150°,180°

b sin x(cos x + cosec x) = 2cos2 x

Þ sin xcos x +1= 2cos2 x

Þ in xcos x = 2cos2 x -1

Þ1

2sin2x = cos2x (using the double-angle formulae for sin2x and cos2x)

Þ tan2x = 2, for - 360° £ 2x £ 360°

So 2x = 63.43° - 360°, 63.43° -180°, 63.43°, 63.43° +180°

Solution set: -148.3°, - 58.3°, 31.7°,121.7° (1 d.p.)


26 a Rsin(x +a) = Rsin xcosa + Rcos xsina

So Rcosa = 3, Rsina = 2

R2 cos2 a + R2 sin2 a = 32 + 22 = 9 + 4 = 13

Þ R = 13 (as cos2 a + sin2 a º 1)

tana =

2

3Þa = 0.588 (3 d.p.)

b R4 = 13( )

4

= 169 since the maximum value the sin function can take is 1

c 13sin(x + 0.588) = 1

sin(x + 0.5880) =1

13= 0.27735…

x + 0.588 = p - 0.281, 2p + 0.281

x = 2.273, 5.976 (3 d.p.)

27 a LHS º cotq - tanq

ºcosq

sinq-

sinq

cosq

ºcos2 q - sin2 q

sinq cosq

ºcos2q12sin2q

(using the double angle formulae for sin2q and cos2q )

º 2cot 2q º RHS

b 2cot 2q = 5Þ cot2q =

5

2Þ tan2q =

2

5, for - 2p < 2q < 2p

So 2q = 0.3805 - 2p, 0.3805 - p, 0.3805, 0.3805 + p

Solution set: - 2.95, -1.38, 0.190,1.76 (3 s.f.)

28 a LHS º cos3q

º cos(2q +q )

º cos2q cosq - sin2q sinq

º (cos2q - sin2 q )cosq - 2sinq cosq( )sinq

º cos3q - 3sin2 q cosq

º cos3q - 3(1- cos2q )cosq

º 4cos3q - 3cosq º RHS

b From part a cos3q = 4

2 2

27- 2 = -

19 2

27

So

sec3q = -27

19 2= -

27 2

38


29 sin4q = sin2q( ) sin2q( )

Use the double-angle formula to write sin2q in terms of cos2q

cos2q = 1- sin2q Þ sin2q =

1- cos2q

2

Now substitute the expression for sin2q and expand the brackets

So sin4q =1- cos2q

2

æ

èçö

ø÷1- cos2q

2

æ

èçö

ø÷

=1

41- 2cos2q + cos2 2q( )

Again use the double-angle formula to write cos2 2q in terms of cos4q

So sin4q =1

41- 2cos2q +

1+ cos4q

2

æ

èçö

ø÷

=3

8-

1

2cos2q +

1

8cos4q

30 a Rsin(q +a) = Rsinq cosa + Rcosq sina = 6sinq + 2cosq

So Rcosa = 6, Rsina = 2

R2 cos2 a + R2 sin2 a = 62 + 22 = 36 + 4 = 40

Þ R = 40 (as cos2 a + sin2 a º 1)

tana =2

6Þa = 0.32175…= 0.32 (2 d.p.)

So 6sinq + 2cosq » 40 sin(q + 0.32)

b i 40 , since the maximum value the sin function can take is 1

ii Maximum occurs when sin(q + 0.322) = 1

Þq + 0.322 = p

2

Þq = 1.25 (2 d.p.)

Note that you should use a value of a to 3 decimal places in the model and then give your

answers to 2 decimal places.

c

T = 9 + 40 sinpt

12+ 0.322

æ

èçö

ø÷

So minimum value of T is 9 - 40 = 2.68°C (2 d.p.)

Occurs when sinpt

12+ 0.322

æ

èçö

ø÷= -1

Þpt

12+ 0.322 =

3p

2

Þ t = 16.77 hours


30 d

9 + 40 sinpt

12+ 0.322

æ

èçö

ø÷= 14

Þ 40 sinpt

12+ 0.322

æ

èçö

ø÷= 5

Þ sinpt

12+ 0.322

æ

èçö

ø÷=

5

40

Þpt

12+ 0.322 = 0.9117,2.2299

Þ t = 2.25, 7.29 (2 d.p.)

0.25h »15 minutes and 0.29 h »17 minutes

So times are 11:15am and 4:17 pm

31 a As

4

t¹ 0, x ¹ 1

The equation for y can be rewritten as

y = t -3

2

æ

èçö

ø÷

2

-5

4

So y ³ -1.25

b t =

4

1- x

So y =4

1- x

æ

èçö

ø÷

2

- 34

1- x

æ

èçö

ø÷+1

=16

1- x( )2

-12 1- x( )

1- x( )2

+1- x( )

2

1- x( )2

=16 -12 +12x +1- 2x + x2

1- x( )2

=x2 +10x + 5

1- x( )2

So a =1, b =10, c = 5

32 a x = ln(t + 2) Þex = t + 2 Þ t = ex - 2

y =

3t

t + 3=

3ex - 6

ex +1

t > 4 Þex - 2 > 4 Þex > 6 Þ x > ln6

So the solution is y =

3ex - 6

ex +1, x > ln6


32 b When x®¥, y®3

When x = ln6, y =3eln6 - 6

eln6 +1=

(3´ 6) - 6

6 +1=

12

7

So range is 12

7< y < 3

33 x =

1

1+ tÞ t =

1

x-1=

1- x

x

y =1

1-1- x

x

=x

x - 1- x( )=

x

2x -1

34 a y = cos3t = cos 2t + t( ) = cos2tcost - sin2tsint

= cos2 t -1( )cost - 2sin2 t cost

= 2cos3 t - cos t - 2 1- cos2 t( )cost

= 4cos3 t - 3cos t

x = 2cost Þ cos t =x

2

y = 4x

2

æ

èçö

ø÷

3

- 3x

2

æ

èçö

ø÷=

1

2x3 -

3

2x =

x

2x2 - 3( )

b 0 £ t £

p

2

So 0 £ cost £1 and -1£ cos3t £1

So 0 £ x £ 2, -1£ y £1

35 a

y = sin t +p

6

æ

èçö

ø÷= sin tcos

p

6+ cost sin

p

6

=3

2sin t +

1

2cos t

=3

2sin t +

1

21- sin2 t

=3

2x +

1

21- x2

As -

p

2£ t £

p

2, -1£ sin t £1Þ -1£ x £1


35 b

At A, sin t +p

6

æ

èçö

ø÷= 0Þ t = -

p

6

x = sin -p

6

æ

èçö

ø÷= -

1

2

Coordinates of A are -1

2, 0

æ

èçö

ø÷

At B, x = sin t = 0Þ t = 0

y = sin t +p

6

æ

èçö

ø÷= sin

p

6=

1

2

Coordinates of B are 0,1

2

æ

èçö

ø÷

36 a y = cos2t = 2cos2 t -1

y = 2x

3

æ

èçö

ø÷

2

-1, - 3£ x £ 3

b Curve is a parabola, with a minima and y-intercept at (0,–1) and x-intercepts when

2x

3

æ

èçö

ø÷

2

= 1Þx

3= ±

1

2Þ x = ±

3

2

Coordinates -3

2, 0

æ

èçö

ø÷,

3

2, 0

æ

èçö

ø÷

37 y = 3x + c would intersect curve C if

8t 2t -1( ) = 3 4t( ) + c

16t2 - 20t - c = 0

Using the quadratic formula, this equation has no real solutions if

-20( )2

- 4 16( ) -c( ) < 0

Þ 64c < -400 Þ c < -25

4


38 a The curve intersects the x-axis when 2cos t +1= 0 Þ cos t = -

1

2

Solutions in the interval are t =2p

3,4p

3

Þ x = 3sin4p

3

æ

èçö

ø÷, 3sin

8p

3

æ

èçö

ø÷

So coordinates are -3 3

2, 0

æ

èç

ö

ø÷ and

3 3

2, 0

æ

èç

ö

ø÷

b 3sin2t = 1.5 Þ sin2t =

1

2

In the interval p £ 2t £ 3p solutions are

2t =13p

6,17p

6

Þ t =13p

12,17p

12

39 a Find the time the ball hits the ground by solving -4.9t2 + 25t +50 = 0

t =-25± 252 - 4 -4.9( ) 50( )

2 -4.9( )t ³ 0, so only valid solution is t = 6.64s (2 d.p.)

Þ k = 6.64 (2 d.p.)

b

t =x

25 3

y = 25x

25 3

æ

èçö

ø÷- 4.9

x

25 3

æ

èçö

ø÷

2

+ 50

=x

3-

49

18750x2 + 50

Domain of the function is from where the ball is hit at x = 0 to where

it hits the ground when t = 6.64 seconds.

When t = 6.64, x = 25 3(6.64) = 287.5 (1 d.p.)

So domain is 0 £ x £ 287.5


Challenge

1 Angle of minor arc =

p

2 because it is a quarter circle

Let the chord meet the circle at R and T. The area of P is the area of sector formed by O, R and T less

the area of the triangle ORT.

So area of P = 1

2r 2 p

2-

1

2r 2 sin

p

2= r 2 p

4-

1

2

æ

èçö

ø÷=

r 2

4(p - 2)

Area of Q = pr 2 - area of P

= r 2 p -p

4+

1

2

æ

èçö

ø÷= r 2 3p

4+

1

2

æ

èçö

ø÷=

r 2

4(3p + 2)

So ratio = (p - 2) : (3p + 2) =p - 2

3p + 2:1

2 a sin x

b cosx

c ÐCOA =

p

2- x Þ ÐCAO = x

OA =1¸ sinx = cosec x

d 1 tan cotAC x x

e tan x

f 1 cos secOB x x


3 a sin t =

x - 3

4, cost =

y +1

4

As sin2 t + cos2 t =1

x - 3

4

æ

èçö

ø÷

2

+y +1

4

æ

èçö

ø÷

2

= 1

Þ x - 3( )2

+ y +1( )2

= 16

The curve is a circle centre (3, –1) and radius 4.

Endpoints when t = -p

2, x = -1, y = -1

and when t =p

4, x = 2 2 + 3, y = 2 2 -1

b C is 3

8ths of a circle, radius 4

So length =

3

8´8p = 3p

Documents

Review Exercise 2 x = - 2 2 Þ sec x = - 2 Þ cos x = - 1 2 x = 3p 4, 5 p 4, 11p 4, 13p 4 14 a sinq cos q + cos q sinq = sin2 q + cos 2 q cos qsinq = 1 s inq cos q (us ing cos 2 q