© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 1
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.)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 2
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.)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 3
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 4
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( )
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 5
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 6
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 7
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 8
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( )
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 9
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 10
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 11
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.)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 12
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 13
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 14
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 15
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 16
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 17
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 18
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
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 19
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