A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES
TRIGONOMETRY 1
1. a) need values of 2x for 0 2x720
2x = 30
2x = 150, 390, 510
x = 15, 75, 195, 255
B1 cao
M1(150) M1(+360)
A2 (A1 at least 2;
-1 for extras)
[5]
b) need -30 x30330
x  30 = 30
x30 = -30, 330
x =0, 60, 360
M1
M1 ( 60o)
M1 A1 (0o or 360o)
A1 (both)
[5]
2. a) sinsin + 1) =0
sin = 0 or sin = -1
3
 = 0, , 2 or  = 2
M1
A1 (either)
M1 (sin-1, 1 ans)
A2 (all correct)
[5]
b) (2cos + 1)(cos  1) = 0
cos = - 12 or cos = 1

2
3
,
2
3

M1
A1 (either)
or 0
A2 (all) (A1 for 2)
[4]
3. a) cos2x = 1  sin2x
3sin2x  (1  sin2x) 4sin2x –1
a=4
b = -1
B1
M1 A1 cao
A1 ft
[4]
b) 4sin2x = 2
sinx = 
M1
1
A1 cao
2
x = 45o, 135o
4. sin x =

3
5
 cos x =
A3 (-1 eeoo)
[5]

4
5
M1 A1
90<x < 270 and sinx negative  180<x<270
cosx =
B1
4

5
A1
[4]
Page 1
A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES
TRIGONOMETRY 1
5. a) tanx = 2
x = 1.11 or 4.25
M1
A2
[3]
b) cos2x  1 sin2x
sin2x  sin x  1=0
M1
M1 A1
1 1 4
2
sin x =
M1
sinx = -1.618.. or 0.618..
-1.618… gives no solutions
x = 0.666 or 2.48
A1 cao
A2
[7]
6. a) i) cos2x  1  sin2x
3sin x + 2cos2x  3s + 2(1s2)
 2 + 3s  2s2
 (2s)(1+2s)
ii) tan2x =
B1
M1
A1
A1
[4]
sin 2 x
cos 2 x
M1
cos2x = 1  s2
tan2x =
M1
s2
1s 2
A1
[3]
b) i) (2s)(1+2s)=0
s=2, - 12
M1
A1 (both)
7  11
,
6 6
x=
A2
[4]
ii) 
s
1s 2
=2
M1
s = 2(1  s2)
s2 = 4(1  s2)
5s2 = 4
s2 =
s =
M1 A1
4
5
2
A2
5
[5]
Page 2
A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES
TRIGONOMETRY 1
7. a) -270 < 90  x < 90
90  x = 60, -240
x = 30 o, 330o
M1
A1 (either)
A1 ft
3
2
OR: cosx =
M1
x = 30o, 330o
b) -30
x
3
x
3
A2
[3]
30
M1
= -30
A1
x = -90
A1 cao
[3]
c) -5402x180
1
sin 2 x
=2
sin2x =
1
2
M1
2x = 30, -210, -330
x = 15, -105, -165
A1 (15) M1 A2
[5]
8. a) 1 + tan2x = sec2x
tan2x + 2tanx  3 = 0
(tanx + 3)(tanx  1) = 0
tanx = -3, tanx = 1
x = -71.6, 108.4, 45, 225
b)
M1
A1
A1
A3 (-1 eeoo)
[6]
1
 2 tan x
tan x
tan2x = 12
tanx =

M1
1
A1
2
x = 35.3, 144.7, 215.3
A3 (-1 eeoo)
[5]
Page 3
A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES
TRIGONOMETRY 1
9. a) cos4x  sin4x  (cos2x  sin2x)(cos2x + sin2x)
But sin2x + cos2x = 1
So cos4x  sin4x  cos2x  sin2x
b) cos2: 5 + 2tan2 = 4sec2
1 + tan2  sec2
5 + 2tan2 = 4(1 + tan2)
tan2 = 12




M1
B1
A1
[3]














M1 A1
M1
A1
OR: 5cos2 + 2sin24cos2 + 4sin2 
cos2= 2sin2       
tan2 = 12
































M1 A1
M1
A1
[4]
10. a) cos =

2

 1
3
2
 

12 2
13

5
13
M1
and sin negative     
cos must be negative  cos =

3
2
B1
5
13
A1 cao
[3]
b) tan = sincos
12
= 5






















A1 ft
[2]
c) sin( = -sin for all  
sin(








12
13










M1
A1
[2]
Page 4
A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES
TRIGONOMETRY 1
11. a) segment = 12 r2 (x  sinx)
M1 A1
sector = 12 r2(  x)
1 r2
2
(x  sinx) =
1 r2(
2
B1
 x)
M1
x  sinx =   x
 + sinx = 2x
1 ( + sinx) = x
2
A1
[5]
b) x = 12 ( + sinx)  2x  sinx   = 0
x = 2:
2x  sinx   = -0.051
x = 2.1: 2x  sinx   = 0.195
sign change  root
M1 A1
x = 2.05: 2x  sinx   = 0.071
M1 (looking for
further sign changes)
 between 2 and 2.05
x = 2.03: 2x  sinx   = 0.022
 between 2 and 2.03
x = 2.015: 2x  sinx   = -0.0146
 between 2.015 and 2.03
x = 2.025: 2x  sinx   = 0.0098
 between 2.015 and 2.025
 x = 2.02 (2 DP)
A1 (2 correct values)
M1
A1
[6]
Page 5
A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES
TRIGONOMETRY 1
12. a) QTS=90o, since it is angle in semicircle
QS = (82 + 42)
= 80
= 45
B1
M1
A1
[3]
b) RSQ is isosceles, since RS = RQ
R
60o
P
Q
25
2 5
= tan60 = 3
RP
RP =
2 5
2 15
3
=
3
2
RQ =
M1
A1
 
 2 15 

 2 5
 3 


2
240 4 15

3
9

M1 A1
[4]
c) Arc of C1 =
2  4 15 8 15


3
3
9
M1 A1
Arc of C2 = 25
Perimeter =
2 
9
9 
B1 ft
5  4 15 
A1

Area required = area of semicircle C2 area ofsegment C1
area of C1 = 12 r2(  sin)
M1
M1
2
1  4 15   2
3 

= 

2  3   3
2 
=
area of C2
20 
4  3
9 
A1
3 
A1

= 12 r2
=
1
2
(25)2
=10 

Required area = 10
=
M1



20 
4  3
9 







3 










A1 ft
M1

103 3
9
A1
[12]
Page 6
A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES
TRIGONOMETRY 1
13. a) Area of sector = 12 r2 =
r 2
6
B1
r2 3
4
Area of triangle = 12 r2sin =
Ratio

6
:
3
4
which is
3

:
2
3
M1 A1
or 2 : 33
M1 A1 cao
[5]
b) Perimeter of sector = 2r +
Length of AB = r
Perimeter of triangle = 3r
Ratio 2r + 13 r : 3r
1
3
r
M1 A1
B1
B1
M1
which is 6+: 9
14. a) i) maximum value of cosx is 1, so cosx can’t be
A1
[6]
4
3
B1
[1]
ii) maximum value of sinx is also 1, but the two maxima can’t occur for same value of x B2
OR:
sinx + cosx 2sin(x+45o)  2
M1 A1
[2]
b) i) maximum is 4, minimum is 2
at 90o, 270o
M1 A1
A1 (both)
[3]
ii) maximum is 3, minimum is 1
maximum when cos 2x = -1; minimum when cos2x=1
maximum at x=90; minimum at x = 180
15 a) 2(1  sin2x) + sinx = 1
2sin2x  sinx 1 = 0
sinx = - 12 or 1
M1 A1
A1 (both)
[3]
M1
A1
A1 ft
x = 210, 330, 90
A2 (-1eeoo)
[5]
b) (2cosx  sinx)(cosx  sinx) = 0
tanx = 2, tanx = 1
x = 63.4, 243.4 or 45, 225
M1 A1
M1 A1
A2 (-1 eeoo)
[6]
Page 7
A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES
TRIGONOMETRY 1
16.a) -2<  0
M1
11
7
 
or  6
6
5

  or 
6
6
A2
A1
[4]
b) 2(sec2  1) = 3sec  
2sec2  3sec 2 = 0
(2sec + 1)(sec  2) = 0
sec = - 12 or 2



















M1
M1
A1
No solutions for sec = - 12
sec = 2  cos =

1
2
5
3
 3 or
A2
[5]
17 a) Let BP = x, so PC = 2x
AP2 = 62  x2
also AP2 = 82  (2x)2
So 36  x2 = 64  4x2
x2 =
28
3
28
3
x=
BP = 3x = 6
7
3
=6
=2
21
3
M1
A1
7
3
A1 (
=221
28
3
M1 A1
[5]
62 82 ( 2 21) 2
268
16
1
=
=
96
6
b) cos BÂC =
M1 A1
A1
[3]
c) Area =
1
2
86sin BÂC
M1 A1
2
= 23.7cm
A1
[3]
Page 8
)
A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES
TRIGONOMETRY 1
18 a) Require period = 30 days =
2
a
M1
2


30 15
a=
A1
[2]

b) Require sin  7000    b  = 1
15
 


2
sin((466 3 ) + b) = 1
B1
But sin((466 23 ) + b)=sin ( 23  + b)
So sin( 23 + b)=1
2
3

=6

b
+b=


2






M1
















M1
A1
[4]
c)
1
7000
7015
7030
7045
7060
G1 shape
G1 2 cycles
G1 1, -1 on y-axis
[3]
-1
d) Require 7030t7045
sin is at half its maximum value when  =
5

and
6
6
Relating these to when the maximum occurs:
max occurs when  = 12 ,
so it is at half max value

3
before and

3
B1
M1
after this
A1
The period is 30 days, which corresponds to 2
So

3
corresponds to 5 days
B1
so half max value is attained 5 days before and five days after max value
B1
Since previous max is t = 7030, exams start when t = 7035
So exams start on 5th June
A1
B1
End date is t = 7045 (from graph)
15th June
B1
B1
[9]
Page 9
A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES
TRIGONOMETRY 1
19. a) (120, 6) is a maximum
sin(120k) = 1
120  k = 90  k = 30
B1
M1
A1
[3]
b) Graph goes through (0, 3)
A + B sin(30)=3
A  12 B = 3
M1
M1
A1
Graph goes through (120, 6)
A + Bsin90 = 6
A+B = 6
M1
A1
B1
[6]
c) A+B = 6
A  12 B = 3
1 12 B = 3  B=2 and A = 4
M1 (solving) A1 A1
[3]
d) y = 4 + 2sin(t30)
min when sin(t30) = -1
t  30 = 270, 630
t = 300, 660
Points are (300, 2) and (660, 2)
B1
M1 A1
A1
A1
[5]
20 a)
1
y=sin2x
0
90
180
270
y=cos2x
-1
b) 4
360
G1 (both shapes)
G1 (two cycles each)
G2 (“start” points)
G1 (1 and -1)
-2 if unclear
which is which
[5]
B1 ft
[1]
c) sin2x = cos2x
tan2x = 1
2x = 45, 225
x = 22 12 , 112 12
M1
A2
A1
[4]
d) They are the solutions to tan2x =1; this is periodic
Period is 900  d=90; a =22 12
S10 =
10
2
(45 + 990) = 4275
B1
B1 B1
M1 A1 ft
[5]
Page 10