Double angle Formula revision mark scheme
1.
2sin x cos x + cos x = 0 oe
M1
cos x = 0 or sin x = – 0.5
Either one
A1
cos x = 0 => x = 2nπ ± ... oe
Condone degrees
m1
sin x = – 0.5 => x = nπ + (–1)nα oe
Condone degrees
m1
x = 2nπ ± π/2 oe and
x = nπ +(–1)n(–π/6) oe
Need both in rads.
sc If m0m0 award B1 for four particular
solutions ‘covering all positions’ or
general solution(s) for two positions
(condone degrees)
A1
5
[5]
2.
(a)
(ii)
(ii)
A = B = θ gives
sin 2θ (+ sin 0) = 2 sin θ cos θ
A = 2θ, B = θ⇒
sin 3θ+ sin θ = 2 sin 2θ cos θ
Use of (i) ‘s result: = 2 . 2 sin θ cos2θ
Use of c2 = 1 – s2: = 4 sin θ– 4 sin3θ
ag
South Wolds Comprehensive School
B1
1
M1
A1
M1
A1
4
1
(b)
f(½) = 0 shown or (2s – 1)(4s2 + 2s –3)
B1
(c)
4s – 4s3 =
B1
⇒ 8s3 – 8s + 3 = 0
3
2
(2s – 1)(4s2 + 2s – 3) = 0
sin θ = ½ ⇒ θ =

(or 0.524 rads)
6
cao first answer, ft second answer
sin  
 2  52
 0.6514 ....
8
Ignore sin θ= – 1.1514
⇒ θ = 0.709 rads
cao first answer, ft second answer
And  
5
(or 2.62 rads), 2.43 rads
6
To at least 3 s.f.
1
M1A1
M1
A1ft
M1A1
A1ft
A1
9
[15]
3.
6 sin 2 θ cos θ  2 cos 3 θ
cos θ  0, θ 
tan 2 θ 
θ
π 3π
,
2 2
1
3
B1
M1A1
3
1

2
2
 or cos θ  ; sin θ  
4
4

π 5π 7π 11π
, , ,
6 6 6 6
M1
A1 (any two)
A1 (all rest +
no others)
6
[lose first A1 earned only for answers not in terms of ]
[6]
South Wolds Comprehensive School
2
4.
3(2cos² – 1) – cos + 1 = 0
Correct use of cos2 = 2cos² – 1
B1
6c² – c – 2 = 0
 cos  =
2
1
, –
3
2
M1
Attempt to factorise or solve quadratic
2
cos  ,  48
3
A1
awrt 48
360° – 48° = 312°
cos  = –
A1ft
1
, 120 
2
A1
360° – 120° = 240°
A1ft
Withhold final mark for extra solutions in interval, radians etc
6
[6]
5.
8t =
t2 =
2t
1 t2
where t = tan  (or equivalent in sin  and cos )
B1
t = 0   = 0°
B1
3
3
4
(or sin2 = ; or cos2 = )
4
7
7
 = –40.9° , 40.9°
M1 A1
A1 A1 ft
6
(ignore extra solutions outside range –90° <  < 90°)
(withhold final A1 for extra solutions within range)
[6]
South Wolds Comprehensive School
3
6.
Use of cos2x = 1 – 2 sin2 x
B1
Quadratic in sin x and attempt to solve
M1
10 sin2 × – 3 sin x – 1 = 0  sin x
x=
1
1
, –
2
5
A1

, (0.524)
6
A1
5
(2.62)
6
( – previous answer)
Handling minus sign ; either  + sin –1 (*) or 2 – sin –1 (*)
3.34c , 6.08 c
final root correct
A1 ft
M1
A1
[7]
Use of degrees instead of radians loses 2 marks. Award last four marks as
A0, A1 ft, M1, A0 for example for 30°, 150°, 192°, 348°
7.
(a)
sec x  2  cos x 

1
2
π 5π
,
3 3
M1
A1
2
AWRT 1.05c and 5.24c
(b)
cos2x  cos(x + x)
 cos x × cos x – sin x × sin x
 cos2x – sin2x
 cos2x – (1 – cos2x)
 2 cos2x – 1
AG
South Wolds Comprehensive School
M1
A1
2
4
(c)
(2 cos2x – 1) + 3 cos x – 1  0
M1
2 cos2x + 3 cos x – 2  0
(2cos x – 1)(cos x +2)  0
for factorisation attempted
(or use of formula or completing the square)
m1
cos x  –2  no solutions
B1
M1
for cos x 
cos x 
1
2
1
π 5π
 x ,
2
3 3
ft on their (a)
A1 ft
5
[9]
8.
(a)
cos  = –
5
13
M1A1
5
M0A0 unless from s2 + c2 = 1
13
in which case M1A0
cos  =
(b)
sin ( + ) = sin  cos  + cos  sin 
use of
 12 4   5 3 
=   –  
 13 5   13 5 
M1
A1F
f.t. cos  from (a)
Non-exact value gets M1A0A0 or
possibly M1A1A0
=
33
65
A1F
3
[5]
9.
(a)
tan 45  tan 
1  tan 45 tan 
Use of correct formula for tan (A + B)
tan (45° + ) =
=
1  tan 
1  tan 
M1
A1
2
Replace tan 45° = 1
(b)
Put  = 60° : tan 105° =
South Wolds Comprehensive School
1 3
1 3
M1 A1
5
use of  = 60° i.e. tan105 =
use of tan 60 =
1  tan 60
1  tan 60
M1
3 in correct formula for
tan(A + B) or equiv
3
3
=
(1  3 ) 2
(1  3 )(1  3)
M1
attempt at rationalisation
=
1 2 3  3
2
= –2 –
3
A1F
4
ft if required form
[6]
10.
(a)
(b)
cos(2x + x) = cos 2x cos x – sin 2x sin x
OE
M1
= (2 cos²x – 1) cos x – 2cos x (1 – cos²x)
All in terms of cosx
m1
cos 3x = 4cos³x – 3cosx
AG
A1
8cos³x – 7cosx = 0
Use of (a) attempted. OE
M1
cosx (8cos²x – 7) = 0
Factorisation PI
m1
cosx = 0, cos²x = 0.875
OE
A1
South Wolds Comprehensive School
3
6
cosx = 0 gives x = /2 = 1.57
accept /2 but apply RA rule
A1
cosx = 0.9354.. gives x = 0.36
Dep on M only. ft one minor slip
A1F
cosx = – 0.9354.. gives x = 2.78
RA rule applies
NMS: give 0/6
If cancelled cosx and lost /2, max 3/6
If used degrees, max penalty of 1 mark
A1F
6
[9]
11.
(a)
(b)
Attempt to factorise
c4 – s4 = (c² + s²)(c² – s²)
= cos 2x
convincing completion (AG)
M1
A1
A1
Need cos 2x < 0 and an attempt to use
accept < replaced by =
M1
Inequalities relating x to  4 and
  x  3
4
4
m1
A1
3
4
3
3
degrees lose final mark
[6]
12.
(a)
R = 5
B1
OE
1
oe leading to
2
 = 26.565º
any suitable accuracy
tan a =
(b)
(i)
B1
Solving 5 cos(x + ) = 1 to obtain
accept their R
M1
x +  = 63.4º (and 296.6º)
A1 for either
A1
x = 36.9º and 270º
ft slip; condone rounding errors
A1
South Wolds Comprehensive School
2
3
7
(ii)
(0º) x  36.9 or c’s first value
any suitable accuracy
B1ft
270º  x (< 360º) or c’s second value
36.9º  x  270º not allowed on its won (award B1 only)
B1ft
2
[7]
13.
No mark scheme available.
14.
(a)
sin x cos x
= OE

cos x sin x
changing to sines and cosines or (tan x + 1/tan x)
tan x + cot x =

sin ² x  cos ² x
OE
sin x cos x
writing with single denom
Use of sin²x + cos²x = 1 or 1 + tan²x = sec²x or sin x cos x = ½ sin 2x
Full valid proof to tan x + cot x = 2/(sin 2x)
AG; must be convincing {accept equiv in reverse order}
(b)
B1
B1
M1
A1
Use of (a) to at least sin 2x = k (with k < 1)
M1
1 3
x   ;  {accept 0.392(6..), 1.17(8..)}
A1A1
8 8
Both 22.5º and 67.5º A1 only (ignore ans. outside range)
4
3
[7]
15.
(a)
(b)
cos2x = cos2x – sin2x
= (cos x – sin x)(cos x + sin x)
difference of two squares
B1
M1
Hence result
A1
R = 2  = 45°
B1B1
2 sin (x + 45) = ½ }
x = 114°
}
x = 336°
}
AWRT these are OK
–1 for any extra solutions
M1
A1f.t.
A1f.t.
3
5
[8]
South Wolds Comprehensive School
8