Form 5: Chapter 16 (Trigonometric Functions) SPM Practice Fully-Worked Solutions
Paper 1 1 When cos x = – , 2 x = 120°, 240°
1 1 + t 2
t
tan θ = t
θ
When cos x = 1, x = 0°, 360°
1
(a) cot (– θ) 1 = tan (– θ) 1 = – tan θ 1 = – t
∴ x = 0°, 120°, 240°, 360°
4
(b) cos (90° – θ) = sin θ t = 1 + t 2 2
6 sec2 θ – 20 tan θ = 0 6(1 + tan2 θ) – 20 tan θ = 0 6 + 6 tan2 θ – 20 tan θ = 0 6 tan2 θ – 20 tan θ + 6 = 0 3 tan2 θ – 10 tan θ + 3 = 0 (3 tan θ – 1) (tan θ – 3) = 0 1 tan θ = or 3 3
When sin θ = –1, θ = 270°
∴ θ = 210°, 270°, 330°
5
1 When tan θ = , 3 θ = 18.43°, 198.43° When tan θ = 3, ∴ θ = 71.57°, 251.57°
∴ θ = 18.43°, 71.57°, 198.43°, 251.57° 3
cos 2θ – 3 sin θ = 2 1 – 2 sin2 θ – 3 sin θ – 2 = 0 – 2 sin2 θ – 3 sin θ – 1 = 0 2 sin2 θ + 3 sin θ + 1 = 0 (2 sin θ + 1)(sin θ + 1) = 0 1 sin θ = – or sin θ = –1 2 1 When sin θ = – , 2 basic ∠ = 30° θ = 210°, 330°
15 cos2 x – 7 cos x = 4 cos 60° 15 cos2 x – 7 cos x = 4(0.5) 15 cos2 x – 7 cos x = 2 15 cos2 x – 7 cos x – 2 = 0 (3 cos x – 2)(5 2)(5 cos x + 1) = 0 2 1 cos x = or cos x = – 3 5 2 When cos x = , 3 basic ∠ = 48.19° ∴ x = 48.19°, 311.81°
2
2 sin x + cos x = 1 2(1 – cos 2 x) + cos x – 1 = 0 2 – 2 cos2 x + cos x – 1 = 0 – 2 cos2 x + cos x + 1 = 0 2 cos2 x – cos x – 1 = 0 (2 cos x + 1)(cos x – 1) = 0 1 cos x = – or cos x = 1 2
1 When cos x = – , 5 basic ∠ = 78.46° ∴ x = 101.54°, 258.46°
∴ x = 48.19°, 101.54°, 258.46°, 311.81°
64
6
7
tan x + 2 sin x = 0 sin x + 2 sin x = 0 cos x sin x + 2 sin x cos x = 0 sin x (1 + 2 cos x) = 0 1 sin x = 0 or cos x = – 2
1
k
θ
1 – k 2
When sin x = 0, x = 0°, 180°, 360° 1 When cos x = – , 2 basic ∠ = 60° ∴ x = 120°, 240°
1 sin 2θ 1 = 2 sin θ cos θ 1 = 2k 1 – k 2
cosec 2θ =
∴ x = 0°, 120°, 180°, 240°, 360°
65
Paper 2 1 (a) LHS = tan
x
+ cot
2
sin
x
cos
=
sin
2 x
2
sin
2
+
x
sin2
x
cos
2
=
x
2
cos
2
x
+ cos2
x
Hence, the equation of the straight line for solving the equation 3 3 3 sin x = x – 1 is y = x – 2. 2 π 2 4 π
2 x
2
0
2 π
y
–2
1
Number of solutions = Number of intersection points =3
x
2
1
= sin
x
2
cos
2 (a) The sketch of the graph of y = cos 2 x for 0 ≤ x ≤ π is as shown below:
x
2
y
2
= 2 sin
x
2
cos
x
2
1
2 sin x = 2 cosec x = RHS =
(b)
x
O 1
(i)
(b) 2 sin2 x = 2 –
This is a y = sin θ graph with 1 1 1 cycles because θ = 1 x. 2 2
π
4
2
x
π
x π
1– cos 2 x = 2 –
2 1
3π 4
cos 2 x = 1– 2 sin 2 x ∴ 2 sin2 x = 1– cos 2 x
3 y = 2 sin – x 2
y
π
x π
x
– cos 2 x = 1– π
O
2π
x
cos 2 x =
1 2
x π
π
–1
The straight line that has to be drawn is x y = – 1.
3 x – 2 y = –– 2π
π
(ii)
3 3 sin x = x – 1 2 4 π 3 3 2 sin x = x – 2 2 2 π
x
0
π
y
–1
0
y y = cos 2x
The solutions to the equation 3 3 2 sin x = x – 2 are given by the 2 2 π x-coordinates of the intersection 3 points of the graphs of y = 2 sin x 2 3 and y = x – 2. 2 π
1
π
3 4
4
O
π 2
1
π x
π x y = – – 1
π
66
4 (a), (b)
Hence, the number of solutions to the equation 2 sin2 x = 2 –
x π
y
for 0 ≤ x ≤ π
= Number of intersection points =2
O
|sin 2 x | + 2 cos x = 0 |sin 2 x | = – 2 cos x Number of solutions = Number of intersection points =2
(i) The sketch of the graph of y = – cos 2 x is as shown below. y
(2π,
x
3 π 2π 2
π
2
= – (2 cos2 x – 1) = – cos 2 x = RHS
1
π 2
If cot2 x + 1 = cosec 2 x, then cosec2 x – cot2 x = 1.
x 1 y = –– – – 2π 2
y = |sin 2x |
1
3 (a) LHS = –2 cos2 x + cosec2 x – cot2 x = – 2 cos2 x + 1
(b)
y = – 2 cos x
2
5 (a), (b) The graph of y = |3 sin 2 x | is as shown below.
1 ) 2
y 1 (0, – –) 2
O
π
π
3 π 2
2
1
2π
x
y = |3 sin 2x | 3
y = –cos 2x
x
(ii) 2(–2 cos2 x + cosec2 x – cot2 x) =
π
x
2(–cos 2 x) = – cos 2 x =
π
O
–1
3π 2 1 π 2
x
2 π
y
0 –
1 2
x
π y = 3 –
–1
3
π x
–3
–
1 2
3 3 – |3 sin 2 x | = x
The straight line that has to be x 1 sketched is y = – . 2 π 2 x
2π
π
3 –
3
x = |3 sin 2 x |
π
Therefore, the equation of the straight line 3 that should be drawn is y = 3 – x.
2 π 1 2
π
Number of solutions = Number of intersection points =4
x
0
2 π
y
3
–3
Number of solutions = Number of intersection points =4
67
6 (a) LHS =
=
2 cot x 2 – cosec2 x cos x 2 sin x
y
2– 2 =
(b) (i), (ii)
1 sin2 x O
cos x sin x
1 π 4
3 π 4
1π 2
2 sin2 x – 1 sin2 x
π ×
x (π, –5)
5x y = – π
2 cos x sin2 x × sin x – (1 – 2 sin2 x) 2 sin x cos x = – (1 – 2 sin2 x) sin 2 x = – cos 2 x = – tan 2 x = RHS =
2 cot x 5 x – =0 2 2 – cosec x π 5 x – tan 2 x – =0 π
tan 2 x = –
5 x π
Sketch the straight line y = – Number of solutions = 3
68
5 x π
.
