Trigonometry Practice (No
Calculator) [95 marks]
Let sin θ =
√5
,
3
where θ is acute.
1a. Find cos θ.
[3 marks]
Markscheme
evidence of valid approach
eg right triangle,
correct working
cos2 θ
(M1)
= 1 − sin2 θ
(A1)
eg missing side is 2, √1 − (
cos θ =
2
3
A1
√5 2
)
3
N2
[3 marks]
1b. Find cos 2θ.
[2 marks]
Markscheme
correct substitution into formula for cos 2θ
2
eg 2 × ( 23 ) − 1, 1 − 2(
cos 2θ = − 19
A1
√5
)
3
2
, ( 23 ) − (
2
(A1)
√5 2
)
3
N2
[2 marks]
Let f(x) = 3 sin( π2 x), for 0 ⩽ x ⩽ 4.
2a. (i)
(ii)
Write down the amplitude of f .
Find the period of f .
[3 marks]
Markscheme
(i)
3
(ii)
valid attempt to find the period
eg
A1
N1
(M1)
2π 2π
, π
b
2
period = 4
A1
N2
[3 marks]
2b. On the following grid sketch the graph of f .
[4 marks]
Markscheme
A1A1A1A1
N4
[4 marks]
The following diagram shows a triangle ABC and a sector BDC of a circle with centre B and
The following diagram shows a triangle ABC and a sector BDC of a circle with centre B and
radius 6 cm. The points A , B and D are on the same line.
^ is obtuse.
AB = 2√3 cm, BC = 6 cm, area of triangle ABC = 3√3 cm2 , ABC
^ .
3a. Find ABC
[5 marks]
Markscheme
METHOD 1
correct substitution into formula for area of triangle
eg
1
(6) (2√3) sin B,
2
correct working
6√3 sin B,
1
2
π
(30∘ )
6
^ =
ABC
1
(6) (2√3) sin B
2
= 3 √3
(A1)
eg 6√3 sin B = 3√3, sin B =
sin B =
(A1)
3√3
1
(6)2√3
2
(A1)
(A1)
5π
(150∘ )
6
A1
N3
METHOD 2
(using height of triangle ABC by drawing perpendicular segment from C to AD)
correct substitution into formula for area of triangle
eg
1
(2√3) (h)
2
(A1)
= 3√3, h√3
correct working
(A1)
eg h√3 = 3√3
height of triangle is 3
^ = π (30∘ )
CBD
6
^ =
ABC
5π
(150∘ )
6
A1
(A1)
A1
N3
[5 marks]
3b. Find the exact area of the sector BDC.
[3 marks]
Markscheme
recognizing supplementary angle
^ =
eg CBD
π
,
6
sector =
1
(180
2
(M1)
2
^
− ABC)(6
)
correct substitution into formula for area of sector
eg
1
2
×
π
6
2
× 6 , π(6
area = 3π (cm2 )
2
(A1)
30
) ( 360
)
A1
N2
[3 marks]
Let f(x) = 6x√1 − x2 , for −1 ⩽ x ⩽ 1, and g(x) = cos(x), for 0 ⩽ x ⩽ π.
Let h(x) = (f ∘ g)(x).
4a. Write h(x) in the form a sin(bx), where a, b ∈ Z.
[5 marks]
Markscheme
attempt to form composite in any order
(M1)
eg f (g(x)) , cos(6x√1 − x2 )
correct working
(A1)
eg 6 cos x√1 − cos2 x
correct application of Pythagorean identity (do not accept sin2 x + cos2 x = 1)
(A1)
eg sin2 x = 1 − cos2 x, 6 cos x sin x, 6 cos x |sin x|
valid approach (do not accept 2 sin x cos x = sin 2x)
(M1)
eg 3(2 cos x sin x)
h(x) = 3 sin 2x
A1
N3
[5 marks]
4b. Hence find the range of h.
[2 marks]
Markscheme
valid approach
(M1)
eg amplitude = 3, sketch with max and min y-values labelled, −3 < y < 3
correct range
A1
N2
eg −3 ⩽ y ⩽ 3, [−3, 3] from −3 to 3
Note:
Do not award A1 for −3 < y < 3 or for “between −3 and 3”.
[2 marks]
Let f(x) = 3 sin(πx).
5a. Write down the amplitude of f .
[1 mark]
Markscheme
amplitude is 3
A1
N1
5b. Find the period of f .
[2 marks]
Markscheme
valid approach
(M1)
eg period =
2π 360
, π
π
period is 2
A1
N2
5c. On the following grid, sketch the graph of y = f(x), for 0 ≤ x ≤ 3.
[4 marks]
Markscheme
A1
A1A1A1
Note:
N4
Award A1 for sine curve starting at (0, 0) and correct period.
Only if this A1 is awarded, award the following for points in circles:
A1 for correct x-intercepts;
A1 for correct max and min points;
A1 for correct domain.
The following diagram shows a circle with centre O and a radius of 10 cm. Points A, B and C
lie on the circle.
Angle AOB is 1.2 radians.
6a. Find the length of arc ACB.
[2 marks]
Markscheme
correct substitution
(A1)
eg 10(1.2)
ACB is 12 (cm)
A1
N2
[2 marks]
6b. Find the perimeter of the shaded region.
[3 marks]
Markscheme
valid approach to find major arc
(M1)
eg circumference −AB , major angle AOB × radius
correct working for arc length
(A1)
eg 2π(10) − 12, 10(2 × 3.142 − 1.2), 2π(10) − 12 + 20
perimeter is 20π + 8 (= 70.8) (cm)
A1
N2
[3 marks]
Total [5 marks]
Given that sin x =
3
,
4
where x is an obtuse angle,
7a. find the value of cos x;
[4 marks]
Markscheme
valid approach
(M1)
, sin2 x + cos2 x = 1
eg
correct working
(A1)
eg 42 − 32 , cos2 x = 1 − ( 34 )
correct calculation
eg
√7
,
4
cos2 x =
cos x = −
√7
4
2
(A1)
7
16
A1
N3
[4 marks]
7b. find the value of cos 2x.
[3 marks]
Markscheme
correct substitution (accept missing minus with cos)
3 2
eg 1 − 2( 4 ) , 2(−
correct working
√7
)
4
2
− 1, (
√7
)
4
2
− ( 34 )
(A1)
2
A1
7
eg 2 ( 16
) − 1, 1 −
18
, 7
16 16
2
cos 2x = − 16
(= − 18 )
−
A1
9
16
N2
[3 marks]
Total [7 marks]
The following diagram shows a right-angled triangle,
ABC, where
5
sin A = 13 .
8a. Show that cos A =
12
.
13
[2 marks]
Markscheme
METHOD 1
approach involving Pythagoras’ theorem
eg
(M1)
52 + x2 = 132 , labelling correct sides on triangle
finding third side is 12 (may be seen on diagram)
cos A =
12
13
AG
N0
METHOD 2
approach involving sin2 θ + cos2 θ = 1
eg
2
5
( 13
) + cos2 θ = 1, x2 +
correct working
eg
A1
cos2 θ =
144
169
12
13
AG
cos A =
[2 marks]
N0
25
169
=1
(M1)
A1
8b. Find cos 2A.
[3 marks]
Markscheme
correct substitution into cos 2θ
eg
2
5
1 − 2( 13
) ,
correct working
eg
2
2( 12
)
13
119
169
cos 2A =
2
5
− 1, ( 12
) − ( 13
)
13
2
(A1)
50
, 288
169 169
1−
(A1)
− 1,
A1
144
169
−
25
169
N2
[3 marks]
Let
f(x) = sin(x + π4 ) + k. The graph of f passes through the point
( π4 , 6).
9a. Find the value of k.
[3 marks]
Markscheme
METHOD 1
attempt to substitute both coordinates (in any order) into f
eg
= 6,
f ( π4 )
correct working
eg
sin
k=5
π
2
π
4
= sin(6 +
π
)
4
(M1)
+k
(A1)
= 1, 1 + k = 6
A1
N2
[3 marks]
METHOD 2
recognizing shift of
π
4
left means maximum at 6
R1)
recognizing k is difference of maximum and amplitude
eg
(A1)
6−1
k=5
A1
N2
[3 marks]
9b. Find the minimum value of f(x).
[2 marks]
Markscheme
evidence of appropriate approach
eg
(M1)
minimum value of sin x is −1, − 1 + k, f ′ (x) = 0, ( 5π
, 4)
4
minimum value is 4
A1
N2
[2 marks]
9c.
Let g(x) = sin x. The graph of g is translated to the graph of f by the vector ( ).
p
q
Write down the value of p and of q.
Markscheme
p = − π4 , q = 5 (accept (
− π4
5
))
A1A1
[2 marks]
The diagram below shows part of the graph of a function
N2
[2 marks]
The diagram below shows part of the graph of a function
f.
The graph has a maximum at A(
1,
5) and a minimum at B(
3,
−1) .
The function
f can be written in the form
f(x) = p sin(qx) + r . Find the value of
10a. (a)
p
(b)
q
(c)
r.
[6 marks]
Markscheme
(a)
valid approach to find p
eg amplitude =
p=3
A1
max−min
2
(M1)
,p=6
N2
[2 marks]
(b)
valid approach to find q
eg period = 4 , q =
q=
π
2
A1
(M1)
2π
period
N2
[2 marks]
(c)
valid approach to find r
max+min
2
eg axis =
r=2
A1
(M1)
, sketch of horizontal axis, f(0)
N2
[2 marks]
Total [6 marks]
[2 marks]
10b. p
Markscheme
valid approach to find p
eg amplitude =
p=3
A1
max−min
2
(M1)
,p=6
N2
[2 marks]
10c. q
[2 marks]
Markscheme
valid approach to find q
eg period = 4 , q =
q=
π
2
A1
(M1)
2π
period
N2
[2 marks]
10d. r .
[2 marks]
Markscheme
valid approach to find r
eg axis =
r=2
max+min
2
A1
(M1)
, sketch of horizontal axis, f(0)
N2
[2 marks]
Total [6 marks]
11a. Let sin 100∘ = m. Find an expression for cos 100∘ in terms of m.
[3 marks]
Markscheme
Note: All answers must be given in terms of m. If a candidate makes an error that means
there is no m in their answer, do not award the final A1FT mark.
METHOD 1
valid approach involving Pythagoras
(M1)
e.g. sin2 x + cos2 x = 1 , labelled diagram
correct working (may be on diagram)
(A1)
e.g. m2 + (cos 100)2 = 1 , √1 − m2
cos 100 = −√1 − m2
A1
N2
[3 marks]
METHOD 2
valid approach involving tan identity
sin
cos
e.g. tan =
correct working
e.g. cos 100 =
cos 100 =
(M1)
(A1)
sin 100
tan 100
m
tan 100
A1
N2
[3 marks]
11b. Let sin 100∘ = m . Find an expression for tan 100∘ in terms of m.
[1 mark]
Markscheme
METHOD 1
tan 100 = −
m
√1−m2
(accept
m
)
−√1−m2
A1
N1
[1 mark]
METHOD 2
tan 100 =
m
cos 100
A1
N1
[1 mark]
11c. Let sin 100∘ = m. Find an expression for sin 200∘ in terms of m.
[2 marks]
Markscheme
METHOD 1
valid approach involving double angle formula
(M1)
e.g. sin 2θ = 2 sin θcosθ
sin 200 = −2m√1 − m2 (accept 2m (−√1 − m2 ))
A1
N2
Note: If candidates find cos 100 = √1 − m2 , award full FT in parts (b) and (c), even
though the values may not have appropriate signs for the angles.
[2 marks]
METHOD 2
valid approach involving double angle formula
e.g. sin 2θ = 2 sin θ cos θ , 2m ×
sin 200 =
2m2
(=
tan 100
2m cos 100)
(M1)
m
tan 100
A1
N2
[2 marks]
The diagram below shows part of the graph of
f(x) = a cos(b(x − c)) − 1 , where
a>0.
The point
P ( π4 , 2) is a maximum point and the point
Q ( 34π , −4) is a minimum point.
12a. Find the value of a .
[2 marks]
Markscheme
evidence of valid approach
e.g.
max y value−min y value
2
a=3
A1
(M1)
, distance from y = −1
N2
[2 marks]
12b. (i)
(ii)
Show that the period of f is π .
[4 marks]
Hence, find the value of b .
Markscheme
(i) evidence of valid approach
(M1)
e.g. finding difference in x-coordinates,
evidence of doubling
e.g. 2 ×
π
2
A1
( π2 )
period = π
AG
N0
(ii) evidence of valid approach
e.g. b =
2π
π
b=2
A1
(M1)
N2
[4 marks]
12c. Given that 0 < c < π , write down the value of c .
[1 mark]
Markscheme
c=
π
4
A1
N1
[1 mark]
Let
f(x) = (sin x + cos x)2 .
13a. Show that f(x) can be expressed as 1 + sin 2x .
[2 marks]
Markscheme
attempt to expand
(M1)
e.g. (sin x + cos x)(sin x + cos x) ; at least 3 terms
correct expansion
A1
2
e.g. sin x + 2 sin x cos x + cos2 x
f(x) = 1 + sin 2x
AG
N0
[2 marks]
13b. The graph of f is shown below for 0 ≤ x ≤ 2π .
[2 marks]
Let g(x) = 1 + cos x . On the same set of axes, sketch the graph of g for 0 ≤ x ≤ 2π .
Markscheme
A1A1
N2
Note: Award A1 for correct sinusoidal shape with period 2π and range [0, 2], A1 for
minimum in circle.
The graph of g can be obtained from the graph of f under a horizontal stretch of scale
13c. The graph of g can be obtained from the graph of f under a horizontal stretch of scale [2 marks]
factor p followed by a translation by the vector ( ) .
k
0
Write down the value of p and a possible value of k .
Markscheme
p = 2 , k = − π2
A1A1
N2
[2 marks]
The following diagram shows the graph of
f(x) = a cos(bx) , for
0≤x≤4.
There is a minimum point at P(2, − 3) and a maximum point at Q(4, 3) .
14a. (i)
(ii)
Write down the value of a .
Find the value of b .
[3 marks]
Markscheme
(i) a = 3
A1
N1
(ii) METHOD 1
attempt to find period
e.g. 4 , b = 4 ,
b=
2π
(= π2 )
4
(M1)
2π
b
A1
N2
[3 marks]
METHOD 2
attempt to substitute coordinates
(M1)
e.g. 3 cos(2b) = −3 , 3 cos(4b) = 3
b=
2π
(= π2 )
4
A1
N2
[3 marks]
14b. Write down the gradient of the curve at P.
[1 mark]
Markscheme
0
A1
N1
[1 mark]
14c. Write down the equation of the normal to the curve at P.
[2 marks]
Markscheme
recognizing that normal is perpendicular to tangent
(M1)
e.g. m1 × m2 = −1 , m = − 10 , sketch of vertical line on diagram
x = 2 (do not accept 2 or y = 2 )
A1
N2
[2 marks]
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