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TRIGONOMETRY
1 (a) Write down the values of k for which the line y = k is a tangent to the curve y = 4 sin^ b x^ + 4 rl^ + 10.
[2]
(b) (i) Show that
cos
tan
cos
tan
sin
sin
i^2
i i
i i
i
(ii) Hence solve the equation cos
tan
cos
tan
i
i i
i
= 3, for 0 ° G i G360 .° [4]
2 It is given that tan
i = , for 0 1 i 1 r 2. Find cosec^2 i in the form a + b 3, where a and
b are constants. [5]
y
0 x
The diagram shows the graph of y = a sin bx + c , where a , b and c are integers, for
- 180 ° G x G180. Find the values of° a , b and c. [3]
2 Given that x = sec^2 i and y + 2 = cot^2 i , find y in terms of x. [4]
1
5 Solve the equation^ cot 2 x^ 3
b rl
of r.
3 =0, where - r 1 x 1 rradians. Give your answers in terms
[4]
6 Solve the equation cot^2 b 2 x - r 3 l= 31 ,^ where^ x^ is in radians and 0^ G^ x^^1 r.^ [5]
3 (a) Show that se c 1 x^ - 1 +sec^1 x + 1 = 2 cot x cosec x. [4]
(b) Hence solve the equation se c x^1 - 1 +sec x^1 + 1 = 3 sec x for 0 ° 1 x 1 360 .° [4]
17
y
r
r
(a) The curve has equation y = a cos bx + c where a , b and c are integers. Find the values of a , b
and c. [3]
(b) Another curve has equation y = 2 sin 3 x +4. Write down
(i) the amplitude, [1]
(ii) the period in radians. [1]
5 (a) Show that cose c^1 x - 1 +cosec^1 x + 1 = 2 tan x sec x. [4]
(b) Hence solve the equation cose c^1 x - 1 +cosec^1 x^ + 1 = 5 cosec x for 0 ° 1 x 1 360 .° [4]
1
10 (a) (i) Write 6 x y + 3 y + 4 x +2 in the form( ax + b ) ( c y + d ) , where a , b , c and d are positive integers.
(ii) Hence solve the equation 6 sin i cos i + 3 cos i + 4 sin i + 2 = 0 for 0 ° 1 i 1 360 .° [4]
(b) Solve the equation
1 sec r
2 z^1
b + l= for - r 1 z 1 r, where z is in radians. Give your
answers in terms of r. [5]
[1]
The diagram shows the trapezium ABCD , where AB = 5 3 - 3, DC = 5 3 - 1 and AD = 3 + 2.
(a) Find the area of ABCD , giving your answer in the form a + b 3, where a and b are integers. [3]
(b) Given that angle BCD = i radians, find the value of cot i in the form c + d
integers.
3 , where c and d are
[3]
(c) Using your answer to part (b) , find the value of cosec 2 i in the form e + f 3, where e and f are
integers. [2]
14 The graph of y = a + 2 tan bx , where a and b are constants, passes through the point (0, - 4) and has
period 480°.
(a) Find the value of a and of b. [3]
(b) Sketch the graph of y for values of x between 0° and 480°. [2]
5 The function f is defined, for 0 ° G x G810 , by° f ( ) x =- 2 + cos^23^ x^.
(a) Write down the amplitude of f. [1]
[2]
(c) On the axes, sketch the graph of y =f ( ). x [2]
0° 90° 180° 270° 360° 450° 540° 630° 720° 810° x
y
(b) Find the period of f.
16 (a) (i) Show that sin
cos
x sin
x
1 2 x
2
+ =^ -^.^ [2]
(ii) Hence solve
sin
cos
x
x
= 1 for 0 ° G x G 9 0 .° [4]
(b) Solve cote y
o 3 for 0^ G^ y G^ r^ radians.^ [3]
17 (a) Write down the amplitude of 2 cos 3^ x^ - 1. [1]
(b) Write down the period of 2 cos
x - 1. [1]
(c) Sketch the graph of y = 2 cos -
x 1 for - r G x G 3 r radians. [3]
18 (a) Given that 2 cos x = 3 tan x , show that 2 si n 2 x + 3 sin x - 2 = 0. [3]
(b) Hence solve 2 cos^ b 2 a + r 4 l^ = 3 tanb 2 a + 4 rl^ for 0 1 a 1 r radians, giving your answers in
terms of r. [4]
19 (a) Show that sin 1 -^ x cos^^ tan xx^ = 1 + sec x. [4]
(b) Solve the equation 5 tan x - 3 cot x = 2 sec x for 0 ° G x G360 .° [6]
[1]
(b) Sketch the graph of y = 2 cos -
x 1 f
or
- 360 ° G x G360 .° (^) [3]
20 (a) Write down the period of 2 cos 3^ x^ - 1.
10 (a) (i) Show that 1
se c i - 1 - sec 1 cot
i + =^ i.^ [3]
(ii) Hence solve
se - 1
c 2 x sec 2 x
= 6 for - 90° 1 x 1 90°. [5]
(b) Solve cosec b y + r 3 l= 2 for 0 G y G 2 r radians, giving your answers in terms of r. [4]
22 (a) Solve tan 3 x =- 1 for - r 2 G x G r 2 radians, giving your answers in terms of r. [4]
(b) - r^ x r
Use your answers to part (a) to sketch the graph of y = 4 tan 3 x + 4 for^ G^ G^ radians
Show the coordinates of the points where the curve meets the axes. [3]
31 Solve (a) 6 si n 2 x - 13 cos x = 1 for 0 ° G x G360 .° [4]
(b) (i) Show that, for - r 2 1 y 1 r 2 , tan
tan y
y 1
+^2
integer.
can be written in the form si a n y , where a is an [3]
(ii) Hence solve tan
tan y
y 1
2 +^3
= for - r 2 1 y 1 r 2 radians. [1]
32 Solve 1 + 2 sin ( x + 50 °) = 0 for - 180 ° G x G 180 °. [4]
s in x
8 (i) Show that 1 tan x sec x 1 - sin x 1
=. [4]
(ii) Hence solve the equation 1 - sin x
in
s x
(^1) cosec x 1
for 0 ° G x G360 .° [4]
1
34 (a) Solve 2 sin b x + r 4 l= 3 for 0 1 x 1 r radians. [3]
(b) Solve 3 se c y = 4 cosec y for 0 ° 1 y 1 360 .° [3]
r
35 The curve (i) y = a + b sin cx has an amplitude of 4 and a period of 3. Given that the curve passes
through the point
r
J
L
KK
N
P
OO, find the value of each of the constants a , b and c. [4]
(ii) Using your values of a , b and c , sketch the graph of y = a + b sin cx for 0 G x G r radians. [3]
36 It is given that y = 1 +tan 3 x.
(i) State the period of y. [1]
(ii) Sketch the graph of y = 1 +tan 3 x for 0 ° G x °G 180 °. [3]
37 (a) Solve 3 co s 2 i + 4 sin i = 4 for 0 ° G i G 180 °. [4]
(b) Solve si n 2 z = 3 cos 2 z for - 2 r G z G 2 r^ radians. [4]
38 (a) Solve 10 co s 2 x + 3 sin x = 9 for 0 ° 1 x 1360 °. [5]
(b) Solve 3 tan 2 y = 4 sin 2 y for 0 1 y 1 r radians. [5]
39 (i) Show that cos i cot i + sin i =cosec i. [3]
(ii) Hence solve cos i cot i + sin i = 4 for 0 ° G i G 90 °. [2]
40 The graph of y = a sin( bx ) + c has an amplitude of 4, a period of r
and passes through the point
r
J
L
KK
N
P
OO. Find the value of each of the constants a , b and c. [4]
2 Use of cosec^2 = 1 + cot 2 B1 May be implied by later work
2
2
cosec 2 1
M1 For attempting to deal with tan correctly, forming a single fraction and simplifying, with sufficient detail – at least 4 terms in the numerator
cosec 2
A
cosec 2
M1 For attempt to rationalise their expression, with sufficient detail in the simplification of the numerator – at least 3 terms
A
Alternative 1
Use of cosec^2 = 1 + cot 2
(B1)
May be implied by later work
cot 3 1
(2) M1 for attempting to rationalise cot or
tan with sufficient detail in the simplification of the numerator – at least 3 terms
2
2
cosec 2 1
(2) M1 for expressing as a single fraction and attempt to simplify to required form.
2
Alternative 2
Use of cosec^2 = 1 + cot^2
and cot 2
tan
(B1)
May be implied by later work
tan 2
(2) M1 for attempting to rationalise tan^2
with sufficient detail in the simplification of the numerator – at least 3 terms
2 1
cosec 2 1
(2) M1 for attempting to rationalise cot^2
with sufficient detail in the simplification of the numerator – at least 3 terms and expressing as a single fraction and attempt to simplify to required form.
Alternative 3
Use of right-angled triangle Hyp 2 = 11 + 2 3
M1 For attempt to calculate the square of hypotenuse
cosec 2
(B1) for correct use of cosec^2 with their
squared hypotenuse
cosec 2
(M1) For attempt to rationalise their expression, with sufficient detail in the simplification of the numerator – at least 3 terms
A
3 a 5 B
b (^4) B
c 3 B
(^6) tan π
(^2 ) 3
− =±
x soi
or sin 2 ±
x soi
B1 B0 if negative root is rejected Allow truncated decimals May be implied by subsequent work From use of cosec 2 2 π^1 cot 2 2 π 3 3
(^) − − = −^
x ^ x
π π 2π ,
4π , 3 3 3 3 3
x − =
π −
2 x = 0 , 2π, π, 5π 3 3
0, ,
π π ,
5π 3 2 6
x =
or 0, 1.05, 1.57, 2.62 or greater accuracy
4 M1 for correct order of operations to obtain one solution in the range using − π^ = (^3)
tan 2
x k
, 3
− π^ ^ = < or (^) sin (^) 2 m x^ m^^1 Dep M1 for correct order of operations to obtain a second solution in the range using
2 3
(^) −
π = x^ ^ tan −^1 ( k ) ±π
or (^2) 3
(^) = π − < − π ^ x^ sin^
− (^1) ( m (^) ), m 1 oe
2 π sin −^1 ( m^ ),^1 3
(^) = − < −^
^ x^ m^ oe A1 for any pair of correct solutions A1 for remaining pair of solutions, with no extra solutions within the range
7(a) cos cos
1 − cos x 1 + cos
x x x
s ec x + 1 + sec x − 1 x −
M
2 2
cos x + cos^2 x + cos x − cos 1 −cos
x x
or (^2)
2sec tan
x x
A
2
2cos sin
x x
or
2 2
2cos cos x sin
x x
oe
A
Fully correct justification of given answer: 2cot x cosec x A
7(b) (^) 3tan 2 x = 2 oe or better, soi
or 5cos 2 x = 3 oe or better, soi or 5sin 2 x = 2 oe or better, soi
B
tan (^) [ ]
x = ± oe
or cos (^) [ ]
x = ±
or sin (^) [ ]
x = ±
or [±] 0.816[4…]
oe or [±] 0.774[5…]
oe or [±] 0.632[4…]
M1 FT an equation of the form a tan 2 x = b , a > 0, b > 0 or p sin 2 x = q or p cos 2 x = q where p > 0, q > 0 and p > q
or 39.2315… rot to 2 or more dp or 140.7684… rot to 2 or more dp or 219.2315… rot to 2 or more dp or 320.7684… rot to 2 or more dp
A2 no extras in range
A1 for any two correct answers
8(a) a = 3 B
b = 2 B
c = − 1 B
8(b)(i) 2 B
8(b)(ii) (^) 2π
3
oe or 2.09 or 2.094[395…] rot to 4 or more sf
B
9(a) sin sin 1 −sin
1 +sin
x x x x
or cosec^ x^ +^1 +^ cosec^ x^ −^1 cosec 2 x − 1
oe
M
2 2 2
sin x + sin + sin sin 1 −sin
x x − x x
or (^2)
2cosec cot
x x
oe
A
2
2sin cos
x x
or
2 2
2sin sin x cos
x x
oe
A
Fully correct justification of given answer: 2sin x ×
cos x cos x
= 2 tan x sec x
1 or 2 tan^ cos
x × x
or
2sin cos
x × sec x x
= 2 tan x sec x
= 2 tan x sec x
or equivalent
A
9(b) (^) 2 tan 2 x = 5 or better, soi or 7 cos 2 x = 2 or better, soi or (^) 7 sin 2 x = 5 or better, soi
B
tan (^) [ ]
x = ± (^) oe or [±] 1.58[1…]
or cos (^) [ ]
x = ±
or sin (^) [ ]
x = ±
oe or [±] 0.534[5…]
oe ±or] 0.845[1…] [
M1M1 an equation of the form a tan 2 x = b a > 0, b > 0 or p sin 2 x = q or p cos 2 x = q where p > 0, q > 0 and p > q
57.7 or 57.6884… rot to 2 or more dp
237.7 or 237.6884… rot to 2 or more dp
122.3 or 122.3115… rot to 2 or more dp
302.3 or 302.3115… rot to 2 or more dp
A2 no extras in range
A1 for any two correct answers
11(b)(i) ( )
cos 2 θ+ 1 − 2sinθ + sin^2
cos θ 1 − sin
M1 For dealing with the fractions correctly and expansion of (^) ( 1 −sin θ)^2
( )
1 + 1 − 2sin
cos θ 1 − sin
or better
M1M1 D for use of identity, may be implied
by
2 1( − sin θ) cos θ (^) ( 1 − sin θ)
( ) ( )
2 1 − sin
cos θ 1 − sin
M1M1 De on previous M mark for simplification
= 2sec
cos θ
A1 Need to see this detail for A
Need to have had θ in every
trigonometric ratio.
Alternative 1
cos ×
θ 1 + sin θ θ
1 − sin θ θ cos θ
1 −sin
1 +sin
(M1)
( ) 2
cos θ 1 1 −sin
cos cos
+ sin θ θ
(M1)(M1) for use of identity
1 + sin cos
θ 1 −sinθ
θ cos θ
(M1)(M1) on previous M mark for simplification
= 2sec
cos θ
(A1) Need to see this detail for A
Need to have had θ in every
trigonometric ratio.
Alternative 2
( ) ( ( )
1 − sin^2 θ) cos θ 1 −sin θ
θ + 1 −sin
(M1)
For dealing with the fractions and using cos 2 θ= 1 − sin^2 θ.
( )( ) +^ ( ) ( )
1 − sin^2
cos θ 1 − sin
1 − sinθ 1 +sin θ θ
(M1) (^) Dep for factorising 1 −sin 2 θ
cos
1 + sinθ + 1 −sinθ
(M1)(M1) for simplification
= 2sec
cos θ
(A1) Need to see this detail for A
Need to have had θ in every
trigonometric ratio.
11(b)(ii)
cos3 φ =
B
φ = 20 , 100 ,o^ o^140 o^33 M for one correct solution of^ their
cos3 φ = k using a correct order of
operations A1 for 2 correct solutions A1 for a third correct solution with no extra solutions in the range
12(a) 33 B1 for shape, must have implied symmetry about x = 180 o^ , 1 complete maximum point above the x -axis, 2 minimum points below the x -axis, starting and finishing at the same positive y value. 5 and – 7 not necessary for this mark.
B1 for –7⩽ y ⩽5, may be implied by numbers in a table if not seen on the graph or by coordinates. Must have maximum point(s) at y = 5 only and minimum point(s) at y = − 7 only.
B1 for a completely correct curve with maximum points implied at the end points.
12(b) a = 0 b = 4 c = 1
22 B1 for 2 correct.
Alternative a = 0 b = − 4 c = − 1
(2)(2) for 2 correct.
13(a) )( )
2 (^5 3 −^3 +^5 3 −^1 3 +^2
B1 For the area of complete trapezium
= (^) ( 5 3 − (^2) )( 3 + (^2) ) oe
= 15 − 2 3 + 10 3 − 4 oe = 11 + 8 3
2 M1 for attempt to simplify showing expansion of the brackets with at least 3 correct terms
Alternative 1 × 2 ×( 3 + (^2) )+ (^) ( 5 3 − (^3) ) (^) ( 3 + (^2) ) 2
(B1) For the area of a triangle and a rectangle
= 3 + 2 + (^) ( 15 − 3 3 + 10 3 − (^6) )
= 11 + 8 3
(2) M1 for attempt to simplify showing expansion of the brackets with at least 3 correct terms.
13(b) (^2) cot 2 + 3
B
Allow
tan θ
×
M1 For attempt to rationalise their cot θ.
4 − 2 3 A
13(c) (^) cosec 2 θ^ = cot^2 θ+ 1
( )
2 = 4 − 2 3 + 1
= 29 − 16 3
22 M for use of identity with their result from (b) ,
16(a)(i) Uses correct Pythagorean identity in the left-hand side of the given identity,
e.g.
1 sin 2 2 1 +sin 2
x x
M
( 1 − sin 2 x )( 1 +sin 2 x ) 1 + sin 2 x
oe
and completion to given answer
A
16(a)(ii) (^2) sin 2 3
x =
M
x = sin−^1
soi
M1 dep on first M
20.9 or 20.905… rounded or truncated to 4 or more figures and 69.1 or 69.094… rounded or truncated to 4 or more figures
A2 with no incorrect values in range
A1 for either angle correct, ignoring extra values
16(b) (^) π 1 tan (^2 )
(^) y − =
soi
M
y = +
π π 6 2
M1 dep on first M
π 3
oe
or 2.09 or 2.094[39…] rot to 4 or more significant figs
A1 with no incorrect values in range
17(a) 2 B
17(b) (^) 6π or 1080 o^ B
17(c) (^3) B1 for passing through ( −π, 0) and
(3π, − ) 3 – must be a curve B1 for correct shape with max on y -axis and a min at x = 3π B1 for passing through ( 0, 1) and ( π, 0) only on the positive x -axis
18(a) (^) sin 2cos x = 3 cos
x x
2cos 2 x = 3sin x
M1 (^) x and x
For use of tan x =
sin cos multiplying by cos x
2 1( − sin 2 x )= 3sin x M1^ For use of correct identity
2sin 2 x + 3sin x − 2 = 0 A1^ For correct rearrangement to obtain the given answer
Alternative 2sin 2 x + 3sin x − 2 = 2 1 ( − cos 2 x (^) ) + 3sin x − 2
(M1) For use of correct identity
= −2cos x cos x + 3sin x = −3 tan x cos x + 3sin x
(M1) For use of 2 cos x = 3 tan x
−3sin x + 3sin x = 0 (A1) For use of tan x cos x = sin x and answer 0
18(b)
sin 2 α
π +^ =^
4 ^2
only
B1 For solution of quadratic from (a) to
obtain sin 2 α
π +^ =^
only
π π 2 ,
5π 13 ,
π 4 6 6 6
7π 2 ,
23π 12 12
M1 For correct order of operations in
attempt to solve sin 2 α
π +^ =^
4 ^2
may be implied by one correct solution
7π 24
A
23π 24
A
19(a) sin x ×
sin
LHS cos 1 cos
x x x
M
Uses tan x^ =^
sin cos
x x
( )
1 −co s^2 cos x 1 − cos
x x
M1 (^) Dep Uses sin 2 x = 1 − cos^2 x to eliminate sin x
( )( ) ( )
1 − cos x 1 +cos cos x 1 −cos
x x cos x
1 +co s x = sec x + 1
2 M1Dep Factorise correctly and cancel correctly.
A1 Uses
= sec cos
x x
