Mathematics · May / June 2025 · Paper 31 · Question 8
YMA01
As printed in the paper
The question as text
8. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[3]
tan 2
x
+
tan
x
tan3
x
tan 2
(
x
+
x
)
1
−
tan 2 tan
x
x
or e.g.
sin 2
(
)
sin3
x
x
+
x
sin 2 cos
x
x
+
cos2 sin
x
x
tan 2
x
+
tan
x
tan3
x
cos3
x
cos 2
(
x
+
x
)
cos2 cos
x
x
−
sin 2 sin
x
x
1
−
tan 2 tan
x
x
2tan
x
+
tan
x
tan 2
x
+
tan
x
1
−
tan
2
x
1
−
tan 2 tan
x
x
2tan
x
1
−−
tan
x
1
tan
2
x
2tan
x
+
tan
x
1
tan
2
2tan
x
+
tan
x
−
tan
3
x
−
x
2tan
x
1
−
tan
2
x
−
2tan
2
x
1
−−
tan
x
1
tan
2
x
3tan
x
−
tan
3
x
*
1 3tan
−
2
x
Condone use of a different variable e.g. θ for x.
Condone the use of mixed variables for the M marks as long as the intention is clear.
Condone clear use of “tan” written differently e.g. “tg”
M1: Writes tan 3x as tan (2x + x) which may be implied and applies the addition formula for tan (A + B) to
obtain an expression in terms of tan x and tan2x only.
Condone one sign slip only.
May also reach this point by writing tan3x as sin 3
x
and then uses the addition formulae for
cos3
x
sin (A + B) and cos(A + B) to obtain an expression in terms of tan x and tan2x only.
Condone one sign slip only.
2tan
x
dM1: Replaces
tan 2
x
with 1
to express tan 3x in terms of tan x only.
−
tan
2
x
A1*: Reaches the right hand side with sufficient working shown.
For this mark condone minor notational errors (e.g. a missing x) but not consistent poor notation
throughout such as mixed variables or writing
tan x as
2
tan x
2
(
)
Attempts to use higher multiples of x e.g.
tan3
x
tan 4
x
−
x
should be sent to review.
Way 2
(
)
sin3
x
sin 2
x
+
x
sin 2 cos
x
x
+
cos2 sin
x
x
tan3
x
(
)
cos3
x
cos 2
x
+
x
cos2 cos
x
x
−
sin 2 sin
x
x
(
)
2sin cos
x
2
x
cos
2
x
sin
2
x
sin
x
+
−
(
)
cos
2
x
−
sin
2
x
cos
x
−
2sin
2
x
cos
x
2sin cos
x
2
x
+
cos
2
x
sin
x
−
sin
3
x
3sin cos
x
2
x
−
sin
3
x
cos
3
sin
2
cos
2sin
2
cos
cos
3
3sin
2
cos
x
−
x
x
−
x
x
x
−
x
x
3tan
x
−
tan
3
x
*
1 3tan
−
2
x
Condone use of a different variable e.g. θfor x.
Condone the use of mixed variables for the M marks as long as the intention is clear.
M1: Writes tan3x as sin 3
x
and then uses the addition formulae for sin (A + B) and cos(A + B) condoning
cos3
x
one sign slip and then applies sin2
2sin cos
x
=
x
x
and
cos2
x
±
cos
2
x
±
sin
2
x
or equivalent e.g.
cos2
x
±±
1
2sin
2
x
or
cos2
x
±
2cos
2
x
± to obtain an expression in terms of sin x and cosx only.
1
dM1: Expands and collects terms to obtain 2 terms in the numerator and 2 terms in the denominator. There
must have been no errors in their identities for sin2x or cos2x
A1*: Reaches the right hand side with sufficient working shown.
For this mark condone minor notational errors (e.g. a missing x) but not consistent poor notation
throughout such as mixed variables.
8(b)[5]
3tan
θ
−
tan
3
θ
=
2sec 3
2
θ
−
8
tan3
θ
=
2sec 3
2
θ
−
8
1 3tan
−
2
θ
(
)
2
tan3
θ
=
2 1
+
tan 3
θ
−
8
2tan 3
2
tan3
6
0
or e.g.
2tan 3
2
tan3
6
θ
−
θ
−
=
θ
−
θ
=
3
(
)(
)
2tan3
θ
+
3
tan3
θ
−
2
=
0
tan3
θ
= −
, 2
2
3
tan3
θ
= −
3
θ
=
...
θ
=
...
or tan3
θ
=
2
3
θ
=
...
θ
=
...
2
θ=
0.37, 0.72, 1.42
Allow if a different variable used (such as x).
For mixed variables allow the M’s but only allow the first A
(and final A’s) if recovered or interpreted correctly as the same variable.
tan3
M1: Writes the lhs as
k
θ and applies
sec 3
2
θ
= ±±
1
tan 3
2
θ
on the rhs.
A1:
Reaches the correct 3TQ in tan3θ with terms collected but not necessarily all on one side.
dM1: Solves their 3TQ in tan3θby any method (usual rules) including a calculator (you may need to check)
to obtain at least one value for tan3θand uses correct processing to obtain at least one value for θ e.g.
finds inverse tan and divides by 3.
A1:
Awrt at least one of these answers. Must be in radians and must come from the correct 3TQ.
A1:
Awrt these answers and no extras in range. Must be in radians.
sin3
Note that there may be more convoluted methods which include writing the lhs as
k
cos3
3tan
θ
−
tan
3
θ
sin3
θ
2
=
2sec 3
2
θ
−
8
=
2sec 3
2
θ
−
8
=
−
8
1 3tan
−
2
θ
cos3
θ
cos 3
2
θ
1
cos6
θ
+
1
sin3 cos3
θ
θ
=
2
−
8cos 3
2
θ
sin 6
θ
=
2
−
8
2
2
sin6
θ
+
8cos6
θ
= −
4
4
65sin 6
(
1.4464
)
4
sin 6
(
1.4464
)
...
θ
+
= −
θ
+
= −
θ
=
65
θ=
0.37, 0.72, 1.42
In such cases, the first M mark will only be scored when they reach an equation they can solve directly
(
)
e.g. in the above example when they reach
65sin 6
θ+
1.4464
= −
4
If you are in doubt whether an approach deserves any credit use Review.
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.