Mathematics · Oct / Nov 2024 · Paper 31 · Question 5

sin3

sin 2

(

)

sin2 cos

cos2 sin

x

=

x

+

x

=

x

x

+

x

x

(

)

(

)

=

2sin cos

x

x

cos

x

+

1 2sin

−

2

x

sin

x

2

=

2sin cos

+

sin

−

2sin

3

x

x

x

x

(

)

=

2sin

1 sin

−

2

+

sin

−

2sin

3

=

3sin

−

4sin

3

x

x

x

x

x

x

(

)

3

2sin3

θ

=

5sin 2

θ

2 3sin

θ

−

4sin

θ

=

10sin

θ

cos

θ

Divides or takes out sinθas a factor and uses

sin

2

=1 cos

2

θ to set up

θ

−

and solve a 3TQ in cosθ

E.g.

6sin

8sin

3

10sin

cos

6

(1 cos

2

10cos

θ

θ

−

θ

=

θ

θ

−

−

θ)

=

2

4cos

θ−

5cos

θ−

1= 0

5

−

41

(

)

cos

θ

=

= −

0.175...

8

Any two of the following four answers

sin

0

180 , 360

θ

=

θ

=

°

°

5

−

41

cos

=

θ

=

awrt 100 or awrt 260

°

°

θ

8

All of 180 , 360 , awrt 100.1 , awrt 259.9

°

°

°

°

(a)

(

)

(

M1: Attempts

sin3

=

sin 2

+

and achieves sin2 cos

±

x

x

±

cos2 sin

x

x

. The

=

sin 2x

x

x

x

implied

dM1: Dependent upon the previous mark. It is for using sin2

=

2sin cos

and

cos2

x

x

x

x

or (

)

2

2

2

to form an expression in just sinx and cosx.

±±

1

2cos x

±

cos

x

−

sin

x

A1: Fully correct AND expanded expression in just sinx and cosx

Examples of correct expanded expressions are;

2

3

2sin cos

x

x

+

sin

x

−

2sin

x

2

2

2sin cos

+

2sin cos

−

sin

x

x

x

x

x

2

2

3

2sin cos

x

x

+

sin cos

x

x

−

sin

x

2

or

4sin cos

x

x

−

sin

x

Condone a missing x on a sin x, for example. Condone a slip in notation, e.g. writing

sin x as long as it is subsequently corrected.

3

A1: Proceeds correctly to

3sin

−

4sin

3

using

cos

2

= −

1 sin

2

x

x

x

x

Penalise notational and bracketing errors on this mark only

2

2

E.g. Writing

cos x for

cos x , mixing variables, missing x on a sin x etc

(b) Allow if solved in terms of another variable, say x

The question states hence or otherwise solve 2sin3

θ

=

5sin2

θ

In almost all cases they will use their part (a) and proceed by the ‘hence’ route.

‘Otherwise’ routes could start with earlier versions of 2sin3θsuch as (

2

2 2sin cos

x

x

+

In almost all cases the main mark scheme provides a framework of how the marks are awarded

M1: Uses their part (a) and a double angle for sin2θ of the form

G

sin cos

θ

θto form an equation in sinθ

and cosθ. Look for

sin

θ

+

sin

3

θ

=

sin cos

θ

θ o.e. Condone slips such as forgetting to 2

R

S

T

2

dM1: The mark is scored for dividing or taking out a factor of sinθand using

sin

θ

=

±±

and solve a 3TQ in cosθ. To solve you should expect to see a valid method used, leading to at least

one value forcos

θ

= ,

k

−<

1

k

<

1,

k

≠

0

If a calculator is used to solve the 3TQ in cosθ the value(s) must be correct for their equation.

Candidates cannot go straight from an equation e.g.

4cos

2

θ−

5cos

θ−

1= 0

to a solution e.g. 100.1°

without sight of the value for cosθin between.

A1: Finds two angles of 180 , 360 , awrt 100 , awrt 260

°

°

°

°following at least one previous M1

°must be found via a correct method with sin

= 0

The 180 and/or 360

°

θ

being seen or implied and also

produced from a factorised

P

sin

θ

+

Q

sin

3

θ

=

R

sin cos

θ

θ

Likewise, the 100 and 260

°

°must have been achieved via solving a correct equation.

Allow this mark if the answers are given in radians. So, for two angles of , 2 , awrt 1.75, awrt 4.54

π

π

A1: All of 180 , 360 , awrt 100.1 , awrt 259.9

°

°

°

° with no extra values within the range following M1, dM1

There are no marks for answers appearing without valid method and working

There are other (usually more complicated) ways of attempting part (b) so you must look carefully through

each response. If you think a response deserves credit and you are unsure of how to mark, please send to

review.