Mathematics · May / June 2025 · Paper 41 · Question 10

cos2

x

Assume there is an angle x, 90

°<

x

<

180

°,

1

cos

x

−

sin

x

(

) (

)

+

−

cos

x

sin

x

cos

x

sin

x

cos2

x

cos

x

−

sin

x

−

cos

x

sin

x

+

(

)

2

cos

x

sin

x

1

cos

x

+

sin

x

1

+

1

2sin

x

cos

x

1

sin 2

x

0

This is a contradiction, as sin α is negative for all angles 180

°<α<

° ,

360

hence true *

Condone strict inequalities for the first 3 marks in this question

B1:

Sets up the argument.

It requires

°<

x

<

180

°’

cos2

x

cos2

x

x >

1

but condone

1

−

−

cos

x

sin

x

cos

x

sin

cos2

x

to (

cos

x

+

sin

x

M1:

Uses the double angle formula for cos 2x and simplifies

−

cos

x

sin

x

It may be scored when prematurely removing the modulus before proceeding to (

cos

dM1: Squares (

)

cos

x

+

sin

x

and uses suitable identities in proceeding to sin2

1

x

0,

condoning sin2

It must be in a form in which the contradiction can easily be proven

A1:

Deduces contradiction from opening statement and concludes. May consider the signs of

sin x and cos x in the given range and deduce that the product is negative leading to the contradiction of the

original assumption. Withold this mark if the modulus signs are not correctly dealt with. This mark is

withheld for using strict inequalities

There will be many alternative approaches so you will need to look at what they have done to see if there is any

credit worthy work. If in doubt please use review.

Alternative using Ralpha method

B1 M1:

As above

(

)

dM1: Proceeds to the form

2 cos

x −

45

°

1

with the modulus seen

(

)

Alternatively, to the form

2 sin

x +

45

°

1

with the modulus seen

(

)

(

)

(

)

∈

°

°

−

°∈

°

°

A1*: Considers

x

90 ,180

x

45

45 ,135

1

(

)

(

)

Deduces that

cos

x −

45

°<

which means that

2 cos

x −

45

°

<

1

so meaning that the given

2

statement is true.

Alternatively

(

)

(

)

(

)

∈

°

°

+

°∈

°

°and deduces that

(

x +

°

)

Considers

x

90 ,180

x

45

135 ,225

sin

45

(

)

that

2 sin

x +

45

°

<so meaning that the given statement is true

1

Alternative that proceeds via an identity in cos2x and sin2x

cos2

x

M1: Squares

1

and uses suitable identities leading to an inequality in sin2x and cos2x

cos

x

−

sin

x

(

)

2

−

2

2

−

cos 2

x

cos

x

sin

x

cos 2

x

1

sin 2

x

−

2

−

2

dM1: Then uses Pythagoras identity

1 sin 2

x

1 sin2

x

sin2

x

sin 2

x

A1*: As 90

°<

x

<

180

°,

180

°<

2

x

<

360

°and sin2

x <

0

sin2

x

1

which is impossible and therefore you

have a contradiction and the original statement is true

Alternative using logic for the last two marks

dM1 Attempts to explain the contradiction by referencing the domain 90

°<

x

<

180

°and corresponding range of

values of (say) sin x, cos x. Must involve the modulus in their explanation or have squared previously.

°<

<

°", followed by

Stating that "0 < sin x < 1 and −1 < cos x < 0 for 90

x

180

" so sin

x

+

cos

x

<" would be a minimal but acceptable argument for dM1.

1

Then the A1* is given for a completely rigorous explanation (plus conclusion)

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