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Date May 2021 Marks available 2 Reference code 21M.1.SL.TZ1.6
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 1
Command term Show that Question number 6 Adapted from N/A

Question

Show that sin2x+cos2x-1=2sinxcosx-sinx.

[2]
a.

Hence or otherwise, solve sin2x+cos2x-1+cosx-sinx=0 for 0<x<2π.

 

[6]
b.

Markscheme

Note: Do not award the final A1 for proofs which work from both sides to find a common expression other than 2sinxcosx-2sin2x.

 

METHOD 1 (LHS to RHS)

attempt to use double angle formula for sin2x or cos2x          M1

LHS =2sinxcosx+cos2x-1  OR

sin2x+1-2sin2x-1  OR

2sinxcosx+1-2sin2x-1

=2sinxcosx-2sin2x          A1

sin2x+cos2x-1=2sinxcosx-sinx= RHS          AG

 

METHOD 2 (RHS to LHS)

RHS =2sinxcosx-2sin2x

       attempt to use double angle formula for sin2x or cos2x          M1

       =sin2x+1-2sin2x-1          A1

       =sin2x+cos2x-1= LHS          AG

 

[2 marks]

a.

attempt to factorise          M1

cosx-sinx2sinx+1=0          A1

recognition of cosx=sinxsinxcosx=tanx=1  OR  sinx=-12        (M1)

one correct reference angle seen anywhere, accept degrees        (A1)

π4  OR  π6 (accept -π6,7π6)

 

Note: This (M1)(A1) is independent of the previous M1A1.

 

x=7π6,11π6,π4,5π4          A2

 

Note: Award A1 for any two correct (radian) answers.
Award A1A0 if additional values given with the four correct (radian) answers.
Award A1A0 for four correct answers given in degrees.

 

[6 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.6—Pythagorean identity, double angles
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