Question 7:
--- 7(a) ---
7(a) | (2x−1)( 4x2 +2x−1 ) =8x3+4x2 −2x−4x2 −2x+1 =8x3−4x+1 | B1 | AG
Six correct terms leading to the correct answer.
1
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | sin2
+1
cos2  sin2+ cos2
Starting with the LHS = 
sin2 sin2−cos2

−1
cos2 | *M1 | sin
For use of tan= in the numerator and denominator.
cos
1
= need to see clear evidence of this step
1−cos2−cos2 | DM1 | For use of sin2+cos2=1 twice, in a correct expression,
resulting in an expression in cos2.
1
=
1−2cos2 | A1 | AG
Alternative method 1 for Question 7(b)
sin2+ cos2  sin2+ cos2
Starting with the RHS = 
sin2+cos2−2cos2 sin2−cos2
 | *M1 | For use of sin2+cos2=1 twice.
sin2
+1
cos2
= need to see clear evidence of this step
sin2
−1
cos2 | DM1 | Dividing throughout by cos2.
tan2+1
=
tan2−1 | A1 | AG
Question | Answer | Marks | Guidance
7(b) | Alternative method 2 for Question 7(b)
sec2
Starting with the LHS
sec2−2 | *M1 | For use of 1+ tan2=sec2 twice.
1
Clear statement 
1−2cos2 | DM1 | AG For multiplying throughout by cos2 to give the RHS.
A1
3
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | 1 =4cos leading to 1 = 4cos ( 1−2cos2 )
1−2cos2
8cos3−4cos+1 =0
  | B1 | Replace LHS with RHS from (b) and clear fractions.
(2cos−1)( 4cos2+2cos−1 )=0  | *B1 | Use of the expression from (a) with x=cos.
1 −2 4+16
x or cos= and OR 0.31,−0.81 AWRT
2 8 | DB1 | OE
For all three values.
= 60,72,144 | B2,1,0 | B2 for three correct answers only, B1 for two correct answers
and no others (but allow 36 instead of 144) in the given
range or 3 correct answers plus other values in the given
range. Ignore answers outside of the given range.
Accept AWRT 72.0, 144.0 .
SC B1 for all 3 correct answers in radians and no others:
π 2π 4π
, and .
3 5 5
5
Question | Answer | Marks | Guidance