Standard +0.8 This question requires expanding two squared binomials, collecting like terms using sin²θ + cos²θ = 1, forming a linear equation in sin θ and cos θ, then using the Pythagorean identity again to solve. While the algebraic manipulation is substantial for 6 marks, the techniques are all standard A-level content with no novel insight required. The 'obtuse' condition adds a minor selection step at the end. Slightly above average due to the multi-step algebraic complexity and need to recognize the identity application twice.
It is given that
$$(2 \sin \theta + 3 \cos \theta)^2 + (6 \sin \theta - \cos \theta)^2 = 30$$
and that \(\theta\) is obtuse.
Find the exact value of \(\sin \theta\).
Fully justify your answer.
[6 marks]
Question 6:
6 | Expands brackets, at least one
completely correct | 1.1a | M1 | 4sin2θ+12sinθcosθ+9cos2θ
+36sin2θ−12sinθcosθ+cos2θ=30
40sin2θ+10cos2θ=30
( )
40sin2θ+10 1−sin2θ =30
30sin2θ+10=30
2
sin2θ=
3
6
sinθ=±
3
6
since θ is obtuse sinθ=
3
Expands both brackets correctly | 1.1b | A1
sin2θ+cos2θ=1
Uses
cos2θ
correctly to eliminate or
sin2θ
from their equation
30sin2θ+10=30
PI by | 3.1a | M1
Obtains an equation of the form
sin2θ=k cos2θ=k
or
where 0 < k < 1
PI by sinθ= k or
sinθ=− k | 1.1a | M1
Obtains sinθ=± | 2
3 | 1.1b | A1
Completes reasoned argument
2
to obtain sinθ= and
3
2
explains why sinθ=
3
2
Must come from sinθ=±
3
sin2θ=k
FT their
where 0 < k < 1 | 2.4 | R1F
Question 6 Total | 6
Q | Marking instructions | AO | Marks | Typical solution
It is given that
$$(2 \sin \theta + 3 \cos \theta)^2 + (6 \sin \theta - \cos \theta)^2 = 30$$
and that $\theta$ is obtuse.
Find the exact value of $\sin \theta$.
Fully justify your answer.
[6 marks]
\hfill \mbox{\textit{AQA Paper 2 2024 Q6 [6]}}