| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove algebraic trigonometric identity |
| Difficulty | Standard +0.8 This is a structured proof requiring algebraic manipulation of the given binomial expansion, substitution using sin²θ + cos²θ = 1, and the double angle formula sin 2θ = 2sinθcosθ. Part (b) adds optimization using the range of sin 2θ. While the 'hence' scaffolds the approach, it still requires multiple non-trivial steps and careful algebraic manipulation beyond routine identity work, placing it moderately above average difficulty. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - \dfrac{3}{4}\sin^2 2\theta = 1 - \dfrac{3}{4}(2\sin\theta\cos\theta)^2 = 1 - 3\sin^2\theta\cos^2\theta\) | M1 | AO 3.1a; Use of double angle formula; allow 1 error |
| \((\sin^2\theta + \cos^2\theta)^3 = \sin^6\theta + \cos^6\theta + 3\sin^4\theta\cos^2\theta + 3\sin^2\theta\cos^4\theta\) | M1 | AO 3.1a; Use of given result with sin and cos; both sides seen but might not be equated |
| \(\sin^6\theta + \cos^6\theta = 1 - (3\sin^4\theta\cos^2\theta + 3\sin^2\theta\cos^4\theta)\) | M1 | AO 2.2a; Use of \(\sin^2\theta + \cos^2\theta = 1\) |
| \(\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta(\sin^2\theta + \cos^2\theta) = 1 - 3\sin^2\theta\cos^2\theta\); So LHS = RHS as required | E1 [4] | AO 2.1; Convincing completion |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - \dfrac{3}{4}\sin^2 2\theta\) has min value when \(\sin^2 2\theta = 1\) oe | M1 | AO 1.1 |
| Min value is \(\dfrac{1}{4}\) | A1 [2] | AO 2.2a; \(\dfrac{1}{4}\) unsupported does not score |
## Question 10(a):
$1 - \dfrac{3}{4}\sin^2 2\theta = 1 - \dfrac{3}{4}(2\sin\theta\cos\theta)^2 = 1 - 3\sin^2\theta\cos^2\theta$ | M1 | AO 3.1a; Use of double angle formula; allow 1 error
$(\sin^2\theta + \cos^2\theta)^3 = \sin^6\theta + \cos^6\theta + 3\sin^4\theta\cos^2\theta + 3\sin^2\theta\cos^4\theta$ | M1 | AO 3.1a; Use of given result with sin and cos; both sides seen but might not be equated
$\sin^6\theta + \cos^6\theta = 1 - (3\sin^4\theta\cos^2\theta + 3\sin^2\theta\cos^4\theta)$ | M1 | AO 2.2a; Use of $\sin^2\theta + \cos^2\theta = 1$
$\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta(\sin^2\theta + \cos^2\theta) = 1 - 3\sin^2\theta\cos^2\theta$; So LHS = RHS as required | E1 [4] | AO 2.1; Convincing completion
## Question 10(b):
$1 - \dfrac{3}{4}\sin^2 2\theta$ has min value when $\sin^2 2\theta = 1$ oe | M1 | AO 1.1
Min value is $\dfrac{1}{4}$ | A1 [2] | AO 2.2a; $\dfrac{1}{4}$ unsupported does not score10
\begin{enumerate}[label=(\alph*)]
\item You are given that $\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 } = x ^ { 6 } + 3 x ^ { 4 } y ^ { 2 } + 3 x ^ { 2 } y ^ { 4 } + y ^ { 6 }$.\\
Hence, or otherwise, prove that $\sin ^ { 6 } \theta + \cos ^ { 6 } \theta = 1 - \frac { 3 } { 4 } \sin ^ { 2 } 2 \theta$ for all values of $\theta$.
\item Use the result from part (a) to determine the minimum value of $\sin ^ { 6 } \theta + \cos ^ { 6 } \theta$.
The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2023 Q10 [6]}}