June 2023 Paper 3 Q10
10
(a) You are given that \((x^2 + y^2)^3 = x^6 + 3x^4y^2 + 3x^2y^4 + y^6\).
Hence, or otherwise, prove that \(\sin^6\theta + \cos^6\theta = 1 - \frac{3}{4}\sin^2 2\theta\) for all values of \(\theta\). [4]
Hence, or otherwise, prove that \(\sin^6\theta + \cos^6\theta = 1 - \frac{3}{4}\sin^2 2\theta\) for all values of \(\theta\). [4]
(b) Use the result from part (a) to determine the minimum value of \(\sin^6\theta + \cos^6\theta\). [2]
| Scheme | Marks | AO |
|---|---|---|
| \(1 - \dfrac{3}{4}\sin^2 2\theta = 1 - \dfrac{3}{4}(2\sin\theta\cos\theta)^2\) \(\qquad = 1 - 3\sin^2\theta\cos^2\theta\) | M1 | 3.1a |
| \(\left(\sin^2\theta + \cos^2\theta\right)^3 = \sin^6\theta + \cos^6\theta + 3\sin^4\theta\cos^2\theta + 3\sin^2\theta\cos^4\theta\) | M1 | 3.1a |
| \(\sin^6\theta + \cos^6\theta = 1 - \left(3\sin^4\theta\cos^2\theta + 3\sin^2\theta\cos^4\theta\right)\) | M1 | 2.2a |
| \(\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta\left(\sin^2\theta + \cos^2\theta\right)\) \(\qquad = 1 - 3\sin^2\theta\cos^2\theta\) So LHS = RHS as required | E1 | 2.1 |
| [4] |
Notes
M1: Use of double angle formula
Allow 1 error
M1: Use of given result with sin and cos
Both sides seen but might not be equated
M1: Use of \(\sin^2\theta + \cos^2\theta = 1\)
E1: Convincing completion
| Scheme | Marks | AO |
|---|---|---|
| \(1 - \frac{3}{4}\sin^2 2\theta\) has min value when \(\sin^2 2\theta = 1\) oe | M1 | 1.1 |
| Min value is \(\dfrac{1}{4}\) | A1 | 2.2a |
| [2] |
Notes
A1: \(\dfrac{1}{4}\) unsupported does not score