Standard +0.3 Part (a) is a standard bookwork proof using addition and double angle formulae that appears in most textbooks. Part (b) requires substituting the identity, converting cos²θ using Pythagoras, and solving a cubic in sin θ, but the approach is straightforward once the identity is established. This is slightly easier than average as it's a well-practiced question type with clear signposting.
Using angle addition and double angle formulae: M1 Attempts to use identity for \(\sin(2A + A) = \sin 2A \cos A \pm \cos 2A \sin A\); dM1 Uses correct double angle identities for \(\sin 2A\) and \(\cos 2A\); dDM1 Reaches expression in terms of \(\sin A\) only by use of \(\cos^2 A = 1 - \sin^2 A\); A1 Correct solution only \(\sin 3A \equiv 3\sin A - 4\sin^3 A\)
M1 Attempts to produce equation just in \(\sin\theta\) using both part (a) and identity \(\cos^2\theta = 1 - \sin^2\theta\); dM1 Uses \(\sin 3\theta \equiv 3\sin\theta - 4\sin^3\theta\) and obtains cubic equation in \(\sin\theta\) and attempts solve. This could include factorisation or division of \(\sin\theta\) term followed by attempt to solve 3 terms quadratic equation in \(\sin\theta\) to reach at least one non-zero value for \(\sin\theta\); A1 Correct solution only \(0\), \(\frac{\pi}{2}\), \(-0.848\)
**Part (a):**
Show that $\sin 3A \equiv 3\sin A - 4\sin^3 A$ | (4) | Using angle addition and double angle formulae: M1 Attempts to use identity for $\sin(2A + A) = \sin 2A \cos A \pm \cos 2A \sin A$; dM1 Uses correct double angle identities for $\sin 2A$ and $\cos 2A$; dDM1 Reaches expression in terms of $\sin A$ only by use of $\cos^2 A = 1 - \sin^2 A$; A1 Correct solution only $\sin 3A \equiv 3\sin A - 4\sin^3 A$
**Part (b):**
Solve $1 + \sin 3\theta = \cos^2\theta$ for $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ | (3) | M1 Attempts to produce equation just in $\sin\theta$ using both part (a) and identity $\cos^2\theta = 1 - \sin^2\theta$; dM1 Uses $\sin 3\theta \equiv 3\sin\theta - 4\sin^3\theta$ and obtains cubic equation in $\sin\theta$ and attempts solve. This could include factorisation or division of $\sin\theta$ term followed by attempt to solve 3 terms quadratic equation in $\sin\theta$ to reach at least one non-zero value for $\sin\theta$; A1 Correct solution only $0$, $\frac{\pi}{2}$, $-0.848$
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