Moderate -0.3 This is a straightforward double angle equation requiring standard substitutions (sin 2θ = 2sin θ cos θ, cos 2θ = 1 - 2sin²θ or similar) and factorization. It's slightly easier than average because it's a routine C4 question with a limited range and standard technique, but not trivial since it requires choosing appropriate identities and solving the resulting equation.
Use of correct double angle formulae: \(\sin2\theta \equiv 2\sin\theta\cos\theta\) and any one of \(\cos2\theta \equiv \cos^2\theta - \sin^2\theta\) or \(1-2\sin^2\theta\) or \(2\cos^2\theta-1\)
\(2\cos\theta(2\sin\theta - \cos\theta) = 0\)
A1
Correct equation in solvable form e.g. \(2\sin\theta - \cos\theta = 0\) (oe) or \(5\sin^4\theta - 6\sin^2\theta + 1 = 0\) or \(5\cos^4\theta - 4\cos^2\theta = 0\) but not \(4\sin\theta\cos\theta = 2\cos^2\theta\)
\(\Rightarrow \tan\theta = \frac{1}{2}\)
M1dep*
Use of \(\frac{\sin\theta}{\cos\theta} = \tan\theta\) on \(\alpha\sin\theta + \beta\cos\theta = 0\), or correct method for solving quadratic in \(\sin^2\theta\) or \(\cos^2\theta\)
\(\theta = 26.6°\)
A1
www (26.6 or better)
\(\theta = 90°\)
B1
Not from incorrect working. Ignore additional solutions outside range. If any additional solutions given inside \(0 \leq \theta \leq 180°\) and full marks would have been awarded then remove last mark (so 4/5). Both in radians: 0.464 (or better) and \(\pi/2\) scores B1. Answers with no working scores B1 B1 (max 2/5)
# Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4\sin\theta\cos\theta = 1 + 2\cos^2\theta - 1$ | M1* | Use of correct double angle formulae: $\sin2\theta \equiv 2\sin\theta\cos\theta$ **and** any one of $\cos2\theta \equiv \cos^2\theta - \sin^2\theta$ or $1-2\sin^2\theta$ or $2\cos^2\theta-1$ |
| $2\cos\theta(2\sin\theta - \cos\theta) = 0$ | A1 | Correct equation in solvable form e.g. $2\sin\theta - \cos\theta = 0$ (oe) or $5\sin^4\theta - 6\sin^2\theta + 1 = 0$ or $5\cos^4\theta - 4\cos^2\theta = 0$ but not $4\sin\theta\cos\theta = 2\cos^2\theta$ |
| $\Rightarrow \tan\theta = \frac{1}{2}$ | M1dep* | Use of $\frac{\sin\theta}{\cos\theta} = \tan\theta$ on $\alpha\sin\theta + \beta\cos\theta = 0$, or correct method for solving quadratic in $\sin^2\theta$ or $\cos^2\theta$ |
| $\theta = 26.6°$ | A1 | www (26.6 or better) |
| $\theta = 90°$ | B1 | Not from incorrect working. Ignore additional solutions outside range. If any additional solutions given inside $0 \leq \theta \leq 180°$ **and** full marks would have been awarded then remove last mark (so 4/5). Both in radians: 0.464 (or better) and $\pi/2$ scores B1. Answers with no working scores B1 B1 (max 2/5) |
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