Standard +0.3 This question requires applying the double angle formula for cotangent and converting between trigonometric functions, then solving a resulting quadratic equation in tan θ. While it involves multiple steps and careful algebraic manipulation, it's a standard application of double angle formulae with no novel insight required—slightly easier than average for A-level.
\(5\tan^2\theta = 1\) or \(\sin^2\theta = \dfrac{1}{6}\) or \(\cos^2\theta = \dfrac{5}{6}\)
Solve for \(\theta\)
DM1
Dependent on the first M1
Obtain answer \(24.1°\) (or \(155.9°\))
A1
One correct in range to at least 3 sf
Obtain second answer
A1
FT \(180° - \textit{their}\ 24.1°\) and no others in range. Correct to at least 3 sf. Accept \(156°\) but not \(156.0\). Ignore values outside range. If working in \(\tan\theta\) or \(\cos\theta\) need to be considering both square roots to score the second A1. Mark \(0.421, 2.72\) as a MR, so A0A1
5
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct trig formulae to obtain an equation in $\tan\theta$ or equivalent (e.g. all in $\sin\theta$ or all in $\cos\theta$) | \*M1 | $\dfrac{1-\tan^2\theta}{2\tan\theta} = 2\tan\theta$. Allow $\dfrac{\cot^2\theta - 1}{2\cot\theta} = \dfrac{2}{\cot\theta}$ |
| Obtain a correct simplified equation | A1 | $5\tan^2\theta = 1$ or $\sin^2\theta = \dfrac{1}{6}$ or $\cos^2\theta = \dfrac{5}{6}$ |
| Solve for $\theta$ | DM1 | Dependent on the first M1 |
| Obtain answer $24.1°$ (or $155.9°$) | A1 | One correct in range to at least 3 sf |
| Obtain second answer | A1 | **FT** $180° - \textit{their}\ 24.1°$ and no others in range. Correct to at least 3 sf. Accept $156°$ but not $156.0$. Ignore values outside range. If working in $\tan\theta$ or $\cos\theta$ need to be considering both square roots to score the second A1. Mark $0.421, 2.72$ as a MR, so A0A1 |
| | **5** | |