Standard +0.3 This is a straightforward double angle equation requiring substitution of cos 2θ = 1 - 2sin²θ, leading to a quadratic in sin θ. The steps are standard: substitute, rearrange to quadratic form, solve, and find angles in the given range. Slightly easier than average due to the routine nature of the technique.
Use double angle formula and obtain an equation in \(\sin\theta\)
M1
Reduce to \(6\sin^2\theta + \sin\theta - 5 = 0\), or 3-term equivalent
A1
Solve a 3-term quadratic in \(\sin\theta\) and calculate \(\theta\)
M1
Obtain answer, e.g. \(56.4°\)
A1
Obtain second and third answers, e.g. \(123.6°\) and \(270°\) and no others in the given interval
A1
Ignore answers outside the interval. Treat answers in radians as a misread.
## Question 5:
| Answer | Mark | Guidance |
|--------|------|----------|
| Use double angle formula and obtain an equation in $\sin\theta$ | M1 | |
| Reduce to $6\sin^2\theta + \sin\theta - 5 = 0$, or 3-term equivalent | A1 | |
| Solve a 3-term quadratic in $\sin\theta$ and calculate $\theta$ | M1 | |
| Obtain answer, e.g. $56.4°$ | A1 | |
| Obtain second and third answers, e.g. $123.6°$ and $270°$ and no others in the given interval | A1 | Ignore answers outside the interval. Treat answers in radians as a misread. |
---