Standard +0.3 This is a straightforward double angle equation requiring substitution of cos 2θ = 1 - 2sin²θ to form a quadratic in sin θ, then solving for θ in a given interval. It's slightly easier than average as it follows a standard technique with no conceptual surprises, though it does require careful algebraic manipulation and finding multiple solutions.
Use correct \(\cos 2A\) formula and obtain an equation in \(\sin\theta\)
M1
Obtain \(4\sin^2\theta + \sin\theta - 3 = 0\), or equivalent
A1
Make reasonable attempt to solve a 3-term quadratic in \(\sin\theta\)
M1
Obtain answer \(48.6°\)
A1
Obtain answer \(131.4°\) and no others in the given range
A1\(\sqrt{}\)
Obtain answer \(270°\) and no others in the given range
A1
[6]
*Note: Treat giving of answers in radians as a misread. Ignore answers outside the given range.*
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use correct $\cos 2A$ formula and obtain an equation in $\sin\theta$ | M1 | |
| Obtain $4\sin^2\theta + \sin\theta - 3 = 0$, or equivalent | A1 | |
| Make reasonable attempt to solve a 3-term quadratic in $\sin\theta$ | M1 | |
| Obtain answer $48.6°$ | A1 | |
| Obtain answer $131.4°$ and no others in the given range | A1$\sqrt{}$ | |
| Obtain answer $270°$ and no others in the given range | A1 | [6] |
*Note: Treat giving of answers in radians as a misread. Ignore answers outside the given range.*
---