Standard +0.3 Part (i) is a straightforward algebraic manipulation proving an identity using basic definitions (tan = sin/cos) and finding a common denominator - standard textbook exercise. Part (ii) involves solving a double angle equation requiring rearrangement and consideration of multiple solutions within a range, which is routine A-level technique. Both parts require only standard methods with no novel insight, making this slightly easier than average.
Uses \(\tan\theta \equiv \frac{\sin\theta}{\cos\theta}\) and \(\frac{1}{\tan\theta} \equiv \frac{\cos\theta}{\sin\theta}\); attempts common denominator \(\sin\theta\cos\theta\) with 2-term numerator, one correct. Depends on first mark
\(= \dfrac{1}{\sin\theta\cos\theta}\)
A1*
Correct proof with no notation errors; allow "\(\equiv\)" instead of "\(=\)". Spurious "\(= 0\)"s score A0
Applies correct order of operations to find \(x\); may need to check values if no working shown
\(x = 22.4°\) or \(x = 57.6°\)
A1
Awrt one of these
\(x = 22.4°\) and \(x = 57.6°\)
A1
Awrt both with no extras in range
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\theta + \frac{1}{\tan\theta} = \frac{\sin\theta}{\cos\theta} + \frac{1}{\frac{\sin\theta}{\cos\theta}}$ | M1 | Uses $\tan\theta \equiv \frac{\sin\theta}{\cos\theta}$ on both terms |
| $\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}$ | dM1 | Uses $\tan\theta \equiv \frac{\sin\theta}{\cos\theta}$ and $\frac{1}{\tan\theta} \equiv \frac{\cos\theta}{\sin\theta}$; attempts common denominator $\sin\theta\cos\theta$ with 2-term numerator, one correct. Depends on first mark |
| $= \dfrac{1}{\sin\theta\cos\theta}$ | A1* | Correct proof with no notation errors; allow "$\equiv$" instead of "$=$". Spurious "$= 0$"s score A0 |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\cos^2(2x+10°) = 1 \Rightarrow \cos^2(2x+10°) = \frac{1}{3} \Rightarrow \cos(2x+10°) = \pm\sqrt{\frac{1}{3}}$ | M1 | Divides by 3 and takes square root of both sides; "$\pm$" not required |
| $2x + 10° = \cos^{-1}\left(\pm\sqrt{\frac{1}{3}}\right) \Rightarrow x = \dfrac{\cos^{-1}\left(\pm\sqrt{\frac{1}{3}}\right) - 10°}{2}$ | M1 | Applies correct order of operations to find $x$; may need to check values if no working shown |
| $x = 22.4°$ or $x = 57.6°$ | A1 | Awrt one of these |
| $x = 22.4°$ and $x = 57.6°$ | A1 | Awrt both with no extras in range |