## Question 7:

### Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos(A-B) = \cos A \cos(-B) - \sin A \sin(-B)$ | M1 | Replace $B$ with $-B$ in given identity |
| State $\cos(-B) = \cos B$ and $\sin(-B) = -\sin B$, conclude $\cos(A-B) = \cos A \cos B - \sin A(-\sin B)$, so $\cos(A-B) = \cos A \cos B + \sin A \sin B$ **A.G.** | A1 | $\cos(-B) = \cos B$, $\sin(-B) = -\sin B$ must be stated, but no justification needed. Condone $-\sin A \sin(-B)$ becoming $\sin A \sin B$ with no intermediate step |

### Part (b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta\right)\left(\frac{\sqrt{3}}{2}\cos\theta + \frac{1}{2}\sin\theta\right)$ | B1 | Use correct identities with exact trig values to obtain a correct expression. Allow BOD for ambiguous positioning of $+$ and $-$ signs |
| $\frac{3}{4}\cos^2\theta - \frac{1}{4}\sin^2\theta$ | M1 | Expand brackets. May be recognised as difference of two squares so no need to see $\frac{\sqrt{3}}{4}\cos\theta\sin\theta - \frac{\sqrt{3}}{4}\cos\theta\sin\theta$ |
| $\frac{3}{4}\cos^2\theta - \frac{1}{4}(1-\cos^2\theta)$, so $\cos^2\theta - \frac{1}{4}$ **A.G.** | A1 | Use Pythagorean identity and simplify to given answer. If middle terms shown for expansion, then these must be correct |

### Part (c)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| max value is $\frac{3}{4}$ | B1 | Correct max value |
| when $\theta$ is $180°$ | B1 | Correct angle. **B0** if any extra angles given. Must be in degrees. Must be 'positive' so **B0** for $0°$ |

### Part (c)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| min value is $-\frac{1}{4}$ | B1 | Correct min value |
| when $\theta$ is $90°$ | B1 | Correct angle. **B0** if any extra angles given. Must be in degrees. **SC** If angles in **both** parts are correct but in radians, penalise only once (mark as **B0** in (i) and **B1** in (ii)) |

---