## Question 9(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct shape for $y = k\cos\!\left(\frac{1}{2}x\right)$ | B1 | Must show intention to pass through $(-\pi, 0)$ and $(\pi, 0)$. Should be roughly symmetrical in the $y$-axis, but condone slightly different $y$-values at $-2\pi$ and $2\pi$. Ignore graph outside of given range. |
| Correct shape for $y = \tan\!\left(\frac{1}{2}x\right)$ | B1 | Must show intention to pass through $(-2\pi, 0)$, $(0, 0)$, $(2\pi, 0)$. Asymptotes need not be marked, but there should be no clear overlap of the limbs, nor significant gaps between them. Ignore graph outside of given range. |
| $(0, 3)$ stated or clearly indicated | B1 | Can still be given if $y = 3\cos\!\left(\frac{1}{2}x\right)$ graph is incorrect or not attempted. If more than one point marked on the $y$-axis then mark the label on the graph intercept. |
| **[3]** | | |

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## Question 9(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{\sin(\frac{1}{2}x)}{\cos(\frac{1}{2}x)} = 3\cos\!\left(\tfrac{1}{2}x\right)$ | M1 | Attempt use of relevant identities to show given equation. Must attempt use of both identities; these must be correct but allow poor notation. e.g. using $\frac{\sin}{\cos}\!\left(\frac{1}{2}x\right)$ and/or $3(1-\sin^2)\!\left(\frac{1}{2}x\right)$ could get M1A0. |
| $\sin\!\left(\tfrac{1}{2}x\right) = 3\cos^2\!\left(\tfrac{1}{2}x\right)$ | | |
| $\sin\!\left(\tfrac{1}{2}x\right) = 3\!\left(1 - \sin^2\!\left(\tfrac{1}{2}x\right)\right)$ | | |
| $3\sin^2\!\left(\tfrac{1}{2}x\right) + \sin\!\left(\tfrac{1}{2}x\right) - 3 = 0$ **AG** | A1 | Obtain given equation, with no errors seen. Use both identities correctly to obtain given equation. Brackets around the $\frac{1}{2}x$ not required. |
| $\sin\!\left(\tfrac{1}{2}x\right) = 0.847,\ -1.18$ | M1 | Attempt to solve given quadratic to find solution(s) for $\sin\!\left(\frac{1}{2}x\right)$. Must use quadratic formula (or completing the square) — M0 if attempting to factorise. Allow variables other than $\sin(\frac{1}{2}x)$, e.g. $y=$, or even $x=$. Allow $-1.18$ to be discarded at any stage. |
| $\tfrac{1}{2}x = 1.01,\ 2.13$ | M1 | Attempt to solve $\sin\!\left(\frac{1}{2}x\right) = k$. Attempt $\sin^{-1}$ (their root) and then double the answer. |
| $x = 2.02,\ 4.26$ | A1 | Obtain one correct angle. Allow in degrees ($116°$ and $244°$). |
| | A1 | Obtain both correct angles, and no others in given range. Must both be in radians (allow equivs as multiples of $\pi$). A0 if extra, incorrect angles in given range of $[-2\pi, 2\pi]$ but ignore any outside of given range. **SR:** if no working shown then allow B1 for each correct solution (max of B1 if in degrees, or extra solns in range). |
| **[6]** | | |

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