# Question 10:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = B \Rightarrow \sin 2A = \sin(A+A) = \sin A\cos A + \cos A\sin A$ or $\sin A\cos A + \sin A\cos A$ | M1 | For the underlined equation in either form $\sin A\cos A + \cos A\sin A$ or $\sin A\cos A + \sin A\cos A$ |
| Hence $\sin 2A = 2\sin A\cos A$ * | A1* | Must see: $\sin 2A$ or $\sin(A+A)$ at start; correct expansion; $= 2\sin A\cos A$ at end |

## Part (b) — Way 1A:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \ln\left[\tan\left(\frac{1}{2}x\right)\right] \Rightarrow \frac{dy}{dx} = \frac{\frac{1}{2}\sec^2\left(\frac{1}{2}x\right)}{\tan\left(\frac{1}{2}x\right)}$ | M1 A1 | M1: expression of form $\frac{\pm k\sec^2(\frac{1}{2}x)}{\tan(\frac{1}{2}x)}$, $k$ constant; A1: correct differentiation |
| $= \frac{1}{2\tan(\frac{1}{2}x)\cos^2(\frac{1}{2}x)} = \frac{1}{2\sin(\frac{1}{2}x)\cos(\frac{1}{2}x)}$ | dM1 | Use both $\tan(\frac{1}{2}x) = \frac{\sin(\frac{1}{2}x)}{\cos(\frac{1}{2}x)}$ and $\sec^2(\frac{1}{2}x) = \frac{1}{\cos^2(\frac{1}{2}x)}$. Depends on previous M |
| $= \frac{1}{\sin x} = \cosec x$ * | A1* | Simplify using double angle formula; completely correct work, no errors |

## Part (b) — Way 1B:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{\frac{1}{2}\sec^2(\frac{1}{2}x)}{\tan(\frac{1}{2}x)}$ | M1 A1 | As Way 1A |
| Use $\sec^2(\frac{1}{2}x) = 1 + \tan^2(\frac{1}{2}x)$ and $\tan(\frac{1}{2}x) = \frac{\sin(\frac{1}{2}x)}{\cos(\frac{1}{2}x)}$ | dM1 | |
| $= \frac{1}{2\sin(\frac{1}{2}x)\cos(\frac{1}{2}x)} = \frac{1}{\sin x} = \cosec x$ * | A1* | |

## Part (b) — Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \ln\left[\sin(\frac{1}{2}x)\right] - \ln\left[\cos(\frac{1}{2}x)\right] \Rightarrow \frac{dy}{dx} = \frac{\frac{1}{2}\cos(\frac{1}{2}x)}{\sin(\frac{1}{2}x)} - \frac{-\frac{1}{2}\sin(\frac{1}{2}x)}{\cos(\frac{1}{2}x)}$ | M1 A1 | M1: split and differentiate; A1: correct |
| $= \frac{\cos^2(\frac{1}{2}x) + \sin^2(\frac{1}{2}x)}{2\sin(\frac{1}{2}x)\cos(\frac{1}{2}x)} = \frac{1}{\sin x} = \cosec x$ | M1; A1* | |

## Part (b) — Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Quotes $\int \cosec x\, dx = \ln(\tan(\frac{1}{2}x))$ | M1 A1 | |
| $\frac{d}{dx}\left[\tan(\frac{1}{2}x)\right] = \cosec x$ | M1 A1* | Differentiation is reverse of integration. Must be completely correct. Rare but acceptable |

## Part (c) — Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \ln\left[\tan(\frac{1}{2}x)\right] - 3\sin x \Rightarrow \frac{dy}{dx} = \cosec x - 3\cos x$ | B1 | Correct differentiation |
| $\frac{dy}{dx} = 0 \Rightarrow \cosec x - 3\cos x = 0 \Rightarrow \frac{1}{\sin x} - 3\cos x = 0$ | M1 | Sets $\frac{dy}{dx} = 0$ and uses $\cosec x = \frac{1}{\sin x}$ |
| $\Rightarrow 1 = 3\sin x\cos x \Rightarrow 1 = \frac{3}{2}(2\sin x\cos x)$, so $\sin 2x = k$, $-1 < k < 1$, $k \neq 0$ | M1 | Rearranges using double angle formula to obtain $\sin 2x = k$ |
| $\sin 2x = \frac{2}{3}$ | A1 | May be implied by correct answer |
| $2x = \{0.729727..., 2.411864...\}$, so $x = \{0.364863..., 1.205932...\}$ | A1 A1 | A1: either awrt 0.365 or awrt 1.206; A1: both. Answers in degrees lose both final marks. Extra answers in range lose last A1 |

## Part (c) — Way 2 (Squaring Method):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \cosec x - 3\cos x$; set $= 0$: $\frac{1}{\sin x} - 3\cos x = 0$ | B1, M1 | |
| $\frac{1}{1-\cos^2 x} = 9\cos^2 x$, so $9\cos^4 x - 9\cos^2 x + 1 = 0$ or $9\sin^4 x - 9\sin^2 x + 1 = 0$ | M1 | Obtain quadratic in $\sin x$ or $\cos x$. Condone $\cosec^2 x - 9\cos^2 x = 0$ as working |
| $\cos^2 x = 0.873$ or $0.127$; or $\sin^2 x = 0.873$ or $0.127$ | A1 | |
| $x = \{0.364863..., 1.205932...\}$ | A1 A1 | |

## Part (c) — Way 3 ($t$-method):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \cosec x - 3\cos x = 0 \Rightarrow \frac{1}{\sin x} - 3\cos x = 0$ | B1, M1 | |
| $\frac{1+t^2}{2t} - 3\cdot\frac{1-t^2}{1+t^2} = 0$, so $t^4 + 6t^3 + 2t^2 - 6t + 1 = 0$ | M1 | Uses $t = \tan(\frac{x}{2})$ |
| $t = 0.1845$ or $0.6885$ | A1 | |
| $x = \{0.364863..., 1.205932...\}$ | A1 A1 | |