3. Given that
$$y = \frac { \cos 2 \theta } { 1 + \sin 2 \theta } , \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$
show that
$$\frac { \mathrm { d } y } { \mathrm {~d} \theta } = \frac { a } { 1 + \sin 2 \theta } , \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$
where \(a\) is a constant to be determined.
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Question 3:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(u = \cos 2\theta\), \(v = 1 + \sin 2\theta\), \(\frac{du}{d\theta} = -2\sin 2\theta\), \(\frac{dv}{d\theta} = 2\cos 2\theta\) M1
Applies quotient rule (or product rule) to \(\frac{\cos 2\theta}{1 + \sin 2\theta}\); if rule not quoted, terms must be written out in correct form
\(\frac{dy}{d\theta} = \frac{-2\sin 2\theta(1 + \sin 2\theta) - 2\cos^2 2\theta}{(1 + \sin 2\theta)^2}\) A1
Any fully correct unsimplified form
\(= \frac{-2\sin 2\theta - 2\sin^2 2\theta - 2\cos^2 2\theta}{(1 + \sin 2\theta)^2} = \frac{-2\sin 2\theta - 2}{(1 + \sin 2\theta)^2}\) M1
Applies \(\sin^2 2\theta + \cos^2 2\theta = 1\) to eliminate squared trig terms, obtaining expression of form \(k\sin 2\theta + \lambda\)
\(= \frac{-2(1 + \sin 2\theta)}{(1 + \sin 2\theta)^2} = \frac{-2}{1 + \sin 2\theta}\) A1 cso
Must see factorisation of numerator then answer; \(\frac{-2}{1+\sin 2\theta}\) or \(\frac{a}{1+\sin 2\theta}\) with \(a = -2\), no previous errors
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# Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = \cos 2\theta$, $v = 1 + \sin 2\theta$, $\frac{du}{d\theta} = -2\sin 2\theta$, $\frac{dv}{d\theta} = 2\cos 2\theta$ | M1 | Applies quotient rule (or product rule) to $\frac{\cos 2\theta}{1 + \sin 2\theta}$; if rule not quoted, terms must be written out in correct form |
| $\frac{dy}{d\theta} = \frac{-2\sin 2\theta(1 + \sin 2\theta) - 2\cos^2 2\theta}{(1 + \sin 2\theta)^2}$ | A1 | Any fully correct unsimplified form |
| $= \frac{-2\sin 2\theta - 2\sin^2 2\theta - 2\cos^2 2\theta}{(1 + \sin 2\theta)^2} = \frac{-2\sin 2\theta - 2}{(1 + \sin 2\theta)^2}$ | M1 | Applies $\sin^2 2\theta + \cos^2 2\theta = 1$ to eliminate squared trig terms, obtaining expression of form $k\sin 2\theta + \lambda$ |
| $= \frac{-2(1 + \sin 2\theta)}{(1 + \sin 2\theta)^2} = \frac{-2}{1 + \sin 2\theta}$ | A1 cso | Must see factorisation of numerator then answer; $\frac{-2}{1+\sin 2\theta}$ or $\frac{a}{1+\sin 2\theta}$ with $a = -2$, no previous errors |
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3. Given that
$$y = \frac { \cos 2 \theta } { 1 + \sin 2 \theta } , \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$
show that
$$\frac { \mathrm { d } y } { \mathrm {~d} \theta } = \frac { a } { 1 + \sin 2 \theta } , \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$
where $a$ is a constant to be determined.\\
\hfill \mbox{\textit{Edexcel C34 2014 Q3 [4]}}