| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Topic | Addition & Double Angle Formulae |
| Type | Derive triple angle then solve equation |
| Difficulty | Challenging +1.2 This is a structured multi-part question requiring the double angle formula, algebraic manipulation to derive the triple angle formula, and inverse trigonometric work. Part (i) is standard bookwork. Part (ii) involves careful calculation with exact values. Part (iii) requires solving a trigonometric equation using the derived formula and working with inverse functions, which is moderately challenging but follows a clear path once the setup is understood. The question is harder than average due to the extended algebraic manipulation and exact value work, but remains within standard A-level techniques. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}\) | B1 | |
| Use of identity for \(\tan(\theta+2\theta)\) | M1 | |
| Substituting for \(\tan 2\theta\) | M1 | |
| NIS for the result \(\tan 3\theta = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\) AG | A1 | [4] |
| (ii) Recognition of a right angled triangle with hypotenuse \(= \sqrt{10}\) | M1 | |
| State \(\tan\theta = \frac{1}{3}\) | B1 | |
| Substituting to achieve the correct result (AG) | A1 | [3] |
| (iii) State or imply \(\theta = \sin^{-1}x\) from the \(3\theta\) link in (ii) | M1 | |
| State \(x = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}\) using the link from \(\frac{13}{9}\) in (ii) (AG) | A1 | |
| State or imply that \(3\sin^{-1}x = 3\sin^{-1}\left(\frac{1}{\sqrt{10}}\right) + \pi\) | M1 | |
| Conclude \(x = \sin\left(\sin^{-1}\left(\frac{1}{\sqrt{10}}\right) + \frac{\pi}{3}\right)\) and use \(\sin(A+B)\) formula to obtain \(x = \frac{1}{2\sqrt{10}} + \frac{3\sqrt{3}}{2\sqrt{10}} = \frac{(1+3\sqrt{3})\sqrt{10}}{20}\) (AG) | A1 | [4] |
(i) $\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}$ | B1 |
Use of identity for $\tan(\theta+2\theta)$ | M1 |
Substituting for $\tan 2\theta$ | M1 |
NIS for the result $\tan 3\theta = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$ AG | A1 | [4]
(ii) Recognition of a right angled triangle with hypotenuse $= \sqrt{10}$ | M1 |
State $\tan\theta = \frac{1}{3}$ | B1 |
Substituting to achieve the correct result (AG) | A1 | [3]
(iii) State or imply $\theta = \sin^{-1}x$ from the $3\theta$ link in (ii) | M1 |
State $x = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$ using the link from $\frac{13}{9}$ in (ii) (AG) | A1 |
State or imply that $3\sin^{-1}x = 3\sin^{-1}\left(\frac{1}{\sqrt{10}}\right) + \pi$ | M1 |
Conclude $x = \sin\left(\sin^{-1}\left(\frac{1}{\sqrt{10}}\right) + \frac{\pi}{3}\right)$ and use $\sin(A+B)$ formula to obtain $x = \frac{1}{2\sqrt{10}} + \frac{3\sqrt{3}}{2\sqrt{10}} = \frac{(1+3\sqrt{3})\sqrt{10}}{20}$ (AG) | A1 | [4]\begin{enumerate}[label=(\roman*)]
\item Write down an identity for $\tan 2\theta$ in terms of $\tan \theta$ and use this result to show that
$$\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}.$$ [4]
\item Given that $0 < \theta < \frac{1}{2}\pi$ and $\theta = \sin^{-1}\left(\frac{1}{\sqrt{10}}\right)$, show that $\tan 3\theta = \frac{13}{3}$. [3]
\item Show that the solutions of the equation
$$\tan(3 \sin^{-1} x) = \frac{13}{3}$$
for $0 < x < 2\pi$ are
$$x = \frac{\sqrt{10}}{10} \quad \text{and} \quad x = \frac{\sqrt{10(1 + 3\sqrt{3})}}{20}.$$ [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2010 Q11 [11]}}