| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Show equation reduces to tan form |
| Difficulty | Standard +0.3 Part (a) requires systematic application of compound angle formulae and algebraic manipulation to reach the given tan form—a standard C3 technique but with multiple steps. Part (b) is routine substitution (x=2θ) followed by calculator work. Overall slightly easier than average due to the 'show that' scaffold and straightforward method. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\cos x\cos 50 - 2\sin x\sin 50 = \sin x\cos 40 + \cos x\sin 40\) | M1 | Expand both expressions using \(\cos(x+50) = \cos x\cos 50 - \sin x\sin 50\) and \(\sin(x+40) = \sin x\cos 40 + \cos x\sin 40\). Condone missing bracket on lhs. May condone sign error on one expression. |
| \(\sin x(\cos 40 + 2\sin 50) = \cos x(2\cos 50 - \sin 40)\) | ||
| \(\div \cos x \Rightarrow \tan x(\cos 40 + 2\sin 50) = 2\cos 50 - \sin 40\) | M1 | Divide by \(\cos x\) to reach equation in \(\tan x\). Independent of first M1. |
| \(\tan x = \dfrac{2\cos 50 - \sin 40}{\cos 40 + 2\sin 50}\) (or numerical answer awrt 0.28) | A1 | Accept \(\tan x =\) awrt 0.28 as long as M1M1 achieved |
| Uses \(\cos 50 = \sin 40\) and \(\cos 40 = \sin 50\) to show \(\tan x^{\circ} = \frac{1}{3}\tan 40^{\circ}\) | A1* (cao) | Given answer — all steps must be shown. Do not allow from \(\tan x =\) awrt 0.28 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \(\tan 2\theta = \frac{1}{3}\tan 40\) | M1 | For linking part (a) with (b) |
| \(2\theta = 15.6\) so \(\theta =\) awrt 7.8(1) one answer | A1 | Solves to find one solution, usually awrt 7.8 |
| Also \(2\theta = 195.6,\ 375.6,\ 555.6 \Rightarrow \theta = \ldots\) | M1 | Correct method for at least one further value. Accept \(\theta = \frac{180 + \text{their}\,\alpha}{2}\), \((or)\ \frac{360+\text{their}\,\alpha}{2}\), \((or)\ \frac{540+\text{their}\,\alpha}{2}\) |
| \(\theta =\) awrt 7.8, 97.8, 187.8, 277.8 — all 4 answers | A1 | All four answers to 1dp. Ignore extras outside range. Withhold for extras inside range. Condone different variable. Answers fully in radians loses first A mark; acceptable in rads awrt 0.136, 1.71, 3.28, 4.85 |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\cos x\cos 50 - 2\sin x\sin 50 = \sin x\cos 40 + \cos x\sin 40$ | M1 | Expand both expressions using $\cos(x+50) = \cos x\cos 50 - \sin x\sin 50$ and $\sin(x+40) = \sin x\cos 40 + \cos x\sin 40$. Condone missing bracket on lhs. May condone sign error on one expression. |
| $\sin x(\cos 40 + 2\sin 50) = \cos x(2\cos 50 - \sin 40)$ | | |
| $\div \cos x \Rightarrow \tan x(\cos 40 + 2\sin 50) = 2\cos 50 - \sin 40$ | M1 | Divide by $\cos x$ to reach equation in $\tan x$. Independent of first M1. |
| $\tan x = \dfrac{2\cos 50 - \sin 40}{\cos 40 + 2\sin 50}$ (or numerical answer awrt 0.28) | A1 | Accept $\tan x =$ awrt 0.28 as long as M1M1 achieved |
| Uses $\cos 50 = \sin 40$ and $\cos 40 = \sin 50$ to show $\tan x^{\circ} = \frac{1}{3}\tan 40^{\circ}$ | A1* (cao) | Given answer — all steps must be shown. Do not allow from $\tan x =$ awrt 0.28 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $\tan 2\theta = \frac{1}{3}\tan 40$ | M1 | For linking part (a) with (b) |
| $2\theta = 15.6$ so $\theta =$ awrt 7.8(1) one answer | A1 | Solves to find one solution, usually awrt 7.8 |
| Also $2\theta = 195.6,\ 375.6,\ 555.6 \Rightarrow \theta = \ldots$ | M1 | Correct method for at least one further value. Accept $\theta = \frac{180 + \text{their}\,\alpha}{2}$, $(or)\ \frac{360+\text{their}\,\alpha}{2}$, $(or)\ \frac{540+\text{their}\,\alpha}{2}$ |
| $\theta =$ awrt 7.8, 97.8, 187.8, 277.8 — all 4 answers | A1 | All four answers to 1dp. Ignore extras outside range. Withhold for extras inside range. Condone different variable. Answers fully in radians loses first A mark; acceptable in rads awrt 0.136, 1.71, 3.28, 4.85 |
---3. Given that
$$2 \cos ( x + 50 ) ^ { \circ } = \sin ( x + 40 ) ^ { \circ }$$
\begin{enumerate}[label=(\alph*)]
\item Show, without using a calculator, that
$$\tan x ^ { \circ } = \frac { 1 } { 3 } \tan 40 ^ { \circ }$$
\item Hence solve, for $0 \leqslant \theta < 360$,
$$2 \cos ( 2 \theta + 50 ) ^ { \circ } = \sin ( 2 \theta + 40 ) ^ { \circ }$$
giving your answers to 1 decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2013 Q3 [8]}}