| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Equation with non-equation preliminary part (sketch/proof/identity) |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C2 techniques: solving a double-angle sine equation (routine but requires careful angle manipulation), applying the sine rule (direct substitution), and finding cos from sin using Pythagoras identity. All parts are textbook exercises with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks |
|---|---|
| \(\arcsin 0.6 = 36.9°\) (awrt) | B1 |
| \(2x + 50 = 36.87\), \(2x = -13.13° + 360° = 346.87°\) | M1 M1 |
| \(2x + 50 + 180 = 36.87\), \(2x = 143.13° - 50° = 93.13°\) | M1 |
| \(x = 46.6, 173.4\) | M1 A1 A1 |
| (7 marks) |
| Answer | Marks |
|---|---|
| \(\sin 60° = \frac{\sqrt{3}}{2}\), \(\frac{BC}{\frac{1}{3}} = \frac{18}{\sin 60°}\) | B1, M1 |
| \(BC = 6 \div \frac{\sqrt{3}}{2} = \frac{12}{\sqrt{3}} = 4\sqrt{3}\) | M1 A1 |
| (4 marks) |
| Answer | Marks |
|---|---|
| \(\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{1}{9}\) | M1 |
| \(\sin \theta = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}\) | A1 |
| (2 marks) | |
| (13 marks total) |
## (i)
$\arcsin 0.6 = 36.9°$ (awrt) | B1 |
$2x + 50 = 36.87$, $2x = -13.13° + 360° = 346.87°$ | M1 M1 |
$2x + 50 + 180 = 36.87$, $2x = 143.13° - 50° = 93.13°$ | M1 |
$x = 46.6, 173.4$ | M1 A1 A1 |
| (7 marks) |
## (ii)(a)
$\sin 60° = \frac{\sqrt{3}}{2}$, $\frac{BC}{\frac{1}{3}} = \frac{18}{\sin 60°}$ | B1, M1 |
$BC = 6 \div \frac{\sqrt{3}}{2} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$ | M1 A1 |
| (4 marks) |
## (ii)(b)
$\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{1}{9}$ | M1 |
$\sin \theta = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$ | A1 |
| (2 marks) |
| **(13 marks total)** |\begin{enumerate}[label=(\roman*)]
\item Solve, for $0° < x < 180°$, the equation $\sin (2x + 50°) = 0.6$, giving your answers to 1 d. p. [7]
\item In the triangle $ABC$, $AC = 18$ cm, $\angle ABC = 60°$ and $\sin A = \frac{1}{3}$.
\begin{enumerate}[label=(\alph*)]
\item Use the sine rule to show that $BC = 4\sqrt{3}$. [4]
\item Find the exact value of $\cos A$. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [13]}}