Edexcel C3 — Question 2 9 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeProduct to sum using compound angles
DifficultyStandard +0.3 Part (a) is a straightforward proof requiring direct expansion of standard compound angle formulae and simple algebraic manipulation. Part (b) requires applying the product-to-sum formula, using reciprocal identities, and solving a trigonometric equation within a given interval—standard C3 techniques with multiple steps but no novel insight required. Slightly above average due to the multi-step nature and need to connect parts (a) and (b).
Spec1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals

2. (a) Use the identities for \(\cos ( A + B )\) and \(\cos ( A - B )\) to prove that $$2 \cos A \cos B \equiv \cos ( A + B ) + \cos ( A - B ) .$$ (b) Hence, or otherwise, find in terms of \(\pi\) the solutions of the equation $$2 \cos \left( x + \frac { \pi } { 2 } \right) = \sec \left( x + \frac { \pi } { 6 } \right) ,$$ for \(x\) in the interval \(0 \leq x \leq \pi\).

Part (a)
AnswerMarks
\(\cos(A+B) = \cos A \cos B - \sin A \sin B\)M1 A1
\(\cos(A-B) = \cos A \cos B + \sin A \sin B\)
adding, \(2 \cos A \cos B \equiv \cos(A+B) + \cos(A-B)\)
Part (b)
AnswerMarks
\(2\cos(x + \frac{\pi}{3})\cos(x + \frac{\pi}{6}) = 1\)M1
\(\cos(2x + \frac{2\pi}{3}) + \cos \frac{\pi}{3} = 1\)M1
\(\cos(2x + \frac{2\pi}{3}) = 1 - \frac{1}{2} = \frac{1}{2}\)A1
\(2x + \frac{2\pi}{3} = 2\pi - \frac{\pi}{3}, 2x + \frac{\pi}{3} = \frac{5\pi}{3}, \frac{7\pi}{3}\)B1
\(2x = \pi, \frac{5\pi}{3}\)M1
\(x = \frac{\pi}{2}, \frac{5\pi}{6}\)A2
(9)
**Part (a)**
$\cos(A+B) = \cos A \cos B - \sin A \sin B$ | M1 A1 |
$\cos(A-B) = \cos A \cos B + \sin A \sin B$ |
adding, $2 \cos A \cos B \equiv \cos(A+B) + \cos(A-B)$ |

**Part (b)**
$2\cos(x + \frac{\pi}{3})\cos(x + \frac{\pi}{6}) = 1$ | M1 |
$\cos(2x + \frac{2\pi}{3}) + \cos \frac{\pi}{3} = 1$ | M1 |
$\cos(2x + \frac{2\pi}{3}) = 1 - \frac{1}{2} = \frac{1}{2}$ | A1 |
$2x + \frac{2\pi}{3} = 2\pi - \frac{\pi}{3}, 2x + \frac{\pi}{3} = \frac{5\pi}{3}, \frac{7\pi}{3}$ | B1 |
$2x = \pi, \frac{5\pi}{3}$ | M1 |
$x = \frac{\pi}{2}, \frac{5\pi}{6}$ | A2 |
| (9) |

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2. (a) Use the identities for $\cos ( A + B )$ and $\cos ( A - B )$ to prove that

$$2 \cos A \cos B \equiv \cos ( A + B ) + \cos ( A - B ) .$$

(b) Hence, or otherwise, find in terms of $\pi$ the solutions of the equation

$$2 \cos \left( x + \frac { \pi } { 2 } \right) = \sec \left( x + \frac { \pi } { 6 } \right) ,$$

for $x$ in the interval $0 \leq x \leq \pi$.\\

\hfill \mbox{\textit{Edexcel C3  Q2 [9]}}