Question 3 - A-Level Maths

Edexcel C3 — Question 3 9 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Addition & Double Angle Formulae
Type	Product to sum using compound angles
Difficulty	Standard +0.3 Part (a) is a standard bookwork proof of a sum-to-product formula using compound angle identities—straightforward substitution with clear guidance. Part (b) applies this result to solve a trigonometric equation, requiring factorization and consideration of multiple solutions within an interval. This is slightly above average difficulty due to the multi-step nature and need to find all solutions systematically, but remains a standard C3 exercise with no novel insight required.
Spec	1.05l Double angle formulae: and compound angle formulae 1.05o Trigonometric equations: solve in given intervals

3. (a) Use the identities for $\sin ( A + B )$ and $\sin ( A - B )$ to prove that $$\sin P + \sin Q \equiv 2 \sin \frac { P + Q } { 2 } \cos \frac { P - Q } { 2 } \text {. }$$ (b) Find, in terms of $\pi$, the solutions of the equation $$\sin 5 x + \sin x = 0$$ for $x$ in the interval $0 \leq x < \pi$.

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Answer	Marks	Guidance
(a)	$\sin(A+B) \equiv \sin A \cos B + \cos A \sin B$
$\sin(A-B) \equiv \sin A \cos B - \cos A \sin B$
Adding: $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$	M1 A1
Let $P = A + B, \quad Q = A - B$
Adding: $P + Q = 2A \Rightarrow A = \frac{P+Q}{2}$	M1
Subtracting: $P - Q = 2B \Rightarrow B = \frac{P-Q}{2}$
$\sin P + \sin Q \equiv 2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}$	A1
(b)	$2\sin 3x \cos 2x = 0$	M1
$\sin 3x = 0$ or $\cos 2x = 0$	A1
$3x = 0, \pi, 2\pi$ or $2x = \frac{\pi}{2}, \frac{3\pi}{2}$	M1
$x = 0, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{3\pi}{4}$	A2
(9)

(a) | $\sin(A+B) \equiv \sin A \cos B + \cos A \sin B$ | |
| $\sin(A-B) \equiv \sin A \cos B - \cos A \sin B$ | |
| Adding: $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$ | M1 A1 |
| Let $P = A + B, \quad Q = A - B$ | |
| Adding: $P + Q = 2A \Rightarrow A = \frac{P+Q}{2}$ | M1 |
| Subtracting: $P - Q = 2B \Rightarrow B = \frac{P-Q}{2}$ | |
| $\sin P + \sin Q \equiv 2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}$ | A1 |

(b) | $2\sin 3x \cos 2x = 0$ | M1 |
| $\sin 3x = 0$ or $\cos 2x = 0$ | A1 |
| $3x = 0, \pi, 2\pi$ or $2x = \frac{\pi}{2}, \frac{3\pi}{2}$ | M1 |
| $x = 0, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{3\pi}{4}$ | A2 |
| | (9) |

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

$$\sin P + \sin Q \equiv 2 \sin \frac { P + Q } { 2 } \cos \frac { P - Q } { 2 } \text {. }$$

(b) Find, in terms of $\pi$, the solutions of the equation

$$\sin 5 x + \sin x = 0$$

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

\hfill \mbox{\textit{Edexcel C3  Q3 [9]}}

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Q1 7 Q2 9 Q3 9 Q4 10 Q5 13 Q6 13 Q7 14

Answer	Marks	Guidance
(a)	\(\sin(A+B) \equiv \sin A \cos B + \cos A \sin B\)
\(\sin(A-B) \equiv \sin A \cos B - \cos A \sin B\)
Adding: \(\sin(A+B) + \sin(A-B) = 2\sin A \cos B\)	M1 A1
Let \(P = A + B, \quad Q = A - B\)
Adding: \(P + Q = 2A \Rightarrow A = \frac{P+Q}{2}\)	M1
Subtracting: \(P - Q = 2B \Rightarrow B = \frac{P-Q}{2}\)
\(\sin P + \sin Q \equiv 2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}\)	A1
(b)	\(2\sin 3x \cos 2x = 0\)	M1
\(\sin 3x = 0\) or \(\cos 2x = 0\)	A1
\(3x = 0, \pi, 2\pi\) or \(2x = \frac{\pi}{2}, \frac{3\pi}{2}\)	M1
\(x = 0, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{3\pi}{4}\)	A2
(9)