Question 7 - A-Level Maths

Edexcel C3 — Question 7 12 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Addition & Double Angle Formulae
Type	Solve with multiple compound angles
Difficulty	Standard +0.3 This is a standard C3 compound angle question with routine application of addition formulae and a trigonometric equation. Part (a)(i) requires expanding sin(x+30) and sin(x-30) using standard formulae, part (a)(ii) is direct substitution, and part (b) uses the standard identity cot²y + 1 = cosec²y to reduce to a quadratic. All techniques are textbook exercises with no novel insight required, making it slightly easier than average.
Spec	1.05g Exact trigonometric values: for standard angles 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs 1.05l Double angle formulae: and compound angle formulae 1.05o Trigonometric equations: solve in given intervals

7. (a) (i) Show that $$\sin ( x + 30 ) ^ { \circ } + \sin ( x - 30 ) ^ { \circ } \equiv a \sin x ^ { \circ }$$ where $a$ is a constant to be found.
(ii) Hence find the exact value of $\sin 75 ^ { \circ } + \sin 15 ^ { \circ }$, giving your answer in the form $b \sqrt { 6 }$.
(b) Solve, for $0 \leq y \leq 360$, the equation $$2 \cot ^ { 2 } y ^ { \circ } + 5 \operatorname { cosec } y ^ { \circ } + \operatorname { cosec } ^ { 2 } y ^ { \circ } = 0$$

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Answer	Marks	Guidance
(a)(i) LHS $= \sin x\cos 30° + \cos x \sin 30° + \sin x \cos 30° - \cos x \sin 30°$	M1 A1
$= 2\sin x \cos 30° = \sqrt{3}\sin x$	A1	$[a = \sqrt{3}]$
(ii) Let $x = 45°$, $\sin 75° + \sin 15° = \sqrt{3}\sin 45° = \sqrt{3} \times \frac{1}{\sqrt{2}} = \frac{1}{2}\sqrt{6}$	M2 A1
(b) $2(\cosec^2 y - 1) + 5\cosec y + \cos^2 y = 0$	M1
$3\cosec^2 y + 5\cosec y - 2 = 0$, $(3\cosec y - 1)(\cosec y + 2) = 0$	M1
$\cosec y = -2$ or $\frac{1}{3}$ (no solutions)	A1
$\sin y = -\frac{1}{2}$	B1 M1
$y = 180° + 30°, 360° - 30°$	A1
$y = 210, 330$	A1	(12 marks)

**(a)(i)** LHS $= \sin x\cos 30° + \cos x \sin 30° + \sin x \cos 30° - \cos x \sin 30°$ | M1 A1 |

$= 2\sin x \cos 30° = \sqrt{3}\sin x$ | A1 | $[a = \sqrt{3}]$

**(ii)** Let $x = 45°$, $\sin 75° + \sin 15° = \sqrt{3}\sin 45° = \sqrt{3} \times \frac{1}{\sqrt{2}} = \frac{1}{2}\sqrt{6}$ | M2 A1 |

**(b)** $2(\cosec^2 y - 1) + 5\cosec y + \cos^2 y = 0$ | M1 |

$3\cosec^2 y + 5\cosec y - 2 = 0$, $(3\cosec y - 1)(\cosec y + 2) = 0$ | M1 |

$\cosec y = -2$ or $\frac{1}{3}$ (no solutions) | A1 |

$\sin y = -\frac{1}{2}$ | B1 M1 |

$y = 180° + 30°, 360° - 30°$ | A1 |

$y = 210, 330$ | A1 | (12 marks)

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7. (a) (i) Show that

$$\sin ( x + 30 ) ^ { \circ } + \sin ( x - 30 ) ^ { \circ } \equiv a \sin x ^ { \circ }$$

where $a$ is a constant to be found.\\
(ii) Hence find the exact value of $\sin 75 ^ { \circ } + \sin 15 ^ { \circ }$, giving your answer in the form $b \sqrt { 6 }$.\\
(b) Solve, for $0 \leq y \leq 360$, the equation

$$2 \cot ^ { 2 } y ^ { \circ } + 5 \operatorname { cosec } y ^ { \circ } + \operatorname { cosec } ^ { 2 } y ^ { \circ } = 0$$

\hfill \mbox{\textit{Edexcel C3  Q7 [12]}}

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Q1 4 Q2 6 Q3 8 Q4 10 Q5 10 Q6 12 Q7 12 Q8 13

Answer	Marks	Guidance
(a)(i) LHS \(= \sin x\cos 30° + \cos x \sin 30° + \sin x \cos 30° - \cos x \sin 30°\)	M1 A1
\(= 2\sin x \cos 30° = \sqrt{3}\sin x\)	A1	\([a = \sqrt{3}]\)
(ii) Let \(x = 45°\), \(\sin 75° + \sin 15° = \sqrt{3}\sin 45° = \sqrt{3} \times \frac{1}{\sqrt{2}} = \frac{1}{2}\sqrt{6}\)	M2 A1
(b) \(2(\cosec^2 y - 1) + 5\cosec y + \cos^2 y = 0\)	M1
\(3\cosec^2 y + 5\cosec y - 2 = 0\), \((3\cosec y - 1)(\cosec y + 2) = 0\)	M1
\(\cosec y = -2\) or \(\frac{1}{3}\) (no solutions)	A1
\(\sin y = -\frac{1}{2}\)	B1 M1
\(y = 180° + 30°, 360° - 30°\)	A1
\(y = 210, 330\)	A1	(12 marks)