Question 4 - A-Level Maths

Edexcel C3 — Question 4 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	Challenging +1.2 Part (a) requires applying standard sum-to-product identities and simplifying to reach a given result—a structured proof with clear steps. Part (b) demands choosing appropriate values (x=67.5°, y=37.5°) to apply the identity, then evaluating exact trigonometric values involving nested surds, which requires careful algebraic manipulation beyond routine exercises. The multi-step reasoning and non-standard exact value make this moderately challenging.
Spec	1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05l Double angle formulae: and compound angle formulae

4. (a) Use the identities for ( $\sin A + \sin B$ ) and ( $\cos A + \cos B$ ) to prove that $$\frac { \sin 2 x + \sin 2 y } { \cos 2 x + \cos 2 y } \equiv \tan ( x + y ) .$$ (b) Hence, show that $$\tan 52.5 ^ { \circ } = \sqrt { 6 } - \sqrt { 3 } - \sqrt { 2 } + 2 .$$

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Answer	Marks	Guidance
(a) $\text{LHS} = \frac{2\sin(x+y)\cos(x-y)}{2\cos(x+y)\cos(x-y)}$	M1 A1
$\equiv \frac{\sin(x+y)}{\cos(x+y)} = \tan(x+y) = \text{RHS}$	M1 A1
(b) let $x = 30°$, $y = 22.5°$ $\therefore \tan(30 + 22.5) = \frac{\sin 60 + \sin 45}{\cos 60 + \cos 45}$	M1
$\tan 52.5 = \frac{\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} = \frac{\sqrt{3} + \sqrt{2}}{1 + \sqrt{2}}$	B1 A1
$= \frac{\sqrt{3} + \sqrt{2}}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}}$	M1
$= \frac{\sqrt{3} - \sqrt{6} + \sqrt{2} - 2}{1 - 2} = \sqrt{6} - \sqrt{3} - \sqrt{2} + 2$	A1	(9 marks)

**(a)** $\text{LHS} = \frac{2\sin(x+y)\cos(x-y)}{2\cos(x+y)\cos(x-y)}$ | M1 A1
$\equiv \frac{\sin(x+y)}{\cos(x+y)} = \tan(x+y) = \text{RHS}$ | M1 A1

**(b)** let $x = 30°$, $y = 22.5°$ $\therefore \tan(30 + 22.5) = \frac{\sin 60 + \sin 45}{\cos 60 + \cos 45}$ | M1
$\tan 52.5 = \frac{\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} = \frac{\sqrt{3} + \sqrt{2}}{1 + \sqrt{2}}$ | B1 A1
$= \frac{\sqrt{3} + \sqrt{2}}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}}$ | M1
$= \frac{\sqrt{3} - \sqrt{6} + \sqrt{2} - 2}{1 - 2} = \sqrt{6} - \sqrt{3} - \sqrt{2} + 2$ | A1 | (9 marks)

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

$$\frac { \sin 2 x + \sin 2 y } { \cos 2 x + \cos 2 y } \equiv \tan ( x + y ) .$$

(b) Hence, show that

$$\tan 52.5 ^ { \circ } = \sqrt { 6 } - \sqrt { 3 } - \sqrt { 2 } + 2 .$$

\hfill \mbox{\textit{Edexcel C3  Q4 [9]}}

This paper (8 questions)

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Q1 5 Q2 6 Q3 8 Q4 9 Q5 10 Q6 11 Q7 12 Q8 14

Answer	Marks	Guidance
(a) \(\text{LHS} = \frac{2\sin(x+y)\cos(x-y)}{2\cos(x+y)\cos(x-y)}\)	M1 A1
\(\equiv \frac{\sin(x+y)}{\cos(x+y)} = \tan(x+y) = \text{RHS}\)	M1 A1
(b) let \(x = 30°\), \(y = 22.5°\) \(\therefore \tan(30 + 22.5) = \frac{\sin 60 + \sin 45}{\cos 60 + \cos 45}\)	M1
\(\tan 52.5 = \frac{\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} = \frac{\sqrt{3} + \sqrt{2}}{1 + \sqrt{2}}\)	B1 A1
\(= \frac{\sqrt{3} + \sqrt{2}}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}}\)	M1
\(= \frac{\sqrt{3} - \sqrt{6} + \sqrt{2} - 2}{1 - 2} = \sqrt{6} - \sqrt{3} - \sqrt{2} + 2\)	A1	(9 marks)