IBDP Maths AHL 3.10 Compound angle identities AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
To find intersection points, set \( f(x) = g(x) \):
\( \cos x = \sin 2x \)
Using identity: \( \sin 2x = 2 \sin x \cos x \)
\( \cos x = 2 \sin x \cos x \)
\( \cos x (1 – 2 \sin x) = 0 \)
Solutions:
1. \( \cos x = 0 \Rightarrow x = \frac{\pi}{2} \) (point B)
2. \( 1 – 2 \sin x = 0 \Rightarrow \sin x = \frac{1}{2} \)
\( \Rightarrow x = \frac{\pi}{6} \) (point A) and \( x = \frac{5\pi}{6} \) (point C)
✅ x-coordinates: A at \( \frac{\pi}{6} \), C at \( \frac{5\pi}{6} \)
METHOD 1
Area \( R = \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} (\cos x – \sin 2x) \, dx \)
\( = \left[ \sin x + \frac{1}{2} \cos 2x \right]_{\frac{\pi}{2}}^{\frac{5\pi}{6}} \)
Evaluating:
At \( x = \frac{5\pi}{6} \): \( \sin \frac{5\pi}{6} = \frac{1}{2} \), \( \cos \frac{5\pi}{3} = \frac{1}{2} \)
At \( x = \frac{\pi}{2} \): \( \sin \frac{\pi}{2} = 1 \), \( \cos \pi = -1 \)
\( = \left( \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \right) – \left( 1 + \frac{1}{2} \cdot (-1) \right) \)
\( = \left( \frac{1}{2} + \frac{1}{4} \right) – \left( 1 – \frac{1}{2} \right) = \frac{3}{4} – \frac{1}{2} = \frac{1}{4} \)
METHOD 2
Separate integrals:
\( \int \cos x \, dx = \sin x \)
\( \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \)
Evaluated between limits:
\( \left( \sin \frac{5\pi}{6} – \sin \frac{\pi}{2} \right) – \left( -\frac{1}{2} \cos \frac{5\pi}{3} + \frac{1}{2} \cos \pi \right) \)
\( = \left( \frac{1}{2} – 1 \right) – \left( -\frac{1}{4} – \frac{1}{2} \right) = -\frac{1}{2} + \frac{3}{4} = \frac{1}{4} \)
✅ Area of R = \( \frac{1}{4} \)
(a) Intersection points occur when \( \cos x = \sin 2x \):
Using double-angle identity: \( \sin 2x = 2 \sin x \cos x \)
Thus: \( \cos x = 2 \sin x \cos x \)
Factorizing: \( \cos x (1 – 2 \sin x) = 0 \)
Solutions in \( [0, \pi] \):
1. \( \cos x = 0 \Rightarrow x = \frac{\pi}{2} \) (point B)
2. \( \sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6} \) (points A and C)
(b) Area calculation between \( x = \frac{\pi}{2} \) and \( x = \frac{5\pi}{6} \):
Since \( \cos x > \sin 2x \) in this interval:
Area \( R = \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} (\cos x – \sin 2x) \, dx \)
Antiderivatives:
\( \int \cos x \, dx = \sin x \)
\( \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \)
Evaluating at limits:
\( \left[ \sin x + \frac{1}{2} \cos 2x \right]_{\frac{\pi}{2}}^{\frac{5\pi}{6}} = \frac{1}{4} \)
▶️ Answer/Explanation
First find the lengths of \( AC \) and \( AB \):
\( AC = \sqrt{AD^2 + CD^2} = \sqrt{3^2 + 4^2} = 5 \) (A1)
\( AB = \sqrt{AD^2 + BD^2} = \sqrt{3^2 + 2^2} = \sqrt{13} \) (A1)
Find trigonometric ratios:
For angle \( \alpha \) in \( \triangle CAD \):
\( \cos \alpha = \frac{AD}{AC} = \frac{3}{5} \), \( \sin \alpha = \frac{CD}{AC} = \frac{4}{5} \) (A1)
For angle \( \beta \) in \( \triangle BAD \):
\( \cos \beta = \frac{AD}{AB} = \frac{3}{\sqrt{13}} \), \( \sin \beta = \frac{BD}{AB} = \frac{2}{\sqrt{13}} \) (A1)
Apply cosine difference identity:
\( \cos(\alpha – \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \) (M1)
\( = \left(\frac{3}{5}\right)\left(\frac{3}{\sqrt{13}}\right) + \left(\frac{4}{5}\right)\left(\frac{2}{\sqrt{13}}\right) \) (M1)
\( = \frac{9 + 8}{5\sqrt{13}} = \frac{17}{5\sqrt{13}} = \frac{17\sqrt{13}}{65} \) A1 N1
First find the lengths:
\( AC = 5 \), \( AB = \sqrt{13} \), \( BC = BD + DC = 6 \) (A1)
Use cosine rule for angle \( (\alpha + \beta) \):
\( \cos(\alpha + \beta) = \frac{AC^2 + AB^2 – BC^2}{2 \times AC \times AB} \) (M1)
\( = \frac{25 + 13 – 36}{2 \times 5 \times \sqrt{13}} = \frac{2}{10\sqrt{13}} = \frac{1}{5\sqrt{13}} \) A1
Using identity:
\( \cos(\alpha + \beta) + \cos(\alpha – \beta) = 2\cos \alpha \cos \beta \) (M1)
We know \( \cos \alpha = \frac{3}{5} \), \( \cos \beta = \frac{3}{\sqrt{13}} \) (A1)
Therefore:
\( \cos(\alpha – \beta) = 2\cos \alpha \cos \beta – \cos(\alpha + \beta) \)
\( = 2 \times \frac{3}{5} \times \frac{3}{\sqrt{13}} – \frac{1}{5\sqrt{13}} \)
\( = \frac{18}{5\sqrt{13}} – \frac{1}{5\sqrt{13}} = \frac{17}{5\sqrt{13}} = \frac{17\sqrt{13}}{65} \) A1 N1
▶️ Answer/Explanation
Using the sine subtraction formula:
\( \sin \left( (2n + 1)x – x \right) = \sin (2n + 1)x \cos x – \cos (2n + 1)x \sin x \) (M1A1)
\( = \sin 2nx \) (AG)
[2 marks]
Base Case (n=1):
LHS \( = \cos x \)
RHS \( = \frac{\sin 2x}{2\sin x} = \frac{2\sin x \cos x}{2\sin x} = \cos x \) (M1)
Thus true for n=1 (R1)
Inductive Step:
Assume true for n=k:
\( \cos x + \cos 3x + \cdots + \cos (2k – 1)x = \frac{\sin 2kx}{2\sin x} \) (M1)
For n=k+1:
LHS \( = \frac{\sin 2kx}{2\sin x} + \cos (2k + 1)x \) (A1)
\( = \frac{\sin 2kx + 2\cos (2k + 1)x \sin x}{2\sin x} \) (M1)
Using part (a):
\( = \frac{\sin (2k + 2)x}{2\sin x} = \frac{\sin 2(k + 1)x}{2\sin x} \) (M1A1)
Thus true for n=k+1 when true for n=k. By induction, true for all \( n \in \mathbb{Z}^+ \). (R1)
[12 marks]
Using part (b) with n=2:
\( \frac{\sin 4x}{2\sin x} = \frac{1}{2} \) ⇒ \( \sin 4x = \sin x \) (M1A1)
Solutions in \( (0, \pi) \):
1. \( 4x = \pi – x \) ⇒ \( x = \frac{\pi}{5} \) (A1)
2. \( 4x = 2\pi + x \) ⇒ \( x = \frac{2\pi}{3} \) (A1)
3. \( 4x = 3\pi – x \) ⇒ \( x = \frac{3\pi}{5} \) (A1)
Note: \( x = 0 \) is excluded from domain (R1)
[6 marks]
▶️ Answer/Explanation
Starting with the given equation:
\( \sin \left( x + \frac{\pi}{3} \right) = \sin x \cos \left( \frac{\pi}{3} \right) + \cos x \sin \left( \frac{\pi}{3} \right) \) (M1)
Substitute known values:
\( \frac{1}{2}\sin x + \frac{\sqrt{3}}{2}\cos x = 2\sin x \left( \frac{\sqrt{3}}{2} \right) \) (A1)
Simplify the equation:
\( \frac{1}{2}\sin x + \frac{\sqrt{3}}{2}\cos x = \sqrt{3}\sin x \)
Divide both sides by \( \cos x \):
\( \frac{1}{2}\tan x + \frac{\sqrt{3}}{2} = \sqrt{3}\tan x \) (M1)
Solve for \( \tan x \):
\( \tan x = \frac{\sqrt{3}}{2\sqrt{3} – 1} \) (A1)
Rationalize the denominator:
Multiply numerator and denominator by \( 2\sqrt{3} + 1 \):
\( \tan x = \frac{\sqrt{3}(2\sqrt{3} + 1)}{(2\sqrt{3})^2 – 1^2} = \frac{6 + \sqrt{3}}{12 – 1} = \frac{6 + \sqrt{3}}{11} \) (M1)
Multiply both sides by 11:
\( 11\tan x = 6 + \sqrt{3} \) (A1)
[6 marks]