IBDP Maths AHL 3.11 Relationships in trigonometric functions AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
The period \( T \) of \( f(x) = \sin(qx) \) is:
\( T = \frac{2\pi}{q} \)
Given the x-intercepts at \( x = 0, 2m, 4m \), we observe that \( 4m \) represents one full period:
\( 4m = \frac{2\pi}{q} \)
Solving for \( m \):
\( m = \frac{2\pi}{4q} = \frac{\pi}{2q} \)
For \( g(x) = 3\sin\left(\frac{2qx}{3}\right) \):
Amplitude = 3
Period \( T = \frac{2\pi}{\frac{2q}{3}} = \frac{3\pi}{q} = 3m \) (using part (a) result)
Key points for one period (0 to \( 3m \)):
- Start at (0,0)
- Maximum at \( \left(\frac{3m}{4}, 3\right) \)
- Midpoint at \( \left(\frac{3m}{2}, 0\right) \)
- Minimum at \( \left(\frac{9m}{4}, -3\right) \)
- End at \( (3m,0) \)
Repeat this pattern for the interval \( 3m \) to \( 6m \).
The graph should show:
- X-intercepts at \( x = 0, 1.5m, 3m, 4.5m, 6m \)
- Maxima at \( x = 0.75m \) and \( 3.75m \)
- Minima at \( x = 2.25m \) and \( 5.25m \)
▶️ Answer/Explanation
Start with the given equation:
\( \tan\theta + \cot\theta = 3 \)
Express \( \cot\theta \) as \( \frac{1}{\tan\theta} \):
\( \tan\theta + \frac{1}{\tan\theta} = 3 \)
Multiply through by \( \tan\theta \):
\( \tan^2\theta – 3\tan\theta + 1 = 0 \)
Solve the quadratic equation:
\( \tan\theta = \frac{3 \pm \sqrt{9 – 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \)
Calculate approximate values:
\( \tan\theta = 0.382 \) or \( 2.618 \)
Find corresponding angles:
\( \theta = \tan^{-1}(0.382) \approx 20.9^\circ \)
\( \theta = \tan^{-1}(2.618) \approx 69.1^\circ \)
Express in terms of sine and cosine:
\( \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = 3 \)
Combine fractions:
\( \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = 3 \)
Simplify numerator:
\( \frac{1}{\sin\theta\cos\theta} = 3 \)
Use double-angle identity:
\( \frac{2}{\sin 2\theta} = 3 \) ⇒ \( \sin 2\theta = \frac{2}{3} \)
Solve for \( \theta \):
\( 2\theta = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.81^\circ \) or \( 138.19^\circ \)
Thus:
\( \theta \approx 20.9^\circ \) or \( 69.1^\circ \)
▶️ Answer/Explanation
Key features of \( y = \cos(4x) \):
- Period: \( \frac{2\pi}{4} = \frac{\pi}{2} \)
- Amplitude: 1
- Zeros at \( x = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \)
Key features of \( y = \sec(4x) \):
- Vertical asymptotes where \( \cos(4x) = 0 \): \( x = \frac{\pi}{8} + \frac{k\pi}{4} \), \( k \in \mathbb{Z} \)
- Range: \( (-\infty, -1] \cup [1, \infty) \)
- Reciprocal of \( \cos(4x) \), so undefined at zeros of cosine
Solve \( \sec(4x) = -1 \):
This occurs when \( \cos(4x) = -1 \).
Within \( 0 \leq x \leq \pi \), solutions are:
\( 4x = \pi \) ⇒ \( x = \frac{\pi}{4} \)
\( 4x = 3\pi \) ⇒ \( x = \frac{3\pi}{4} \)
Solve \( \cos(4x) \leq 0 \):
From the graph, this occurs in the intervals where cosine is negative:
\( \frac{\pi}{8} \leq x \leq \frac{3\pi}{8} \) and \( \frac{5\pi}{8} \leq x \leq \frac{7\pi}{8} \)
▶️ Answer/Explanation
Solution:
For a water depth of 10 cm, set \( y = -10 \) (since depth is measured downward):
\( -10 = 16\sec\left(\frac{\pi x}{36}\right) – 32 \)
Rearrange the equation:
\( 16\sec\left(\frac{\pi x}{36}\right) = 22 \)
\( \sec\left(\frac{\pi x}{36}\right) = \frac{22}{16} = \frac{11}{8} \)
Convert to cosine:
\( \cos\left(\frac{\pi x}{36}\right) = \frac{8}{11} \)
Solve for \( x \):
\( \frac{\pi x}{36} = \pm \arccos\left(\frac{8}{11}\right) \)
\( x = \pm \frac{36}{\pi} \arccos\left(\frac{8}{11}\right) \)
The total width is the distance between the two x-values:
\( \text{Width} = \frac{36}{\pi} \arccos\left(\frac{8}{11}\right) – \left(-\frac{36}{\pi} \arccos\left(\frac{8}{11}\right)\right) \)
\( \text{Width} = \frac{72}{\pi} \arccos\left(\frac{8}{11}\right) \) cm