Home / IBDP Maths AHL 3.9 Reciprocal trigonometric ratios AA HL Paper 1- Exam Style Questions

IBDP Maths AHL 3.9 Reciprocal trigonometric ratios AA HL Paper 1- Exam Style Questions

IBDP Maths AHL 3.9 Reciprocal trigonometric ratios AA HL Paper 1- Exam Style Questions- New Syllabus

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Question
It is given that cosec θ = \(\frac{3}{2}\), where \(\frac{\pi}{2} < \theta < \frac{3\pi}{2}\). Find the exact value of cot θ.
▶️ Answer/Explanation
Solution
METHOD 1

Attempt to use a right-angled triangle:

Correct placement of all three values and θ seen in the triangle

cot θ < 0 (since cosec θ > 0 puts θ in the second quadrant)

\( \cot \theta = -\frac{\sqrt{5}}{2} \)

METHOD 2

Attempt to use \( 1 + \cot^2 \theta = \csc^2 \theta \)

\( 1 + \cot^2 \theta = \frac{9}{4} \)

\( \cot^2 \theta = \frac{5}{4} \)

\( \cot \theta = \pm \frac{\sqrt{5}}{2} \)

cot θ < 0 (since cosec θ > 0 puts θ in the second quadrant)

\( \cot \theta = -\frac{\sqrt{5}}{2} \)

METHOD 3

\( \sin \theta = \frac{2}{3} \)

Attempt to use \( \sin^2 \theta + \cos^2 \theta = 1 \)

\( \frac{4}{9} + \cos^2 \theta = 1 \)

\( \cos^2 \theta = \frac{5}{9} \)

\( \cos \theta = \pm \frac{\sqrt{5}}{3} \)

cos θ < 0 (since θ is in the second quadrant)

\( \cos \theta = -\frac{\sqrt{5}}{3} \)

\( \cot \theta = \frac{\cos \theta}{\sin \theta} = -\frac{\sqrt{5}}{2} \)

✅ Final Answer: \( \cot \theta = -\frac{\sqrt{5}}{2} \)
Question
Show that \(\frac{\cos A + \sin A}{\cos A – \sin A} = \sec 2A + \tan 2A\).
▶️ Answer/Explanation
Solution
METHOD 1

Consider right hand side:

\(\sec 2A + \tan 2A = \frac{1}{\cos 2A} + \frac{\sin 2A}{\cos 2A}\) (M1A1)

\(= \frac{\cos^2 A + 2\sin A \cos A + \sin^2 A}{\cos^2 A – \sin^2 A}\) (A1A1)

Note: Award A1 for recognizing the need for single angles and A1 for recognizing \(\cos^2 A + \sin^2 A = 1\).

\(= \frac{(\cos A + \sin A)^2}{(\cos A + \sin A)(\cos A – \sin A)}\) (M1A1)

\(= \frac{\cos A + \sin A}{\cos A – \sin A}\) (AG)

METHOD 2

\(\frac{\cos A + \sin A}{\cos A – \sin A} = \frac{(\cos A + \sin A)^2}{(\cos A + \sin A)(\cos A – \sin A)}\) (M1A1)

\(= \frac{\cos^2 A + 2\sin A \cos A + \sin^2 A}{\cos^2 A – \sin^2 A}\) (A1A1)

Note: Award A1 for correct numerator and A1 for correct denominator.

\(= \frac{1 + \sin 2A}{\cos 2A}\) (M1A1)

\(= \sec 2A + \tan 2A\) (AG)

 
Question
a. Show that \(\cot \alpha = \tan \left( \frac{\pi}{2} – \alpha \right)\) for \(0 < \alpha < \frac{\pi}{2}\).
b. Hence find \(\int_{\tan \alpha}^{\cot \alpha} \frac{1}{1 + x^2} \, dx\), \(0 < \alpha < \frac{\pi}{2}\).
▶️ Answer/Explanation
Solution (a)
EITHER

Use of a diagram and trig ratios:

Right triangle diagram

\(\tan \alpha = \frac{O}{A} \Rightarrow \cot \alpha = \frac{A}{O}\)

From diagram, \(\tan \left( \frac{\pi}{2} – \alpha \right) = \frac{A}{O}\) (R1)

OR

Use of trigonometric identity:

\(\tan \left( \frac{\pi}{2} – \alpha \right) = \frac{\sin \left( \frac{\pi}{2} – \alpha \right)}{\cos \left( \frac{\pi}{2} – \alpha \right)} = \frac{\cos \alpha}{\sin \alpha}\) (R1)

THEN

\(\cot \alpha = \tan \left( \frac{\pi}{2} – \alpha \right)\) (AG)

[1 mark]

Solution (b)

\(\int_{\tan \alpha}^{\cot \alpha} \frac{1}{1 + x^2} \, dx = [\arctan x]_{\tan \alpha}^{\cot \alpha}\) (A1)

Note: Limits may be ignored at this stage.

\(= \arctan (\cot \alpha) – \arctan (\tan \alpha)\) (M1)

Using part (a):

\(= \arctan \left( \tan \left( \frac{\pi}{2} – \alpha \right) \right) – \alpha\)

\(= \frac{\pi}{2} – \alpha – \alpha\) (A1)

\(= \frac{\pi}{2} – 2\alpha\) A1

[4 marks]

✅ Final Answer: \(\frac{\pi}{2} – 2\alpha\)
Question
a. Find the value of \(\sin \frac{\pi}{4} + \sin \frac{3\pi}{4} + \sin \frac{5\pi}{4} + \sin \frac{7\pi}{4} + \sin \frac{9\pi}{4}\).
b. Show that \(\frac{1 – \cos 2x}{2\sin x} \equiv \sin x\), \(x \ne k\pi\) where \(k \in \mathbb{Z}\).
c. Use the principle of mathematical induction to prove that: \[\sin x + \sin 3x + \cdots + \sin (2n – 1)x = \frac{1 – \cos 2nx}{2\sin x}\] for \(n \in \mathbb{Z}^+\), \(x \ne k\pi\) where \(k \in \mathbb{Z}\).
d. Hence or otherwise solve \(\sin x + \sin 3x = \cos x\) in \(0 < x < \pi\).
▶️ Answer/Explanation
Solution (a)

\(\sin \frac{\pi}{4} + \sin \frac{3\pi}{4} + \sin \frac{5\pi}{4} + \sin \frac{7\pi}{4} + \sin \frac{9\pi}{4}\)

\(= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} – \frac{\sqrt{2}}{2} – \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\) (M1)A1

Note: Award M1 for 5 terms with correct signs.

[2 marks]

Solution (b)

\(\frac{1 – \cos 2x}{2\sin x} \equiv \frac{1 – (1 – 2\sin^2 x)}{2\sin x}\) M1

\(\equiv \frac{2\sin^2 x}{2\sin x}\) A1

\(\equiv \sin x\) AG

[2 marks]

Solution (c)

Base case (n=1):

\(\frac{1 – \cos 2x}{2\sin x} \equiv \sin x\) (from part b) R1

Inductive step:

Assume true for n=k: \(\sum_{i=1}^k \sin(2i-1)x = \frac{1 – \cos 2kx}{2\sin x}\) M1

For n=k+1:

\(\sum_{i=1}^{k+1} \sin(2i-1)x = \frac{1 – \cos 2kx}{2\sin x} + \sin(2k+1)x\) A1

\(= \frac{1 – \cos 2kx + 2\sin x \sin(2k+1)x}{2\sin x}\) M1

Using trigonometric identities:

\(= \frac{1 – [\cos 2kx – 2\sin x \sin(2k+1)x]}{2\sin x}\) M1

\(= \frac{1 – \cos(2kx + 2x)}{2\sin x}\) A1

\(= \frac{1 – \cos 2(k+1)x}{2\sin x}\) A1

Conclusion: True for n=1, and if true for n=k then true for n=k+1.

Therefore true for all positive integers n by induction. R1

[9 marks]

Solution (d)

Using part (c) with n=2:

\(\sin x + \sin 3x = \frac{1 – \cos 4x}{2\sin x} = \cos x\) M1

\(\Rightarrow 1 – \cos 4x = 2\sin x \cos x\) A1

\(\Rightarrow 2\sin^2 2x = \sin 2x\) M1

\(\Rightarrow \sin 2x(2\sin 2x – 1) = 0\) M1

Solutions in \(0 < x < \pi\):

\(\sin 2x = 0 \Rightarrow x = \frac{\pi}{2}\)

\(\sin 2x = \frac{1}{2} \Rightarrow x = \frac{\pi}{12}, \frac{5\pi}{12}\) A1A1

✅ Final solutions: \(x = \frac{\pi}{12}, \frac{\pi}{2}, \frac{5\pi}{12}\)

[6 marks]

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