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IB Mathematics AA Reciprocal trigonometric ratios Study Notes | New Syllabus

IB Mathematics AA Reciprocal trigonometric ratios Study Notes | New Syllabus

IB Mathematics AA Reciprocal trigonometric ratios Study Notes

IB Mathematics AA Reciprocal trigonometric ratios Study Notes Offer a clear explanation of Reciprocal trigonometric ratios , including various formula, rules, exam style questions as example to explain the topics. Worked Out  examples and common problem types provided here will be sufficient to cover for topic Reciprocal trigonometric ratios.

Reciprocal Trigonometric Ratios

Reciprocal trigonometric ratios extend the basic trigonometric functions (\(\sin\), \(\cos\), and \(\tan\)) by defining their reciprocals: \(\sec\theta\), \(\csc\theta\), and \(\cot\theta\). These ratios are essential in understanding advanced trigonometric identities and their applications.

Key Concepts

  • Definitions: The reciprocal trigonometric ratios are defined as:
    • \(\sec\theta = \frac{1}{\cos\theta}\), valid for \(\cos\theta \neq 0\).
    • \(\csc\theta = \frac{1}{\sin\theta}\), valid for \(\sin\theta \neq 0\).
    • \(\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}\), valid for \(\sin\theta \neq 0\).
  • Pythagorean Identities: Reciprocal trigonometric ratios satisfy these key identities:
    • \(1 + \tan^2\theta = \sec^2\theta\).
    • \(1 + \cot^2\theta = \csc^2\theta\).
  • Inverse Trigonometric Functions: The inverse functions allow calculation of angles from given trigonometric values:
    • \(f(x) = \arcsin x\), domain: \([-1, 1]\), range: \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
    • \(f(x) = \arccos x\), domain: \([-1, 1]\), range: \([0, \pi]\).
    • \(f(x) = \arctan x\), domain: \(\mathbb{R}\), range: \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
  •  

Examples

  • Example 1: Find \(\sec\theta\) and \(\cot\theta\) given \(\sin\theta = \frac{3}{5}\), \(\theta\) in the first quadrant.
    • Step 1: Use \(\sin^2\theta + \cos^2\theta = 1\) to find \(\cos\theta = \frac{4}{5}\).
    • Step 2: Calculate \(\sec\theta = \frac{1}{\cos\theta} = \frac{5}{4}\).
    • Step 3: Calculate \(\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{4}{3}\).
  • Example 2: Solve \(1 + \cot^2\theta = \csc^2\theta\) to verify values.
    • Step 1: Substitute \(\cot^2\theta = \frac{\cos^2\theta}{\sin^2\theta}\) and \(\csc^2\theta = \frac{1}{\sin^2\theta}\).
    • Step 2: Verify \(1 + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta}\).
    • Step 3: Simplify to show \(1 = \sin^2\theta + \cos^2\theta\), which holds true.
  •  

IB Mathematics AA SL Reciprocal trigonometric ratios Exam Style Worked Out Questions

Question

It is given that cosec θ \(\frac{3}{2}\) , where \(\frac{\pi }{2}< \Lambda < \frac{3\pi }{2}\) Find the exact value of cot q .

▶️Answer/Explanation

Ans: 

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 >o puts θ in the second quadrant)

\(cot \Theta = – \sqrt{\frac{5}{2}}\)

METHOD 2

Attempt to use 1+ cot2 θ = coses2 θ

 \(1+ cot^2 \Theta = \frac{9}{4}\)

\(cot^2 \Theta = \frac{5}{4}\) 

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

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

cot θ \(\frac{\sqrt{5}}{2}\)

METHOD 3

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

attempt to use sin2 θ + cos2θ =1

\(\frac{4}{9}\)+cos2 θ =1

cos2 θ  = \(\frac{5}{9}\)

cos θ = ± \(\frac{\sqrt{5}}{3}\)

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

cos θ =  \(\frac{\sqrt{5}}{3}\) 

cot θ =  -\(\frac{\sqrt{5}}{2}\)

Question

Show that \(\frac{{\cos A + \sin A}}{{\cos A – \sin A}} = \sec 2A + \tan 2A\) .

▶️Answer/Explanation

Markscheme

METHOD 1

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

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

[6 marks]

Examiners report

Solutions to this question were good in general with many candidates realising that multiplying the numerator and denominator by \((\cos A + \sin A)\) might be helpful.

 

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