1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1

710 questions

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CAIE P1 2020 June Q7
8 marks Standard +0.3
7
  1. Show that \(\frac { \tan \theta } { 1 + \cos \theta } + \frac { \tan \theta } { 1 - \cos \theta } \equiv \frac { 2 } { \sin \theta \cos \theta }\).
  2. Hence solve the equation \(\frac { \tan \theta } { 1 + \cos \theta } + \frac { \tan \theta } { 1 - \cos \theta } = \frac { 6 } { \tan \theta }\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2021 June Q7
5 marks Standard +0.3
7
  1. Prove the identity \(\frac { 1 - 2 \sin ^ { 2 } \theta } { 1 - \sin ^ { 2 } \theta } \equiv 1 - \tan ^ { 2 } \theta\).
  2. Hence solve the equation \(\frac { 1 - 2 \sin ^ { 2 } \theta } { 1 - \sin ^ { 2 } \theta } = 2 \tan ^ { 4 } \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
CAIE P1 2021 June Q10
7 marks Standard +0.3
10
  1. Prove the identity \(\frac { 1 + \sin x } { 1 - \sin x } - \frac { 1 - \sin x } { 1 + \sin x } \equiv \frac { 4 \tan x } { \cos x }\).
  2. Hence solve the equation \(\frac { 1 + \sin x } { 1 - \sin x } - \frac { 1 - \sin x } { 1 + \sin x } = 8 \tan x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
CAIE P1 2021 June Q4
6 marks Standard +0.3
4
  1. Show that the equation $$\frac { \tan x + \sin x } { \tan x - \sin x } = k$$ where \(k\) is a constant, may be expressed as $$\frac { 1 + \cos x } { 1 - \cos x } = k$$
  2. Hence express \(\cos x\) in terms of \(k\).
  3. Hence solve the equation \(\frac { \tan x + \sin x } { \tan x - \sin x } = 4\) for \(- \pi < x < \pi\).
CAIE P1 2022 June Q4
6 marks Standard +0.3
4
  1. Prove the identity \(\frac { \sin ^ { 3 } \theta } { \sin \theta - 1 } - \frac { \sin ^ { 2 } \theta } { 1 + \sin \theta } \equiv - \tan ^ { 2 } \theta \left( 1 + \sin ^ { 2 } \theta \right)\).
  2. Hence solve the equation $$\frac { \sin ^ { 3 } \theta } { \sin \theta - 1 } - \frac { \sin ^ { 2 } \theta } { 1 + \sin \theta } = \tan ^ { 2 } \theta \left( 1 - \sin ^ { 2 } \theta \right)$$ for \(0 < \theta < 2 \pi\).
CAIE P1 2022 June Q11
10 marks Standard +0.3
11 The function f is given by \(\mathrm { f } ( x ) = 4 \cos ^ { 4 } x + \cos ^ { 2 } x - k\) for \(0 \leqslant x \leqslant 2 \pi\), where \(k\) is a constant.
  1. Given that \(k = 3\), find the exact solutions of the equation \(\mathrm { f } ( x ) = 0\).
  2. Use the quadratic formula to show that, when \(k > 5\), the equation \(\mathrm { f } ( x ) = 0\) has no solutions.
    If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
CAIE P1 2023 June Q1
3 marks Moderate -0.3
1 Solve the equation \(4 \sin \theta + \tan \theta = 0\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2023 June Q4
7 marks Standard +0.3
4
  1. Show that the equation $$3 \tan ^ { 2 } x - 3 \sin ^ { 2 } x - 4 = 0$$ may be expressed in the form \(a \cos ^ { 4 } x + b \cos ^ { 2 } x + c = 0\), where \(a , b\) and \(c\) are constants to be found.
  2. Hence solve the equation \(3 \tan ^ { 2 } x - 3 \sin ^ { 2 } x - 4 = 0\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).
CAIE P1 2024 June Q4
6 marks Standard +0.3
4
  1. Show that the equation \(\cos \theta ( 7 \tan \theta - 5 \cos \theta ) = 1\) can be written in the form \(a \sin ^ { 2 } \theta + b \sin \theta + c = 0\), where \(a , b\) and \(c\) are integers to be found.
  2. Hence solve the equation \(\cos 2 x ( 7 \tan 2 x - 5 \cos 2 x ) = 1\) for \(0 ^ { \circ } < x < 180 ^ { \circ }\). \includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-06_2718_35_141_2012} \includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-07_2714_33_144_22}
CAIE P1 2020 March Q5
5 marks Standard +0.8
5 Solve the equation $$\frac { \tan \theta + 3 \sin \theta + 2 } { \tan \theta - 3 \sin \theta + 1 } = 2$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 90 ^ { \circ }\).
CAIE P1 2021 March Q3
4 marks Standard +0.3
3 Solve the equation \(\frac { \tan \theta + 2 \sin \theta } { \tan \theta - 2 \sin \theta } = 3\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2022 March Q7
7 marks Standard +0.8
7
  1. Show that \(\frac { \sin \theta + 2 \cos \theta } { \cos \theta - 2 \sin \theta } - \frac { \sin \theta - 2 \cos \theta } { \cos \theta + 2 \sin \theta } \equiv \frac { 4 } { 5 \cos ^ { 2 } \theta - 4 }\).
  2. Hence solve the equation \(\frac { \sin \theta + 2 \cos \theta } { \cos \theta - 2 \sin \theta } - \frac { \sin \theta - 2 \cos \theta } { \cos \theta + 2 \sin \theta } = 5\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2023 March Q7
8 marks Moderate -0.3
7
  1. By first obtaining a quadratic equation in \(\cos \theta\), solve the equation $$\tan \theta \sin \theta = 1$$ for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
  2. Show that \(\frac { \tan \theta } { \sin \theta } - \frac { \sin \theta } { \tan \theta } \equiv \tan \theta \sin \theta\).
CAIE P1 2024 March Q4
6 marks Moderate -0.3
4
  1. Prove that \(\frac { ( \sin \theta + \cos \theta ) ^ { 2 } - 1 } { \cos ^ { 2 } \theta } \equiv 2 \tan \theta\). \includegraphics[max width=\textwidth, alt={}]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_63_1569_333_328} .............................................................................................................................................. \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_65_1570_511_324} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_62_1570_603_324} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_72_1570_685_324} \includegraphics[max width=\textwidth, alt={}]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_72_1570_776_324} ...................................................................................................................................... . \includegraphics[max width=\textwidth, alt={}]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_76_1572_952_322} ........................................................................................................................................ \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_72_1570_1137_324} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_74_1572_1226_322} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_77_1575_1315_319}
  2. Hence solve the equation \(\frac { ( \sin \theta + \cos \theta ) ^ { 2 } - 1 } { \cos ^ { 2 } \theta } = 5 \tan ^ { 3 } \theta\) for \(- 90 ^ { \circ } < \theta < 90 ^ { \circ }\).
CAIE P1 2020 November Q7
6 marks Standard +0.3
7
  1. Show that \(\frac { \sin \theta } { 1 - \sin \theta } - \frac { \sin \theta } { 1 + \sin \theta } \equiv 2 \tan ^ { 2 } \theta\).
  2. Hence solve the equation \(\frac { \sin \theta } { 1 - \sin \theta } - \frac { \sin \theta } { 1 + \sin \theta } = 8\), for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2020 November Q6
6 marks Moderate -0.3
6
  1. Prove the identity \(\left( \frac { 1 } { \cos x } - \tan x \right) \left( \frac { 1 } { \sin x } + 1 \right) \equiv \frac { 1 } { \tan x }\).
  2. Hence solve the equation \(\left( \frac { 1 } { \cos x } - \tan x \right) \left( \frac { 1 } { \sin x } + 1 \right) = 2 \tan ^ { 2 } x\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).
CAIE P1 2020 November Q3
5 marks Standard +0.3
3 Solve the equation \(3 \tan ^ { 2 } \theta + 1 = \frac { 2 } { \tan ^ { 2 } \theta }\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2020 November Q7
7 marks Standard +0.3
7 The first and second terms of an arithmetic progression are \(\frac { 1 } { \cos ^ { 2 } \theta }\) and \(- \frac { \tan ^ { 2 } \theta } { \cos ^ { 2 } \theta }\), respectively, where \(0 < \theta < \frac { 1 } { 2 } \pi\).
  1. Show that the common difference is \(- \frac { 1 } { \cos ^ { 4 } \theta }\).
  2. Find the exact value of the 13th term when \(\theta = \frac { 1 } { 6 } \pi\).
CAIE P1 2021 November Q1
4 marks Standard +0.3
1 Solve the equation \(2 \cos \theta = 7 - \frac { 3 } { \cos \theta }\) for \(- 90 ^ { \circ } < \theta < 90 ^ { \circ }\).
CAIE P1 2021 November Q7
8 marks Standard +0.3
7
  1. Show that the equation \(\frac { \tan x + \cos x } { \tan x - \cos x } = k\), where \(k\) is a constant, can be expressed as $$( k + 1 ) \sin ^ { 2 } x + ( k - 1 ) \sin x - ( k + 1 ) = 0$$
  2. Hence solve the equation \(\frac { \tan x + \cos x } { \tan x - \cos x } = 4\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
CAIE P1 2022 November Q6
6 marks Standard +0.3
6
  1. Show that the equation $$\frac { 1 } { \sin \theta + \cos \theta } + \frac { 1 } { \sin \theta - \cos \theta } = 1$$ may be expressed in the form \(a \sin ^ { 2 } \theta + b \sin \theta + c = 0\), where \(a\), \(b\) and \(c\) are constants to be found.
  2. Hence solve the equation \(\frac { 1 } { \sin \theta + \cos \theta } + \frac { 1 } { \sin \theta - \cos \theta } = 1\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
CAIE P1 2022 November Q7
7 marks Standard +0.3
7
  1. Prove the identity \(\frac { \sin \theta } { \sin \theta + \cos \theta } + \frac { \cos \theta } { \sin \theta - \cos \theta } \equiv \frac { \tan ^ { 2 } \theta + 1 } { \tan ^ { 2 } \theta - 1 }\).
  2. Hence find the exact solutions of the equation \(\frac { \sin \theta } { \sin \theta + \cos \theta } + \frac { \cos \theta } { \sin \theta - \cos \theta } = 2\) for \(0 \leqslant \theta \leqslant \pi\).
CAIE P1 2022 November Q1
3 marks Moderate -0.3
1 Solve the equation \(8 \sin ^ { 2 } \theta + 6 \cos \theta + 1 = 0\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P1 2022 November Q6
5 marks Moderate -0.3
6 It is given that \(\alpha = \cos ^ { - 1 } \left( \frac { 8 } { 17 } \right)\).
Find, without using the trigonometric functions on your calculator, the exact value of \(\frac { 1 } { \sin \alpha } + \frac { 1 } { \tan \alpha }\).
CAIE P1 2023 November Q5
6 marks Standard +0.3
5
  1. Show that the equation $$4 \sin x + \frac { 5 } { \tan x } + \frac { 2 } { \sin x } = 0$$ may be expressed in the form \(a \cos ^ { 2 } x + b \cos x + c = 0\), where \(a , b\) and \(c\) are integers to be found.
  2. Hence solve the equation \(4 \sin x + \frac { 5 } { \tan x } + \frac { 2 } { \sin x } = 0\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).