1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs

182 questions

Sort by: Default | Easiest first | Hardest first
OCR MEI AS Paper 2 Specimen Q12
3 marks Standard +0.3
12 Given that \(\arcsin x = \arccos y\), prove that \(x ^ { 2 } + y ^ { 2 } = 1\). [Hint: Let \(\arcsin x = \theta\) ] \section*{END OF QUESTION PAPER}
OCR MEI Paper 1 2022 June Q3
8 marks Challenging +1.2
3
  1. Sketch the graph of \(\mathrm { y } = \arctan \mathrm { x }\) where \(x\) is in radians.
  2. In this question you must show detailed reasoning. Find all points of intersection of the curves \(\mathrm { y } = 3 \sin \mathrm { xcos } \mathrm { x }\) and \(\mathrm { y } = \cos ^ { 2 } \mathrm { x }\) for \(- \pi \leqslant x \leqslant \pi\).
OCR MEI Paper 3 2021 November Q12
3 marks Moderate -0.5
12 Show that \(\beta = \arctan \left( \frac { 1 } { 3 } \right)\), as given in line 15 .
OCR MEI Paper 3 2021 November Q13
3 marks Moderate -0.8
13
  1. Use triangle ABE in Fig. C 2 to show that \(\arctan x + \arctan \left( \frac { 1 } { x } \right) = \frac { \pi } { 2 }\), as given in line 29 .
  2. Sketch the graph of \(\mathrm { y } = \arctan \mathrm { x }\).
  3. What property of the arctan function ensures that \(\mathrm { y } > \frac { 1 } { \mathrm { x } } \Rightarrow \arctan y > \arctan \left( \frac { 1 } { \mathrm { x } } \right)\), as given in line 30 ?
OCR MEI Paper 3 2021 November Q14
5 marks Challenging +1.2
14
  1. Show that $$\arctan \left( \frac { 1 } { n + 1 } \right) + \arctan \left( \frac { 1 } { n ^ { 2 } + n + 1 } \right) = \arctan \left( \frac { 1 } { n } \right) \Rightarrow \arctan \left( \frac { 1 } { 2 } \right) + \arctan \left( \frac { 1 } { 3 } \right) = \arctan 1 .$$
  2. Use the arctan addition formula in line 23 to show that $$\arctan \left( \frac { 1 } { n + 1 } \right) + \arctan \left( \frac { 1 } { n ^ { 2 } + n + 1 } \right) = \arctan \left( \frac { 1 } { n } \right) , \text { as given in line } 39 .$$
OCR MEI Paper 3 2021 November Q15
4 marks Challenging +1.8
15 Prove that \(\arctan 1 + \arctan 2 + \arctan 3 = \pi\), as given in line 41 . \section*{END OF QUESTION PAPER} \section*{OCR
Oxford Cambridge and RSA}
AQA C2 2006 June Q8
11 marks Moderate -0.3
8
  1. Describe the single geometrical transformation by which the curve with equation \(y = \tan \frac { 1 } { 2 } x\) can be obtained from the curve \(y = \tan x\).
  2. Solve the equation \(\tan \frac { 1 } { 2 } x = 3\) in the interval \(\mathbf { 0 } < \boldsymbol { x } < \mathbf { 4 } \boldsymbol { \pi }\), giving your answers in radians to three significant figures.
  3. Solve the equation $$\cos \theta ( \sin \theta - 3 \cos \theta ) = 0$$ in the interval \(0 < \theta < 2 \pi\), giving your answers in radians to three significant figures.
    (5 marks)
AQA C3 2007 January Q8
7 marks Standard +0.3
8 The diagram shows the curve \(y = \cos ^ { - 1 } x\) for \(- 1 \leqslant x \leqslant 1\). \includegraphics[max width=\textwidth, alt={}, center]{6890a681-2b7f-4853-a5f0-f88b7b435367-4_492_698_1640_671}
  1. Write down the exact coordinates of the points \(A\) and \(B\).
  2. The equation \(\cos ^ { - 1 } x = 3 x + 1\) has only one root. Given that the root of this equation is \(\alpha\), show that \(0.1 \leqslant \alpha \leqslant 0.2\).
  3. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 3 } \left( \cos ^ { - 1 } x _ { n } - 1 \right)\) with \(x _ { 1 } = 0.1\) to find the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to three decimal places.
AQA C3 2011 January Q3
6 marks Moderate -0.3
3
  1. Given that \(x = \tan ( 3 y + 1 )\) :
    1. find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\);
    2. find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(y = - \frac { 1 } { 3 }\).
  2. Sketch the graph of \(y = \tan ^ { - 1 } x\).
AQA C3 2005 June Q7
9 marks Standard +0.3
7
  1. Sketch the graph of \(y = \tan ^ { - 1 } x\).
    1. By drawing a suitable straight line on your sketch, show that the equation \(\tan ^ { - 1 } x = 2 x - 1\) has only one root.
    2. Given that the root of this equation is \(\alpha\), show that \(0.8 < \alpha < 0.9\).
  2. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 2 } \left( \tan ^ { - 1 } x _ { n } + 1 \right)\) with \(x _ { 1 } = 0.8\) to find the value of \(x _ { 3 }\), giving your answer to two significant figures.
AQA C3 2006 June Q9
7 marks Standard +0.3
9 The diagram shows the curve with equation \(y = \sin ^ { - 1 } 2 x\), where \(- \frac { 1 } { 2 } \leqslant x \leqslant \frac { 1 } { 2 }\). \includegraphics[max width=\textwidth, alt={}, center]{0ab0e757-270b-4c15-9202-9df2f02dddf3-4_790_752_906_644}
  1. Find the \(y\)-coordinate of the point \(A\), where \(x = \frac { 1 } { 2 }\).
    1. Given that \(y = \sin ^ { - 1 } 2 x\), show that \(x = \frac { 1 } { 2 } \sin y\).
    2. Given that \(x = \frac { 1 } { 2 } \sin y\), find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\).
  2. Using the answers to part (b) and a suitable trigonometrical identity, show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { \sqrt { 1 - 4 x ^ { 2 } } }$$
AQA C3 2012 June Q9
14 marks Standard +0.3
9
  1. Given that \(x = \frac { \sin y } { \cos y }\), use the quotient rule to show that $$\frac { \mathrm { d } x } { \mathrm {~d} y } = \sec ^ { 2 } y$$ (3 marks)
  2. Given that \(\tan y = x - 1\), use a trigonometrical identity to show that $$\sec ^ { 2 } y = x ^ { 2 } - 2 x + 2$$
  3. Show that, if \(y = \tan ^ { - 1 } ( x - 1 )\), then $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { x ^ { 2 } - 2 x + 2 }$$ (l mark)
  4. A curve has equation \(y = \tan ^ { - 1 } ( x - 1 ) - \ln x\).
    1. Find the value of the \(x\)-coordinate of each of the stationary points of the curve.
    2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
    3. Hence show that the curve has a minimum point which lies on the \(x\)-axis.
AQA C3 2013 June Q6
4 marks Moderate -0.3
6
  1. Sketch the graph of \(y = \cos ^ { - 1 } x\), where \(y\) is in radians. State the coordinates of the end points of the graph.
  2. Sketch the graph of \(y = \pi - \cos ^ { - 1 } x\), where \(y\) is in radians. State the coordinates of the end points of the graph.
    (2 marks) \includegraphics[max width=\textwidth, alt={}, center]{063bbfa5-df49-44a1-8143-5e076397f63f-05_759_1258_678_431} \includegraphics[max width=\textwidth, alt={}, center]{063bbfa5-df49-44a1-8143-5e076397f63f-05_751_1241_1564_443}
Edexcel C3 Q7
14 marks Standard +0.3
7. (a) Solve the equation $$\pi - 3 \arccos \theta = 0$$ (b) Sketch on the same diagram the curves \(y = \arccos ( x - 1 ) , 0 \leq x \leq 2\) and \(y = \sqrt { x + 2 } , x \geq - 2\). Given that \(\alpha\) is the root of the equation $$\arccos ( x - 1 ) = \sqrt { x + 2 }$$ (c) show that \(0 < \alpha < 1\),
(d) use the iterative formula $$x _ { n + 1 } = 1 + \cos \sqrt { x _ { n } + 2 }$$ with \(x _ { 0 } = 1\) to find \(\alpha\) correct to 3 decimal places. END
Edexcel C3 Q3
11 marks Standard +0.8
3. $$f ( x ) = x ^ { 2 } + 5 x - 2 \sec x , \quad x \in \mathbb { R } , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 } .$$
  1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root in the interval [1,1.5]. A more accurate estimate of this root is to be found using iterations of the form $$x _ { n + 1 } = \arccos \mathrm { g } \left( x _ { n } \right) .$$
  2. Find a suitable form for \(\mathrm { g } ( x )\) and use this formula with \(x _ { 0 } = 1.25\) to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\). Give the value of \(x _ { 4 }\) to 3 decimal places. The curve \(y = \mathrm { f } ( x )\) has a stationary point at \(P\).
  3. Show that the \(x\)-coordinate of \(P\) is 1.0535 correct to 5 significant figures.
AQA FP2 2012 June Q5
6 marks Standard +0.8
5 The function f , where \(\mathrm { f } ( x ) = \sec x\), has domain \(0 \leqslant x < \frac { \pi } { 2 }\) and has inverse function \(\mathrm { f } ^ { - 1 }\), where \(\mathrm { f } ^ { - 1 } ( x ) = \sec ^ { - 1 } x\).
  1. Show that $$\sec ^ { - 1 } x = \cos ^ { - 1 } \frac { 1 } { x }$$
  2. Hence show that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 4 } - x ^ { 2 } } }$$
AQA FP2 2013 June Q6
8 marks Challenging +1.2
6
  1. Show that \(\frac { 1 } { 5 \cosh x - 3 \sinh x } = \frac { \mathrm { e } ^ { x } } { m + \mathrm { e } ^ { 2 x } }\), where \(m\) is an integer.
  2. Use the substitution \(u = \mathrm { e } ^ { x }\) to show that $$\int _ { 0 } ^ { \ln 2 } \frac { 1 } { 5 \cosh x - 3 \sinh x } \mathrm {~d} x = \frac { \pi } { 8 } - \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$$
AQA FP2 2014 June Q7
7 marks Challenging +1.2
7
  1. Given that \(y = \tan ^ { - 1 } \left( \frac { 1 + x } { 1 - x } \right)\) and \(x \neq 1\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }\).
    [0pt] [4 marks]
  2. Hence, given that \(x < 1\), show that \(\tan ^ { - 1 } \left( \frac { 1 + x } { 1 - x } \right) - \tan ^ { - 1 } x = \frac { \pi } { 4 }\).
    [0pt] [3 marks]
OCR MEI Further Pure Core 2021 November Q2
4 marks Standard +0.3
2 In this question you must show detailed reasoning.
Find the gradient of the curve \(y = 6 \arcsin ( 2 x )\) at the point with \(x\)-coordinate \(\frac { 1 } { 4 }\). Express the result in the form \(\mathrm { m } \sqrt { \mathrm { n } }\), where \(m\) and \(n\) are integers.
Edexcel CP2 2021 June Q5
8 marks Standard +0.8
  1. The curve \(C\) has equation
$$y = \arccos \left( \frac { 1 } { 2 } x \right) \quad - 2 \leqslant x \leqslant 2$$
  1. Show that \(C\) has no stationary points. The normal to \(C\), at the point where \(x = 1\), crosses the \(x\)-axis at the point \(A\) and crosses the \(y\)-axis at the point \(B\). Given that \(O\) is the origin,
  2. show that the area of the triangle \(O A B\) is \(\frac { 1 } { 54 } \left( p \sqrt { 3 } + q \pi + r \sqrt { 3 } \pi ^ { 2 } \right)\) where \(p\), \(q\) and \(r\) are integers to be determined.
    (5)
Edexcel CP2 2022 June Q5
6 marks Standard +0.3
  1. (a) Given that
$$y = \arcsin x \quad - 1 \leqslant x \leqslant 1$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$$ (b) $$\mathrm { f } ( x ) = \arcsin \left( \mathrm { e } ^ { x } \right) \quad x \leqslant 0$$ Prove that \(\mathrm { f } ( x )\) has no stationary points.
Edexcel AEA 2012 June Q7
24 marks Hard +2.3
7. \(\left[ \arccos x \right.\) and \(\arctan x\) are alternative notation for \(\cos ^ { - 1 } x\) and \(\tan ^ { - 1 } x\) respectively \(]\) \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fc5d0d07-b750-4646-bdcb-419a290200c9-5_387_935_322_566} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows a sketch of the curve \(C _ { 1 }\) with equation \(y = \cos ( \cos x ) , 0 \leqslant x < 2 \pi\) .
The curve has turning points at \(( 0 , \cos 1 ) , P , Q\) and \(R\) as shown in Figure 2.
  1. Find the coordinates of the points \(P , Q\) and \(R\) . The curve \(C _ { 2 }\) has equation \(y = \sin ( \cos x ) , 0 \leqslant x < 2 \pi\) .The curves \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the points \(S\) and \(T\) .
  2. Copy Figure 2 and on this diagram sketch \(C _ { 2 }\) stating the coordinates of the minimum point on \(C _ { 2 }\) and the points where \(C _ { 2 }\) meets or crosses the coordinate axes. The coordinates of \(S\) are \(( \alpha , d )\) where \(0 < \alpha < \pi\) .
  3. Show that \(\alpha = \arccos \left( \frac { \pi } { 4 } \right)\) .
  4. Find the value of \(d\) in surd form and write down the coordinates of \(T\) . The tangent to \(C _ { 1 }\) at the point \(S\) has gradient \(\tan \beta\) .
  5. Show that \(\beta = \arctan \sqrt { } \left( \frac { 16 - \pi ^ { 2 } } { 32 } \right)\) .
  6. Find,in terms of \(\beta\) ,the obtuse angle between the tangent to \(C _ { 1 }\) at \(S\) and the tangent to \(C _ { 2 }\) at \(S\) .
CAIE P3 2012 June Q7
8 marks Standard +0.3
  1. Express \(u\) in the form \(x + \mathrm { i } y\), where \(x\) and \(y\) are real.
  2. Show on a sketch of an Argand diagram the points \(A , B\) and \(C\) representing the complex numbers \(u , 1 + 2 \mathrm { i }\) and \(1 - 3 \mathrm { i }\) respectively.
  3. By considering the arguments of \(1 + 2 \mathrm { i }\) and \(1 - 3 \mathrm { i }\), show that $$\tan ^ { - 1 } 2 + \tan ^ { - 1 } 3 = \frac { 3 } { 4 } \pi$$
OCR Further Pure Core 1 2023 June Q9
14 marks Challenging +1.8
9 In this question you must show detailed reasoning.
  1. Use de Moivre's theorem to determine constants \(A\), \(B\) and \(C\) such that $$\sin ^ { 4 } \theta \equiv A \cos 4 \theta + B \cos 2 \theta + C .$$ The function f is defined by \(\mathrm { f } ( x ) = \sin \left( 4 \sin ^ { - 1 } \left( x ^ { \frac { 1 } { 5 } } \right) \right) - 8 \sin \left( 2 \sin ^ { - 1 } \left( x ^ { \frac { 1 } { 5 } } \right) \right) + 12 \sin ^ { - 1 } \left( x ^ { \frac { 1 } { 5 } } \right) , \quad x \in \mathbb { R } , 0 \leqslant x < 1\).
  2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 32 } { 5 \sqrt { 1 - x ^ { \frac { 2 } { 5 } } } }\). \includegraphics[max width=\textwidth, alt={}, center]{478c66d2-16a0-41ef-9444-25cfcd47d11d-7_894_842_1000_260} The diagram shows the curve with equation \(\mathrm { y } = \frac { 1 } { \sqrt { 1 - x ^ { \frac { 2 } { 5 } } } }\) for \(0 \leqslant x < 1\) and the asymptote \(x = 1\). The region \(R\) is the unbounded region between the curve, the \(x\)-axis, the line \(x = 0\) and the line \(x = 1\). You are given that the area of \(R\) is finite.
  3. Determine the exact area of \(R\).
OCR H240/01 2018 December Q10
13 marks Standard +0.3
10 \includegraphics[max width=\textwidth, alt={}, center]{a16ab26f-21fb-4a73-8b94-c16bef611bcb-7_524_714_274_246} The diagram shows the graph of \(y = - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x - \frac { 1 } { 3 } \pi \right)\), which crosses the \(x\)-axis at the point \(A\) and the \(y\)-axis at the point \(B\).
  1. Determine the coordinates of the points \(A\) and \(B\).
  2. Give full details of a sequence of three geometrical transformations which transform the graph of \(y = \tan ^ { - 1 } x\) to the graph of \(y = - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x - \frac { 1 } { 3 } \pi \right)\). The equation \(x = - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x - \frac { 1 } { 3 } \pi \right)\) has only one root.
  3. Show by calculation that this root lies between \(x = 0\) and \(x = 1\).
  4. Use the iterative formula \(x _ { n + 1 } = - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x _ { n } - \frac { 1 } { 3 } \pi \right)\), with a suitable starting value, to find the root correct to 3 significant figures. Show the result of each iteration.
  5. Using the diagram in the Printed Answer Booklet, show how the iterative process converges to the root.