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

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1. The sketch shows the graph of \(y = \sin ^ { - 1 } x\). \includegraphics[max width=\textwidth, alt={}, center]{9aac4ee4-2435-4315-a87d-fe9fa8e15665-006_819_824_456_591} Write down the coordinates of the points \(P\) and \(Q\), the end-points of the graph.
2. Sketch the graph of \(y = - \sin ^ { - 1 } ( x - 1 )\).

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1. The sketch shows the graph of \(y = \sin ^ { - 1 } x\). \includegraphics[max width=\textwidth, alt={}, center]{908f530c-076d-47b1-90dd-38dbfe44f898-05_835_834_447_587} Write down the coordinates of the points \(P\) and \(Q\), the end-points of the graph.
2. Sketch the graph of \(y = - \sin ^ { - 1 } ( x - 1 )\).

8 The sketch shows the graph of \(y = \cos ^ { - 1 } x\). \includegraphics[max width=\textwidth, alt={}, center]{59b896ae-60ce-49ea-9c70-0f76fc5fffae-5_593_686_383_683}

1. Write down the coordinates of \(P\) and \(Q\), the end points of the graph.
2. Describe a sequence of two geometrical transformations that maps the graph of \(y = \cos ^ { - 1 } x\) onto the graph of \(y = 2 \cos ^ { - 1 } ( x - 1 )\).
3. Sketch the graph of \(y = 2 \cos ^ { - 1 } ( x - 1 )\).
4. 1. Write the equation \(y = 2 \cos ^ { - 1 } ( x - 1 )\) in the form \(x = \mathrm { f } ( y )\).
   2. Hence find the value of \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) when \(y = 2\).

2 [Figure 1, printed on the insert, is provided for use in this question.]

1. 1. Sketch the graph of \(y = \sin ^ { - 1 } x\), where \(y\) is in radians. State the coordinates of the end points of the graph.
   2. By drawing a suitable straight line on your sketch, show that the equation $$\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1$$ has only one solution.
2. The root of the equation \(\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1\) is \(\alpha\). Show that \(0.5 < \alpha < 1\).
3. The equation \(\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1\) can be rewritten as \(x = \sin \left( \frac { 1 } { 4 } x + 1 \right)\).
   1. Use the iteration \(x _ { n + 1 } = \sin \left( \frac { 1 } { 4 } x _ { n } + 1 \right)\) with \(x _ { 1 } = 0.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   2. The sketch on Figure 1 shows parts of the graphs of \(y = \sin \left( \frac { 1 } { 4 } x + 1 \right)\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.

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1. Sketch, on the axes below, the curve with equation \(y = \sin ^ { - 1 } ( 3 x )\), where \(y\) is in radians. State the exact values of the coordinates of the end points of the graph.
2. Given that \(x = \frac { 1 } { 3 } \sin y\), write down \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) and hence find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\). \section*{Answer space for question 6}
   1. \includegraphics[max width=\textwidth, alt={}, center]{2df59047-3bfe-4b9c-a77f-142bc7506cbc-14_839_1451_813_324}

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1. Find the first three non-zero terms of the binomial expansion of \(\frac { 1 } { \sqrt { 4 - x ^ { 2 } } }\) for \(| x | < 2\).
2. Use this result to find an approximation for \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x\), rounding your answer to
   4 significant figures.
3. Given that \(\int \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x = \arcsin \left( \frac { 1 } { 2 } x \right) + c\), evaluate \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x\), rounding your answer to 4 significant figures.

3.(i)Determine the value of \(k\) such that $$\arctan \frac { 1 } { 2 } - \arctan \frac { 1 } { 3 } = \arctan k$$ (ii)(a)Prove that $$\cos 3 A \equiv 4 \cos ^ { 3 } A - 3 \cos A$$ Given that \(a = \cos 20 ^ { \circ }\) (b)write down,in terms of \(a\) ,an expression for \(\cos 40 ^ { \circ }\) (c)determine,in terms of \(a\) ,a simplified expression for \(\cos 80 ^ { \circ }\) (d)Use part(a)to show that $$4 a ^ { 3 } - 3 a = \frac { 1 } { 2 }$$ (e)Hence,or otherwise,show that $$\cos 20 ^ { \circ } \cos 40 ^ { \circ } \cos 80 ^ { \circ } = \frac { 1 } { 8 }$$

7. \begin{figure}[h] \end{figure} Figure 4 shows a shape \(S ( \theta )\) made up of five line segments \(A B , B C , C D , D E\) and \(E A\) . The lengths of the sides are \(A B = B C = 5 \mathrm {~cm} , C D = E A = 3 \mathrm {~cm}\) and \(D E = 7 \mathrm {~cm}\) . Angle \(B A E =\) angle \(B C D = \theta\) radians. The length of each line segment always remains the same but the value of \(\theta\) can be varied so that different symmetrical shapes can be formed,with the added restriction that none of the line segments cross.

1. Sketch \(S ( \pi )\) ,labelling the vertices clearly. The shape \(S ( \phi )\) is a trapezium.
2. Sketch \(S ( \phi )\) and calculate the value of \(\phi\) . The smallest possible value for \(\theta\) is \(\alpha\) ,where \(\alpha > 0\) ,and the largest possible value for \(\theta\) is \(\beta\) , where \(\beta > \pi\) .
3. Show that \(\alpha = \arccos \left( \frac { 29 } { 40 } \right) \cdot \left[ \arccos ( x ) \right.\) is an alternative notation for \(\left. \cos ^ { - 1 } ( x ) \right]\)
4. Find the value of \(\beta\) . The area,in \(\mathrm { cm } ^ { 2 }\) ,of shape \(S ( \theta )\) is \(R ( \theta )\) .
5. Show that for \(\alpha \leqslant \theta < \pi\) $$R ( \theta ) = 15 \sin \theta + \frac { 7 } { 4 } \sqrt { 87 - 120 \cos \theta }$$ Given that this formula for \(R ( \theta )\) holds for \(\alpha \leqslant \theta \leqslant \beta\)
6. show that \(R ( \theta )\) has only one stationary point and that this occurs when \(\theta = \frac { 2 \pi } { 3 }\)
7. find the maximum and minimum values of \(R ( \theta )\). FOR STYLE, CLARITY AND PRESENTATION: 7 MARKS TOTAL FOR PAPER: 100 MARKS
   END

13 The function f is defined by $$\mathrm { f } ( x ) = \arccos x \text { for } 0 \leq x \leq a$$ The curve with equation \(y = \mathrm { f } ( x )\) is shown below. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-18_842_837_550_603} 13

1. State the value of \(a\) 13
2. (i) On the diagram above, sketch the curve with equation $$y = \cos x \text { for } 0 \leq x \leq \frac { \pi } { 2 }$$ and
   sketch the line with equation $$y = x \text { for } 0 \leq x \leq \frac { \pi } { 2 }$$ 13 (b) (ii) Explain why the solution to the equation $$x - \cos x = 0$$ must also be a solution to the equation $$\cos x = \arccos x$$ Question 13 continues on the next page 13
3. Use the Newton-Raphson method with \(x _ { 0 } = 0\) to find an approximate solution, \(x _ { 3 }\), to the equation $$x - \cos x = 0$$ Give your answer to four decimal places. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-21_2491_1716_219_153}

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1. Use de Moivre's theorem to show that \(\tan 4 \theta = \frac { 4 t \left( 1 - t ^ { 2 } \right) } { 1 - 6 t ^ { 2 } + t ^ { 4 } }\), where \(t = \tan \theta\).
2. Given that \(\theta\) is the acute angle such that \(\tan \theta = \frac { 1 } { 5 }\), express \(\tan 4 \theta\) as a rational number in its simplest form, and verify that $$\frac { 1 } { 4 } \pi + \tan ^ { - 1 } \left( \frac { 1 } { 239 } \right) = 4 \tan ^ { - 1 } \left( \frac { 1 } { 5 } \right)$$

2 Sketch the curve with equation \(y = \tan x\) for \(- \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\).
On the same diagram, sketch the curve with equation \(y = \tan ^ { - 1 } x\) for all \(x\).
State the geometrical relationship between the curves.

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1. (a) Given that \(x \geqslant 1\), show that \(\sec ^ { - 1 } x = \cos ^ { - 1 } \left( \frac { 1 } { x } \right)\), and deduce that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } }\).
   (b) Use integration by parts to determine \(\int \sec ^ { - 1 } x \mathrm {~d} x\).
2. \includegraphics[max width=\textwidth, alt={}, center]{5d526fd9-72f8-42b1-b156-fd4a0c764c82-4_670_1029_1073_596} The diagram shows the curve \(S\) with equation \(y = \sec ^ { - 1 } x\) for \(x \geqslant 1\). The line \(L\), with gradient \(\frac { 1 } { \sqrt { 2 } }\), is the tangent to \(S\) at the point \(P\) and cuts the \(x\)-axis at the point \(Q\). The point \(I\) has coordinates \(( 1,0 )\).
   (a) Determine the exact coordinates of \(P\) and \(Q\).
   (b) The region \(R\), shaded on the diagram, is bounded by the line segments \(P Q\) and \(Q I\) and the \(\operatorname { arc } I P\) of \(S\). Show that \(R\) has area $$\ln ( 1 + \sqrt { 2 } ) - \frac { \pi ( 8 - \pi ) \sqrt { 2 } } { 32 } .$$ {www.cie.org.uk} after the live examination series. }

Find the exact solution of the equation $$\frac{1}{6}\pi + \tan^{-1}(4x) = -\cos^{-1}(\frac{1}{3}\sqrt{3}).$$ [2]

The polynomial \(p(x)\) is defined by $$p(x) = ax^3 + 3x^2 + bx + 12,$$ where \(a\) and \(b\) are constants. It is given that \((x + 3)\) is a factor of \(p(x)\). It is also given that the remainder is 18 when \(p(x)\) is divided by \((x + 2)\).

1. Find the values of \(a\) and \(b\). [5]
2. When \(a\) and \(b\) have these values,
   1. show that the equation \(p(x) = 0\) has exactly one real root, [4]
   2. solve the equation \(p(\sec y) = 0\) for \(-180° < y < 180°\). [3]

\includegraphics{figure_10} The diagram shows the curve \(y = x^2 \cos 2x\) for \(0 \leq x \leq \frac{1}{4}\pi\). The curve has a maximum point at \(M\) where \(x = p\).

1. Show that \(p\) satisfies the equation \(p = \frac{1}{2} \tan^{-1} \left(\frac{1}{p}\right)\). [3]
2. Use the iterative formula \(p_{n+1} = \frac{1}{2} \tan^{-1} \left(\frac{1}{p_n}\right)\) to determine the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
3. Find, showing all necessary working, the exact area of the region bounded by the curve and the \(x\)-axis. [5]

\includegraphics{figure_6} In the diagram, \(A\) is a point on the circumference of a circle with centre \(O\) and radius \(r\). A circular arc with centre \(A\) meets the circumference at \(B\) and \(C\). The angle \(OAB\) is \(\theta\) radians. The shaded region is bounded by the circumference of the circle and the arc with centre \(A\) joining \(B\) and \(C\). The area of the shaded region is equal to half the area of the circle.

1. Show that \(\cos 2\theta = \frac{2\sin 2\theta - \pi}{4\theta}\). [5]
2. Use the iterative formula $$\theta_{n+1} = \frac{1}{2}\cos^{-1}\left(\frac{2\sin 2\theta_n - \pi}{4\theta_n}\right),$$ with initial value \(\theta_1 = 1\), to determine \(\theta\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places. [3]

\includegraphics{figure_3} The curve \(C\), shown in Figure 3, has equation $$y = \frac{x^{-\frac{1}{4}}}{\sqrt{1+x}\left(\arctan\sqrt{x}\right)}$$ The region \(R\), shown shaded in Figure 3, is bounded by \(C\), the line with equation \(x = 3\), the \(x\)-axis and the line with equation \(x = \frac{1}{3}\) The region \(R\) is rotated through \(360°\) about the \(x\)-axis to form a solid. Using the substitution \(\tan u = \sqrt{x}\)

1. show that the volume \(V\) of the solid formed is given by $$k \int_a^b \frac{1}{u^2} du$$ where \(k\), \(a\) and \(b\) are constants to be found. [6]
2. Hence, using algebraic integration, find the value of \(V\) in simplest form. [3]

Given that \(y = \arctan \frac{x}{\sqrt{1 + x^2}}\) show that \(\frac{dy}{dx} = \frac{1}{\sqrt{1 + x^2}}\) [4]

Given that \(y = \arcsin x\) prove that

1. \(\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\) [3]
2. \((1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0\) [4]

(Total 7 marks)

The curve \(y = \cos^{-1}(2x - 1)\) intersects the curve \(y = e^x\) at a single point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.4 and 0.5. [2]
2. Show that the equation \(\cos^{-1}(2x - 1) = e^x\) can be written as \(x = \frac{1}{2} + \frac{1}{2}\cos(e^x)\). [1]
3. Use the iteration \(x_{n+1} = \frac{1}{2} + \frac{1}{2}\cos(e^{x_n})\) with \(x_1 = 0.4\) to find the values of \(x_2\) and \(x_3\), giving your answers to three decimal places. [2]

\includegraphics{figure_7} The diagram shows the curve with equation \(y = \cos^{-1} x\).

1. Sketch the curve with equation \(y = 3 \cos^{-1}(x - 1)\), showing the coordinates of the points where the curve meets the axes. [3]
2. By drawing an appropriate straight line on your sketch in part (i), show that the equation \(3 \cos^{-1}(x - 1) = x\) has exactly one root. [1]
3. Show by calculation that the root of the equation \(3 \cos^{-1}(x - 1) = x\) lies between 1.8 and 1.9. [2]
4. The sequence defined by $$x_1 = 2, \quad x_{n+1} = 1 + \cos(\frac{1}{3}x_n)$$ converges to a number \(\alpha\). Find the value of \(\alpha\) correct to 2 decimal places and explain why \(\alpha\) is the root of the equation \(3 \cos^{-1}(x - 1) = x\). [5]

\includegraphics{figure_1} Each diagram above shows part of a curve, the equation of which is one of the following: $$y = \sin^{-1} x, \quad y = \cos^{-1} x, \quad y = \tan^{-1} x, \quad y = \sec x, \quad y = \cosec x, \quad y = \cot x.$$ State which equation corresponds to

1. Fig. 1, [1]
2. Fig. 2, [1]
3. Fig. 3. [1]