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

2 It is given that \(\mathrm { f } ( x ) = x ^ { 2 } - \tan ^ { - 1 } x\).

1. Show by calculation that the equation \(\mathrm { f } ( x ) = 0\) has a root in the interval \(0.8 < x < 0.9\).
2. Use the Newton-Raphson method, with a first approximation 0.8, to find the next approximation to this root. Give your answer correct to 3 decimal places.

2 \includegraphics[max width=\textwidth, alt={}, center]{15dd10f9-73d4-4107-bb45-7866f5470572-2_577_700_577_721} The diagram shows parts of the curves with equations \(y = \cos ^ { - 1 } x\) and \(y = \frac { 1 } { 2 } \sin ^ { - 1 } x\), and their point of intersection \(P\).

1. Verify that the coordinates of \(P\) are \(\left( \frac { 1 } { 2 } \sqrt { 3 } , \frac { 1 } { 6 } \pi \right)\).
2. Find the gradient of each curve at \(P\).

5 It is given that \(y = \tan ^ { - 1 } 2 x\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } = 0\).
2. Find the Maclaurin series for \(y\) up to and including the term in \(x ^ { 3 }\). Show all your working.
3. The result in part (ii), together with the value \(x = \frac { 1 } { 2 }\), is used to find an estimate for \(\pi\). Show that this estimate is only correct to 1 significant figure.

5.(a)The box below shows a student's attempt to prove the following identity for \(a > b > 0\) $$\arctan a - \arctan b \equiv \arctan \frac { a - b } { 1 + a b }$$ Let \(x = \arctan a\) and \(y = \arctan b\) ,so that \(a = \tan x\) and \(b = \tan y\) $$\begin{aligned} \text { So } \tan ( \arctan a - \arctan b ) & \equiv \tan ( x - y ) \\ & \equiv \frac { \tan x - \tan y } { 1 - \tan ^ { 2 } ( x y ) } \\ & \equiv \frac { a - b } { 1 - ( a b ) ^ { 2 } } \\ & \equiv \frac { a - a b + a b - b } { ( 1 - a b ) ( 1 + a b ) } \\ & \equiv \frac { a ( 1 - a b ) - b ( 1 - a b ) } { ( 1 - a b ) ( 1 + a b ) } \\ & \equiv \frac { a - b } { 1 + a b } \end{aligned}$$ Taking arctan of both sides gives \(\arctan a - \arctan b \equiv \arctan \frac { a - b } { 1 + a b }\) as required. There are three errors in the proof where the working does not follow from the previous line.

1. Describe these three errors.
2. Write out a correct proof of the identity.
   (b)[In this question take \(g\) to be \(9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) ] \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4d5b914c-28b2-4485-a42e-627c95fa16e2-22_244_1267_1870_504} \captionsetup{labelformat=empty} \caption{Figure 3}
   \end{figure} Balls are projected,one after another,from a point,\(A\) ,one metre above horizontal ground. Each ball travels in a vertical plane towards a 6 metre high vertical wall of negligible thickness,which is a horizontal distance of \(10 \sqrt { 2 }\) metres from \(A\) . The balls are modelled as particles and it is assumed that there is no air resistance.
   Each ball is projected with an initial speed of \(28 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and at a random angle \(\theta\) to the horizontal,where \(0 < \theta < 90 ^ { \circ }\) Given that a ball will pass over the wall precisely when \(\alpha \leqslant \theta \leqslant \beta\)
3. find, in degrees, the angle \(\beta - \alpha\)
4. Deduce that the probability that a particular ball will pass over the wall is \(\frac { 2 } { 3 }\)
5. Hence find the probability that exactly 2 of the first 10 balls projected pass over the wall. You should give your answer in the form \(\frac { P } { Q ^ { k } }\) where \(P , Q\) and \(k\) are integers and \(P\) is not a multiple of \(Q\).
6. Explain whether taking air resistance into account would increase or decrease the probability in (b)(iii).
7. find, in degrees, the angle \(\beta - \alpha\)

3.(a)Solve,for \(0 \leq x < 2 \pi\) , $$\cos x + \cos 2 x = 0$$ (b)Find the exact value of \(x , x \geq 0\) ,for which $$\arccos x + \arccos 2 x = \frac { \pi } { 2 }$$ [ \(\arccos x\) is an alternative notation for \(\cos ^ { - 1 } x\) .]

3. (a) Solve, for \(0 \leqslant \theta < 2 \pi\), $$\sin \left( \frac { \pi } { 3 } - \theta \right) = \frac { 1 } { \sqrt { } 3 } \cos \theta$$ (b) Find the value of \(x\) for which $$\begin{aligned} & \arcsin ( 1 - 2 x ) = \frac { \pi } { 3 } - \arcsin x , \quad 0 < x < 0.5 \\ & { \left[ \arcsin x \text { is an alternative notation for } \sin ^ { - 1 } x \right] } \end{aligned}$$

2.Find the value of $$\arccos \left( \frac { 1 } { \sqrt { 2 } } \right) + \arcsin \left( \frac { 1 } { 3 } \right) + 2 \arctan \left( \frac { 1 } { \sqrt { 2 } } \right)$$ Give your answer as a multiple of \(\pi\) . $$\text { (arccos } x \text { is an alternative notion for } \cos ^ { - 1 } x \text { etc.) }$$

10. (a) Given that \(- \frac { \pi } { 2 } < \mathrm { g } ( x ) < \frac { \pi } { 2 }\), sketch the graph of \(y = \mathrm { g } ( x )\) where $$\mathrm { g } ( x ) = \arctan x , \quad x \in \mathbb { R }$$ (b) Find the exact value of \(x\) for which $$3 g ( x + 1 ) - \pi = 0$$ The equation \(\arctan x - 4 + \frac { 1 } { 2 } x = 0\) has a positive root at \(x = \alpha\) radians.
(c) Show that \(5 < \alpha < 6\) The iteration formula $$x _ { n + 1 } = 8 - 2 \arctan x _ { n }$$ can be used to find an approximation for \(\alpha\) (d) Taking \(x _ { 0 } = 5\), use this formula to find \(x _ { 1 }\) and \(x _ { 2 }\), giving each answer to 3 decimal places.

1

1. Given that \(y = \arctan \sqrt { x }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer in terms of \(x\). Hence show that $$\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { x } ( x + 1 ) } \mathrm { d } x = \frac { \pi } { 2 }$$
2. A curve has cartesian equation $$x ^ { 2 } + y ^ { 2 } = x y + 1$$
   1. Show that the polar equation of the curve is $$r ^ { 2 } = \frac { 2 } { 2 - \sin 2 \theta }$$
   2. Determine the greatest and least positive values of \(r\) and the values of \(\theta\) between 0 and \(2 \pi\) for which they occur.
   3. Sketch the curve.

1

1. A curve has polar equation \(r = 1 + \cos \theta\) for \(0 \leqslant \theta < 2 \pi\).
   1. Sketch the curve.
   2. Find the area of the region enclosed by the curve, giving your answer in exact form.
2. Assuming that \(x ^ { 4 }\) and higher powers may be neglected, write down the Maclaurin series approximations for \(\sin x\) and \(\cos x\) (where \(x\) is in radians). Hence or otherwise obtain an approximation for \(\tan x\) in the form \(a x + b x ^ { 3 }\).
3. Find \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 - \frac { 1 } { 4 } X ^ { 2 } } } \mathrm {~d} x\), giving your answer in exact form.

1

1. 1. Differentiate with respect to \(x\) the equation \(a \tan y = x\) (where \(a\) is a constant), and hence show that the derivative of \(\arctan \frac { x } { a }\) is \(\frac { a } { a ^ { 2 } + x ^ { 2 } }\).
   2. By first expressing \(x ^ { 2 } - 4 x + 8\) in completed square form, evaluate the integral \(\int _ { 0 } ^ { 4 } \frac { 1 } { x ^ { 2 } - 4 x + 8 } \mathrm {~d} x\), giving your answer exactly.
   3. Use integration by parts to find \(\int \arctan x \mathrm {~d} x\).
2. 1. A curve has polar equation \(r = 2 \cos \theta\), for \(- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\). Show, by considering its cartesian equation, that the curve is a circle. State the centre and radius of the circle.
   2. Another circle has radius 2 and its centre, in cartesian coordinates, is ( 0,2 ). Find the polar equation of this circle.

1

1. Given that \(\mathrm { f } ( x ) = \arccos x\),
   1. sketch the graph of \(y = \mathrm { f } ( x )\),
   2. show that \(\mathrm { f } ^ { \prime } ( x ) = - \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\),
   3. obtain the Maclaurin series for \(\mathrm { f } ( x )\) as far as the term in \(x ^ { 3 }\).
2. A curve has polar equation \(r = \theta + \sin \theta , \theta \geqslant 0\).
   1. By considering \(\frac { \mathrm { d } r } { \mathrm {~d} \theta }\) show that \(r\) increases as \(\theta\) increases. Sketch the curve for \(0 \leqslant \theta \leqslant 4 \pi\).
   2. You are given that \(\sin \theta \approx \theta\) for small \(\theta\). Find in terms of \(\alpha\) the approximate area bounded by the curve and the lines \(\theta = 0\) and \(\theta = \alpha\), where \(\alpha\) is small.

5

1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( x - \tan ^ { - 1 } x \right) = \frac { x ^ { 2 } } { 1 + x ^ { 2 } }\).
2. Show that \(\int _ { 0 } ^ { \sqrt { 3 } } x \tan ^ { - 1 } x \mathrm {~d} x = \frac { 2 } { 3 } \pi - \frac { 1 } { 2 } \sqrt { 3 }\).

6 \includegraphics[max width=\textwidth, alt={}, center]{00a4be37-c095-4d9c-a1cd-d03b8ab1d411-3_553_579_274_726} The diagram shows the curve \(y = 8 \sin ^ { - 1 } \left( x - \frac { 3 } { 2 } \right)\). The end-points \(A\) and \(B\) of the curve have coordinates ( \(a , - 4 \pi\) ) and ( \(b , 4 \pi\) ) respectively.

1. State the values of \(a\) and \(b\).
2. It is required to find the root of the equation \(8 \sin ^ { - 1 } \left( x - \frac { 3 } { 2 } \right) = x\).
   1. Show by calculation that the root lies between 1.7 and 1.8.
   2. In order to find the root, the iterative formula $$x _ { n + 1 } = p + \sin \left( q x _ { n } \right) ,$$ with a suitable starting value, is to be used. Determine the values of the constants \(p\) and \(q\) and hence find the root correct to 4 significant figures. Show the result of each step of the iteration process.

7

1. By sketching the curves \(y = x ( 2 x + 5 )\) and \(y = \cos ^ { - 1 } x\) (where \(y\) is in radians) in a single diagram, show that the equation \(x ( 2 x + 5 ) = \cos ^ { - 1 } x\) has exactly one real root.
2. Use the iterative formula $$x _ { n + 1 } = \frac { \cos ^ { - 1 } x _ { n } } { 2 x _ { n } + 5 } \text { with } x _ { 1 } = 0.25$$ to find the root correct to 3 significant figures. Show the result of each iteration correct to at least 4 significant figures.
3. Two new curves are obtained by transforming each of the curves \(y = x ( 2 x + 5 )\) and \(y = \cos ^ { - 1 } x\) by the pair of transformations:
   reflection in the \(x\)-axis followed by reflection in the \(y\)-axis.
   State an equation of each of the new curves and determine the coordinates of their point of intersection, giving each coordinate correct to 3 significant figures.

7 Given that \(\arcsin x = \arccos y\), prove that \(x ^ { 2 } + y ^ { 2 } = 1\). [Hint: let \(\arcsin x = \theta\).] Section B (36 marks)

3 Sketch the curve \(y = 2 \arccos x\) for \(- 1 \leqslant x \leqslant 1\).

6 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 2 \arcsin x , - 1 \leqslant x \leqslant 1\).
Fig. 6 also shows the curve \(y = \mathrm { g } ( x )\), where \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\).
P is the point on the curve \(y = \mathrm { f } ( x )\) with \(x\)-coordinate \(\frac { 1 } { 2 }\). \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P , giving your answer in terms of \(\pi\). The point Q is the reflection of P in \(y = x\).
2. Find \(\mathrm { g } ( x )\) and its derivative \(\mathrm { g } ^ { \prime } ( x )\). Hence determine the exact gradient of the curve \(y = \mathrm { g } ( x )\) at the point Q . Write down the exact gradient of \(y = \mathrm { f } ( x )\) at the point P .

6 Solve each of the following equations, giving your answers in exact form.

1. \(6 \arcsin x - \pi = 0\).
2. \(\arcsin x = \arccos x\).

4 The curves \(y = \cos ^ { - 1 } x\) and \(y = \tan ^ { - 1 } ( \sqrt { 2 } x )\) intersect at a point \(A\).

1. Verify that the coordinates of \(A\) are \(\left( \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { 4 } \pi \right)\).
2. Determine whether the tangents to the curves at \(A\) are perpendicular.

3 By first completing the square, find the exact value of \(\int _ { \frac { 1 } { 2 } } ^ { 1 } \frac { 1 } { \sqrt { 2 x - x ^ { 2 } } } \mathrm {~d} x\).

5 It is given that \(y = \sin ^ { - 1 } 2 x\).

1. Using the derivative of \(\sin ^ { - 1 } x\) given in the List of Formulae (MF1), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Show that \(\left( 1 - 4 x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x }\).
3. Hence show that \(\left( 1 - 4 x ^ { 2 } \right) \frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } - 12 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } = 0\).
4. Using your results from parts (i), (ii) and (iii), find the Maclaurin series for \(\sin ^ { - 1 } 2 x\) up to and including the term in \(x ^ { 3 }\).

4 Use the formula for \(\tan ( A - B )\) in the List of Formulae (MF10) to show that $$\tan ^ { - 1 } ( x + 1 ) - \tan ^ { - 1 } ( x - 1 ) = \tan ^ { - 1 } \left( \frac { 2 } { x ^ { 2 } } \right)$$ Deduce the sum to \(n\) terms of the series $$\tan ^ { - 1 } \left( \frac { 2 } { 1 ^ { 2 } } \right) + \tan ^ { - 1 } \left( \frac { 2 } { 2 ^ { 2 } } \right) + \tan ^ { - 1 } \left( \frac { 2 } { 3 ^ { 2 } } \right) + \ldots .$$

8 Functions f and g are defined for \(0 \leqslant x \leqslant 2 \pi\) by \(\mathrm { f } ( x ) = 2 \tan x\) and \(\mathrm { g } ( x ) = \sec x\).

1. 1. State the range of f .
   2. State the range of \(g\).
2. 1. Show that \(\operatorname { fg } ( 0.6 ) = 5.33\), correct to 3 significant figures.
   2. Explain why \(\mathrm { f } ^ { - 1 } \mathrm {~g} ( 0.6 )\) is not defined.
3. In this question you must show detailed reasoning. Solve the equation \(( \mathrm { f } ( x ) ) ^ { 2 } + 6 \mathrm {~g} ( x ) = 0\).

5

1. Sketch the graphs of \(y = 4 \cos x\) and \(y = 2 \sin x\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\) on the same axes.
2. Find the exact coordinates of the point of intersection of these graphs, giving your answer in the form (arctan \(a , k \sqrt { b }\) ), where \(a\) and \(b\) are integers and \(k\) is rational.
3. A student argues that without the condition \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\) all the points of intersection of the graphs would occur at intervals of \(360 ^ { \circ }\) because both \(\sin x\) and \(\cos x\) are periodic functions with this period. Comment on the validity of the student's argument.