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

1. (i) Find the exact value of \(x\) such that

$$3 \tan ^ { - 1 } ( x - 2 ) + \pi = 0$$ (ii) Solve, for \(- \pi < \theta < \pi\), the equation $$\cos 2 \theta - \sin \theta - 1 = 0$$ giving your answers in terms of \(\pi\).

8. (i) Solve the equation $$\pi - 3 \cos ^ { - 1 } \theta = 0$$ (ii) Sketch on the same diagram the curves \(y = \cos ^ { - 1 } ( x - 1 ) , 0 \leq x \leq 2\) and \(y = \sqrt { x + 2 } , x \geq - 2\). Given that \(\alpha\) is the root of the equation $$\cos ^ { - 1 } ( x - 1 ) = \sqrt { x + 2 }$$ (iii) show that \(0 < \alpha < 1\),
(iv) 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.
You should show the result of each iteration.

7 \includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-3_465_748_1133_717} 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.
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.
3. Show by calculation that the root of the equation \(3 \cos ^ { - 1 } ( x - 1 ) = x\) lies between 1.8 and 1.9 .
4. The sequence defined by $$x _ { 1 } = 2 , \quad x _ { n + 1 } = 1 + \cos \left( \frac { 1 } { 3 } x _ { n } \right)$$ 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 \(3 \cos ^ { - 1 } ( x - 1 ) = x\).

6 \includegraphics[max width=\textwidth, alt={}, center]{32f90420-e1eb-47ab-b588-e3806b64813f-3_641_837_1306_657} The diagram shows the graph of \(y = - \sin ^ { - 1 } ( x - 1 )\).

1. Give details of the pair of geometrical transformations which transforms the graph of \(y = - \sin ^ { - 1 } ( x - 1 )\) to the graph of \(y = \sin ^ { - 1 } x\).
2. Sketch the graph of \(y = \left| - \sin ^ { - 1 } ( x - 1 ) \right|\).
3. Find the exact solutions of the equation \(\left| - \sin ^ { - 1 } ( x - 1 ) \right| = \frac { 1 } { 3 } \pi\).

9 \includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-4_424_707_1260_724} The diagram shows the curve \(y = \tan ^ { - 1 } x\) and its asymptotes \(y = \pm a\).

1. State the exact value of \(a\).
2. Find the value of \(x\) for which \(\tan ^ { - 1 } x = \frac { 1 } { 2 } a\). The equation of another curve is \(y = 2 \tan ^ { - 1 } ( x - 1 )\).
3. Sketch this curve on a copy of the diagram, and state the equations of its asymptotes in terms of \(a\).
4. Verify by calculation that the value of \(x\) at the point of intersection of the two curves is 1.54 , correct to 2 decimal places. Another curve (which you are not asked to sketch) has equation \(y = \left( \tan ^ { - 1 } x \right) ^ { 2 }\).
5. Use Simpson's rule, with 4 strips, to find an approximate value for \(\int _ { 0 } ^ { 1 } \left( \tan ^ { - 1 } x \right) ^ { 2 } \mathrm {~d} x\).

6 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 + 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\).

1. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x - 1 } { 2 } \right)\) and state the domain of this function. Fig. 6 shows a sketch of the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{6a4c3f3b-a298-4b13-b97e-b52f8d9d527b-4_506_561_705_751} \captionsetup{labelformat=empty} \caption{Fig. 6}
   \end{figure}
2. Write down the coordinates of the points \(\mathrm { A } , \mathrm { B }\) and C .

3 Fig. 3 shows the curve defined by the equation \(y = \arcsin ( x - 1 )\), for \(0 \leqslant x \leqslant 2\). \begin{figure}[h] \end{figure}

1. Find \(x\) in terms of \(y\), and show that \(\frac { \mathrm { d } x } { \mathrm {~d} y } = \cos y\).
2. Hence find the exact gradient of the curve at the point where \(x = 1.5\).

6 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \arctan x\). \begin{figure}[h] \end{figure}

1. Find the range of the function \(\mathrm { f } ( x )\), giving your answer in terms of \(\pi\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). Find the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the origin.
3. Hence write down the gradient of \(y = \frac { 1 } { 2 } \arctan x\) at the origin.

8 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
P is the point on the curve with \(x\)-coordinate \(\frac { 1 } { 3 } \pi\). \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P .
2. Find \(\mathrm { f } ^ { \prime } ( x )\). Hence find the gradient of the curve at the point P .
3. Show that the derivative of \(\frac { \sin x } { 1 + \cos x }\) is \(\frac { 1 } { 1 + \cos x }\). Hence find the exact area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 3 } \pi\).
4. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)\). State the domain of this inverse function, and add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.

6 The function \(\mathrm { f } ( x )\) is defined by $$f ( x ) = 1 + 2 \sin 3 x , \quad - \frac { \pi } { 6 } \leqslant x \leqslant \frac { \pi } { 6 }$$ You are given that this function has an inverse, \(\mathrm { f } ^ { - 1 } ( x )\).
Find \(\mathrm { f } ^ { - 1 } ( x )\) and its domain.

2 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \arctan x\). \begin{figure}[h] \end{figure}

1. Find the range of the function \(\mathrm { f } ( x )\), giving your answer in terms of \(\pi\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). Find the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the origin.
3. Hence write down the gradient of \(y = \frac { 1 } { 2 } \arctan x\) at the origin.

4 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 2 \arcsin x , - 1 \nless \leqslant 1\).
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 \(\frac { 1 } { 2 }\).

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 .

3 The function \(f ( x ) = \ln \left( t + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = f ( x )\). \begin{figure}[h] \end{figure}

1. Show algebraically that the function is even. State how this property relates to the shape of the curve.
2. Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
3. Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(f ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(f ( x )\) is the function \(g ( x )\).
4. Sketch the curves \(y = f ( x )\) and \(y = g ( x )\) on the same axes. State the domain of the function \(g ( x )\),
   Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
5. Differentiate \(\mathrm { g } ( \mathrm { x } )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).

4 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 + 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\).

1. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x \mathrm { r } } { 2 } \right)\) and state the domain of this function. Fig. 6 shows a sketch of the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-2_499_562_779_785} \captionsetup{labelformat=empty} \caption{Fig. 6}
   \end{figure}
2. Write down the coordinates of the points \(\mathrm { A } , \mathrm { B }\) and C .

5 Given that \(\arcsin x = \frac { 1 } { 6 } \pi\), find \(x\). Find \(\arccos x\) in terms of \(\pi\).

8 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 - 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Fig. 3 shows the curve \(y = \mathrm { f } ( x )\). \begin{figure}[h] \end{figure}

1. Write down the range of the function \(\mathrm { f } ( x )\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\).
3. Find \(\mathrm { f } ^ { \prime } ( 0 )\). Hence write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + \sin 2 x\) for \(- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi\). \begin{figure}[h] \end{figure}

1. State a sequence of two transformations that would map part of the curve \(y = \sin x\) onto the curve \(y = \mathrm { f } ( x )\).
2. Find the area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = \frac { 1 } { 4 } \pi\).
3. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point ( 0,1 ). Hence write down the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
4. State the domain of \(\mathrm { f } ^ { - 1 } ( x )\). Add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

4 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
P is the point on the curve with \(x\)-coordinate \(\frac { 1 } { 3 } \pi\). \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P .
2. Find \(\mathrm { f } ^ { \prime } ( x )\). Hence find the gradient of the curve at the point P .
3. Show that the derivative of \(\frac { \sin x } { 1 + \cos x }\) is \(\frac { 1 } { 1 + \cos x }\). Hence find the exact area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 3 } \pi\).
4. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)\). State the domain of this inverse function, and add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.

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

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

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

5 Fig. 7 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + 2 \arctan x , x \in \mathbb { R }\). The scales on the \(x\) - and \(y\)-axes are the same. \begin{figure}[h] \end{figure}

1. Find the range of f , giving your answer in terms of \(\pi\).
2. Find \(\mathrm { f } ^ { - 1 } ( x )\), and add a sketch of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) to the copy of Fig. 7.

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

8 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \arctan x\). \begin{figure}[h] \end{figure}

1. Find the range of the function \(\mathrm { f } ( x )\), giving your answer in terms of \(\pi\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). Find the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the origin.
3. Hence write down the gradient of \(y = \frac { 1 } { 2 } \arctan x\) at the origin.

9 Given that \(\arcsin x = \frac { 1 } { 6 } \pi\), find \(x\). Find \(\arccos x\) in terms of \(\pi\).

7

1. Find the first three non-zero terms of the binomial expansion of \(\frac { 1 } { \sqrt { 4 - x ^ { 2 } } }\) for \(| x | < 2\). [4]
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.