1.05a Sine, cosine, tangent: definitions for all arguments

6. One end of a light elastic string, of natural length 5 l and modulus of elasticity 20 mg , is attached to a fixed point \(A\). A particle \(P\) of mass \(2 m\) is attached to the free end of the string and \(P\) hangs freely in equilibrium at the point \(B\).

1. Find the distance \(A B\).
   (3) The particle is now pulled vertically downwards from \(B\) to the point \(C\) and released from rest. In the subsequent motion the string does not become slack.
2. Show that \(P\) moves with simple harmonic motion with centre \(B\).
3. Find the period of this motion. The greatest speed of \(P\) during this motion is \(\frac { 1 } { 5 } \sqrt { g l }\)
4. Find the amplitude of this motion. The point \(D\) is the midpoint of \(B C\) and the point \(E\) is the highest point reached by \(P\).
5. Find the time taken by \(P\) to move directly from \(D\) to \(E\).

5. \begin{figure}[h] \end{figure} Figure 3 shows the finite region \(R\) which is bounded by part of the curve with equation \(y = \sin x\), the \(x\)-axis and the line with equation \(x = \frac { \pi } { 2 }\). A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis. Using algebraic integration,

1. show that the volume of \(S\) is \(\frac { \pi ^ { 2 } } { 4 }\)
2. find, in terms of \(\pi\), the \(x\) coordinate of the centre of mass of \(S\).

3. A particle \(P\) is moving in a straight line with simple harmonic motion between two points \(A\) and \(B\), where \(A B\) is \(2 a\) metres. The point \(C\) lies on the line \(A B\) and \(A C = \frac { 1 } { 2 } a\) metres. The particle passes through \(C\) with speed \(\frac { 3 a \sqrt { 3 } } { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Find the period of the motion. The maximum magnitude of the acceleration of \(P\) is \(45 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Find
2. the value of \(a\),
3. the maximum speed of \(P\). The point \(D\) lies on \(A B\) and \(P\) takes a quarter of one period to travel directly from \(C\) to \(D\).
4. Find the distance CD.

7 Two quantities, \(x\) and \(\theta\), vary with time and are related by the equation \(x = 5 \sin \theta - 4 \cos \theta\).

1. Find the value of \(x\) when \(\theta = \frac { \pi } { 2 }\).
2. When \(\theta = \frac { \pi } { 2 }\), its rate of increase (in suitable units) is given by \(\frac { \mathrm { d } \theta } { \mathrm { d } t } = 0.1\). Show that at that moment \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.4\).

3 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 \(( 1,0 )\).

1 Fig. 8 shows part of the curve \(y = x \cos 2 x\), together with a point P at which the curve crosses the \(x\)-axis. \begin{figure}[h] \end{figure}

1. Find the exact coordinates of P .
2. Show algebraically that \(x \cos 2 x\) is an odd function, and interpret this result graphically.
3. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
4. Show that turning points occur on the curve for values of \(x\) which satisfy the equation \(x \tan 2 x = \frac { 1 } { 2 }\).
5. Find the gradient of the curve at the origin. Show that the second derivative of \(x \cos 2 x\) is zero when \(x = 0\).
6. Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x \cos 2 x \mathrm {~d} x\), giving your answer in terms of \(\pi\). Interpret this result graphically.

2 Fig. 8 shows part of the curve \(y = x \sin 3 x\). It crosses the \(x\)-axis at P . The point on the curve with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\) is Q . \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of P .
2. Show that Q lies on the line \(y = x\).
3. Differentiate \(x \sin 3 x\). Hence prove that the line \(y = x\) touches the curve at Q .
4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).

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 .

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.

3 Fig. 8 shows part of the curve \(y = x \cos 2 x\), together with a point P at which the curve crosses the \(x\)-axis. \begin{figure}[h] \end{figure}

1. Find the exact coordinates of P .
2. Show algebraically that \(x \cos 2 x\) is an odd function, and interpret this result graphically.
3. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
4. Show that turning points occur on the curve for values of \(x\) which satisfy the equation \(x \tan 2 x = \frac { 1 } { 2 }\).
5. Find the gradient of the curve at the origin. Show that the second derivative of \(x \cos 2 x\) is zero when \(x = 0\).
6. Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x \cos 2 x \mathrm {~d} x\), giving your answer in terms of \(\pi\). Interpret this result graphically.

3 Fig. 8 shows part of the curve \(y = x \sin 3 x\). It crosses the \(x\)-axis at P . The point on the curve with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\) is Q . \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of P .
2. Show that Q lies on the line \(y = x\).
3. Differentiate \(x \sin 3 x\). Hence prove that the line \(y = x\) touches the curve at Q .
4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).
5. Differentiate \(x \cos 2 x\) with respect to \(x\).
6. Integrate \(x \cos 2 x\) with respect to \(x\).

7 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.

1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
   1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
   2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
   1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
   2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P - 1 } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 - \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
   3. Find the greatest and least values of \(P\) predicted by this model. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{9296c786-a42a-4aa5-b326-39adbb544cbc-05_609_622_301_719} \captionsetup{labelformat=empty} \caption{Fig. 8}
      \end{figure} In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axes shown, the path of C is modelled by the parametric equations $$x = 10 \cos \theta + 5 \cos 2 \theta , \quad y = 10 \sin \theta + 5 \sin 2 \theta , \quad ( 0 \leqslant \theta < 2 \pi )$$ where \(x\) and \(y\) are in metres.
      1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { \cos \theta + \cos 2 \theta } { \sin \theta + \sin 2 \theta }\). Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 3 } \pi\). Hence find the exact coordinates of the highest point A on the path of C .
      2. Express \(x ^ { 2 } + y ^ { 2 }\) in terms of \(\theta\). Hence show that $$x ^ { 2 } + y ^ { 2 } = 125 + 100 \cos \theta$$
      3. Using this result, or otherwise, find the greatest and least distances of C from O . You are given that, at the point B on the path vertically above O , $$2 \cos ^ { 2 } \theta + 2 \cos \theta - 1 = 0$$
      4. Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures. \section*{ADVANCED GCE UNIT MATHEMATICS (MEI)} Applications of Advanced Mathematics (C4) \section*{Paper B: Comprehension} \section*{THURSDAY 14 JUNE 2007} Afternoon
         Time: Up to 1 hour
         Additional materials:
         Rough paper
         MEI Examination Formulae and Tables (MF2) Candidate
         Name □
         Centre
         Number sufficient detail of the working to indicate that a correct method is being used. 1 This basic cycloid has parametric equations $$x = a \theta - a \sin \theta , \quad y = a - a \cos \theta$$
         \includegraphics[max width=\textwidth, alt={}]{9296c786-a42a-4aa5-b326-39adbb544cbc-10_307_1138_445_411}
         Find the coordinates of the points M and N , stating the value of \(\theta\) at each of them. Point M Point N 2 A sea wave has parametric equations (in suitable units) $$x = 7 \theta - 0.25 \sin \theta , \quad y = 0.25 \cos \theta$$ Find the wavelength and height of the wave.
         3 The graph below shows the profile of a wave.
         1. Assuming that it has parametric equations of the form given on line 68 , find the values of \(a\) and \(b\).
         2. Investigate whether the ratio of the trough length to the crest length is consistent with this shape. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-11_312_1141_623_415}
         3. \(\_\_\_\_\)
         4. \(\_\_\_\_\) 4 This diagram illustrates two wave shapes \(U\) and \(V\). They have the same wavelength and the same height. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-12_423_1552_356_205} One of the curves is a sine wave, the other is a curtate cycloid.
         5. State which is which, justifying your answer.
         6. \(\_\_\_\_\) The parametric equations for the curves are: $$x = a \theta , \quad y = b \cos \theta ,$$ and $$x = a \theta - b \sin \theta , \quad y = b \cos \theta .$$
         7. Show that the distance marked \(d\) on the diagram is equal to \(b\).
         8. Hence justify the statement in lines 109 to 111: "In such cases, the curtate cycloid and the sine curve with the same wavelength and height are very similar and so the sine curve is also a good model."
         9. \(\_\_\_\_\)
         10. \(\_\_\_\_\) 5 The diagram shows a curtate cycloid with scales given. Show that this curve could not be a scale drawing of the shape of a stable sea wave. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-13_289_1310_397_331}

3 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC , where C is on the ground. AC is angled at \(\alpha\) to the vertical. The beam produces an oval-shaped area of light on the ground, of length DE . The width of the oval at C is GF . Angles DAC, EAC, FAC and GAC are all \(\beta\). \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} In the following, all lengths are in metres.

1. Find AC in terms of \(\alpha\), and hence show that \(\mathrm { GF } = 10 \sec \alpha \tan \beta\).
2. Show that \(\mathrm { CE } = 5 ( \tan ( \alpha + \beta ) - \tan \alpha )\). $$\text { Hence show that } \mathrm { CE } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 - \tan \alpha \tan \beta } \text {. }$$ Similarly, it can be shown that \(\mathrm { CD } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 + \tan \alpha \tan \beta }\). [You are not required to derive this result.]
   You are now given that \(\alpha = 45 ^ { \circ }\) and that \(\tan \beta = t\).
3. Find CE and CD in terms of \(t\). Hence show that \(\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }\).
4. Show that \(\mathrm { GF } = 10 \sqrt { 2 } t\). For a certain value of \(\beta , \mathrm { DE } = 2 \mathrm { GF }\).
5. Show that \(t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }\). Hence find this value of \(\beta\).

2 In Fig. 6, \(\mathrm { ABC } , \mathrm { ACD }\) and AED are right-angled triangles and \(\mathrm { BC } = 1\) unit. Angles CAB and CAD are \(\theta\) and \(\phi\) respectively. \begin{figure}[h] \end{figure}

1. Find AC and AD in terms of \(\theta\) and \(\phi\).
2. Hence show that \(\mathrm { DE } = 1 + \frac { \tan \phi } { \tan \theta }\).

2 In a game of rugby, a kick is to be taken from a point P (see Fig. 7). P is a perpendicular distance \(y\) metres from the line TOA. Other distances and angles are as shown. \begin{figure}[h] \end{figure}

1. Show that \(\theta = \beta - \alpha\), and hence that \(\tan \theta = \frac { 6 y } { 160 + y ^ { 2 } }\). Calculate the angle \(\theta\) when \(y = 6\).
2. By differentiating implicitly, show that \(\frac { \mathrm { d } \theta } { \mathrm { d } y } = \frac { 6 \left( 160 - y ^ { 2 } \right) } { \left( 160 + y ^ { 2 } \right) ^ { 2 } } \cos ^ { 2 } \theta\).
3. Use this result to find the value of \(y\) that maximises the angle \(\theta\). Calculate this maximum value of \(\theta\). [You need not verify that this value is indeed a maximum.]

2. The point \(P\) with coordinates \(\left( \frac { \pi } { 2 } , 1 \right)\) lies on the curve with equation $$4 x \sin x = \pi y ^ { 2 } + 2 x , \quad \frac { \pi } { 6 } \leqslant x \leqslant \frac { 5 \pi } { 6 }$$ Find an equation of the normal to the curve at \(P\).

8 Fig. 8 shows a sketch of part of the curve \(y = x \sin 2 x\), where \(x\) is in radians.
The curve crosses the \(x\)-axis at the point P . The tangent to the curve at P crosses the \(y\)-axis at Q . \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that the \(x\)-coordinates of the turning points of the curve satisfy the equation \(\tan 2 x + 2 x = 0\).
2. Find, in terms of \(\pi\), the \(x\)-coordinate of the point P . Show that the tangent PQ has equation \(2 \pi x + 2 y = \pi ^ { 2 }\).
   Find the exact coordinates of Q .
3. Show that the exact value of the area shaded in Fig. 8 is \(\frac { 1 } { 8 } \pi \left( \pi ^ { 2 } - 2 \right)\).

3 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 \(( 1,0 )\).

8 The points \(A\) and \(B\) have position vectors relative to the origin \(O\) given by $$\overrightarrow { O A } = \left( \begin{array} { c } 3 \sin \alpha \\ 2 \cos \alpha \\ - 1 \end{array} \right) \text { and } \overrightarrow { O B } = \left( \begin{array} { c } 2 \cos \alpha \\ 4 \sin \alpha \\ 3 \end{array} \right)$$ where \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). It is given that \(\overrightarrow { O A }\) and \(\overrightarrow { O B }\) are perpendicular.

1. Calculate the two possible values of \(\alpha\).
2. Calculate the area of triangle \(O A B\) for the smaller value of \(\alpha\) from part (i).

6 In Fig. 6, OAB is a thin bent rod, with \(\mathrm { OA } = a\) metres, \(\mathrm { AB } = b\) metres and angle \(\mathrm { OAB } = 120 ^ { \circ }\). The bent rod lies in a vertical plane. OA makes an angle \(\theta\) above the horizontal. The vertical height BD of B above O is \(h\) metres. The horizontal through A meets BD at C and the vertical through A meets OD at E . \begin{figure}[h] \end{figure}

1. Find angle BAC in terms of \(\theta\). Hence show that $$h = a \sin \theta + b \sin \left( \theta - 60 ^ { \circ } \right) .$$
2. Hence show that \(h = \left( a + \frac { 1 } { 2 } b \right) \sin \theta - \frac { \sqrt { 3 } } { 2 } b \cos \theta\). The rod now rotates about O , so that \(\theta\) varies. You may assume that the formulae for \(h\) in parts (i) and (ii) remain valid.
3. Show that OB is horizontal when \(\tan \theta = \frac { \sqrt { 3 } b } { 2 a + b }\). In the case when \(a = 1\) and \(b = 2 , h = 2 \sin \theta - \sqrt { 3 } \cos \theta\).
4. Express \(2 \sin \theta - \sqrt { 3 } \cos \theta\) in the form \(R \sin ( \theta - \alpha )\). Hence, for this case, write down the maximum value of \(h\) and the corresponding value of \(\theta\).

4 Sandy is throwing a stone at a plum tree. The stone is thrown from a point O at a speed of \(35 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(\alpha\) to the horizontal, where \(\cos \alpha = 0.96\). You are given that, \(t\) seconds after being thrown, the stone is \(\left( 9.8 t - 4.9 t ^ { 2 } \right) \mathrm { m }\) higher than O . When descending, the stone hits a plum which is 3.675 m higher than O . Air resistance should be neglected. Calculate the horizontal distance of the plum from O .

9 Use the substitution \(x = 2 \sin \theta\) to show that \(\int _ { 1 } ^ { \sqrt { 3 } } \sqrt { 4 - x ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 3 } \pi\).

7

1. Use the result \(\cos ( A + B ) = \cos A \cos B - \sin A \sin B\) to show that \(\cos ( A - B ) = \cos A \cos B + \sin A \sin B\). The function \(\mathrm { f } ( \theta )\) is defined as \(\cos \left( \theta + 30 ^ { \circ } \right) \cos \left( \theta - 30 ^ { \circ } \right)\), where \(\theta\) is in degrees.
2. Show that \(f ( \theta ) = \cos ^ { 2 } \theta - \frac { 1 } { 4 }\).
3. (i) Determine the following.