1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2

6.Figure 1 shows a sketch of part of the curve with equation \(y = x \sin ( \ln x ) , x \geqslant 1\) \begin{figure}[h] \end{figure} For \(x > 1\) ,the curve first crosses the \(x\)-axis at the point \(A\) .

1. Find the \(x\) coordinate of \(A\) .
2. Differentiate \(x \sin ( \ln x )\) and \(x \cos ( \ln x )\) with respect to \(x\) and hence find $$\int \sin ( \ln x ) \mathrm { d } x \text { and } \int \cos ( \ln x ) \mathrm { d } x$$
3. 1. Find \(\int x \sin ( \ln x ) \mathrm { d } x\) .
   2. Hence show that the area of the shaded region \(\boldsymbol { R }\) ,bounded by the curve and the \(x\)-axis between the points \(( 1,0 )\) and \(A\) ,is $$\frac { 1 } { 5 } \left( \mathrm { e } ^ { 2 \pi } + 1 \right)$$

4.A rectangle \(A B C D\) is drawn so that \(A\) and \(B\) lie on the \(x\)-axis,and \(C\) and \(D\) lie on the curve with equation \(y = \cos x , - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }\) .The point \(A\) has coordinates \(( p , 0 )\) ,where \(0 < p < \frac { \pi } { 2 }\) .

1. Find an expression,in terms of \(p\) ,for the area of this rectangle. The maximum area of \(A B C D\) is \(S\) and occurs when \(p = \alpha\) .Show that
2. \(\frac { \pi } { 4 } < \alpha < 1\) ,
3. \(S = \frac { 2 \alpha ^ { 2 } } { \sqrt { } \left( 1 + \alpha ^ { 2 } \right) }\) ,
4. \(\frac { \pi ^ { 2 } } { 2 \sqrt { } \left( 16 + \pi ^ { 2 } \right) } < S < \sqrt { } 2\) .

5. \begin{figure}[h] \end{figure} Figure 1 shows part of a sequence \(S _ { 1 } , S _ { 2 } , S _ { 3 } , \ldots\), of model snowflakes. The first term \(S _ { 1 }\) consists of a single square of side \(a\). To obtain \(S _ { 2 }\), the middle third of each edge is replaced with a new square, of side \(\frac { a } { 3 }\), as shown in Figure 1 . Subsequent terms are obtained by replacing the middle third of each external edge of a new square formed in the previous snowflake, by a square \(\frac { 1 } { 3 }\) of the size, as illustrated by \(S _ { 3 }\) in Figure 1.

1. Deduce that to form \(S _ { 4 } , 36\) new squares of side \(\frac { a } { 27 }\) must be added to \(S _ { 3 }\).
2. Show that the perimeters of \(S _ { 2 }\) and \(S _ { 3 }\) are \(\frac { 20 a } { 3 }\) and \(\frac { 28 a } { 3 }\) respectively.
3. Find the perimeter of \(S _ { n }\).
4. Describe what happens to the perimeter of \(S _ { n }\) as \(n\) increases.
5. Find the areas of \(S _ { 1 } , S _ { 2 }\) and \(S _ { 3 }\).
6. Find the smallest value of the constant \(S\) such that the area of \(S _ { n } < S\), for all values of \(n\). \includegraphics[max width=\textwidth, alt={}, center]{f2290882-b9a4-43ec-a38f-c44d46477242-5_590_1041_283_588} Figure 2 shows a sketch of the curve \(C\) with equation \(y = \tan \frac { t } { 2 } , \quad 0 \leq t \leq \frac { \pi } { 2 }\).
   The point \(P\) on \(C\) has coordinates \(\left( x , \tan \frac { x } { 2 } \right)\).
   The vertices of rectangle \(R\) are at \(( x , 0 ) , \left( \frac { x } { 2 } , 0 \right) , \left( \frac { x } { 2 } , \tan \frac { x } { 2 } \right)\) and \(\left( x , \tan \frac { x } { 2 } \right)\) as shown in Figure 2.

5. $$I = \int \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x , \quad x > 1$$

1. Use the substitution \(x = 1 + u ^ { - 1 }\) to show that $$I = - \left( \frac { x + 1 } { x - 1 } \right) ^ { \frac { 1 } { 2 } } + c$$
2. Hence show that $$\int _ { \sec \alpha } ^ { \sec \beta } \frac { 1 } { ( x - 1 ) \sqrt { } \left( x ^ { 2 } - 1 \right) } \mathrm { d } x = \cot \left( \frac { \alpha } { 2 } \right) - \cot \left( \frac { \beta } { 2 } \right) , \quad 0 < \alpha < \beta < \frac { \pi } { 2 }$$

9

1. Sketch the graph of \(y = \tan \left( \frac { 1 } { 2 } x \right)\) for \(- 2 \pi \leqslant x \leqslant 2 \pi\) on the axes provided.
   On the same axes, sketch the graph of \(y = 3 \cos \left( \frac { 1 } { 2 } x \right)\) for \(- 2 \pi \leqslant x \leqslant 2 \pi\), indicating the point of intersection with the \(y\)-axis.
2. Show that the equation \(\tan \left( \frac { 1 } { 2 } x \right) = 3 \cos \left( \frac { 1 } { 2 } x \right)\) can be expressed in the form $$3 \sin ^ { 2 } \left( \frac { 1 } { 2 } x \right) + \sin \left( \frac { 1 } { 2 } x \right) - 3 = 0$$ Hence solve the equation \(\tan \left( \frac { 1 } { 2 } x \right) = 3 \cos \left( \frac { 1 } { 2 } x \right)\) for \(- 2 \pi \leqslant x \leqslant 2 \pi\).

5

1. Show that the equation \(2 \sin x = \frac { 4 \cos x - 1 } { \tan x }\) can be expressed in the form $$6 \cos ^ { 2 } x - \cos x - 2 = 0 .$$
2. Hence solve the equation \(2 \sin x = \frac { 4 \cos x - 1 } { \tan x }\), giving all values of \(x\) between \(0 ^ { \circ }\) and \(360 ^ { \circ }\).

7

1. Show that \(\frac { \sin ^ { 2 } x - \cos ^ { 2 } x } { 1 - \sin ^ { 2 } x } \equiv \tan ^ { 2 } x - 1\).
2. Hence solve the equation $$\frac { \sin ^ { 2 } x - \cos ^ { 2 } x } { 1 - \sin ^ { 2 } x } = 5 - \tan x$$ for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

3

1. Express \(2 \tan ^ { 2 } \theta - \frac { 1 } { \cos \theta }\) in terms of \(\sec \theta\).
2. Hence solve, for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), the equation $$2 \tan ^ { 2 } \theta - \frac { 1 } { \cos \theta } = 4$$

4 The acute angles \(\alpha\) and \(\beta\) are such that $$2 \cot \alpha = 1 \text { and } 24 + \sec ^ { 2 } \beta = 10 \tan \beta \text {. }$$

1. State the value of \(\tan \alpha\) and determine the value of \(\tan \beta\).
2. Hence find the exact value of \(\tan ( \alpha + \beta )\).

2 Using an appropriate identity in each case, find the possible values of

1. \(\sin \alpha\) given that \(4 \cos 2 \alpha = \sin ^ { 2 } \alpha\),
2. \(\sec \beta\) given that \(2 \tan ^ { 2 } \beta = 3 + 9 \sec \beta\).

4 It is given that \(A\) and \(B\) are angles such that $$\sec ^ { 2 } A - \tan A = 13 \quad \text { and } \quad \sin B \sec ^ { 2 } B = 27 \cos B \operatorname { cosec } ^ { 2 } B$$ Find the possible exact values of \(\tan ( A - B )\).

8 Fig. 8 shows part of the curve \(y = x \cos 3 x\).
The curve crosses the \(x\)-axis at \(\mathrm { O } , \mathrm { P }\) and Q . \begin{figure}[h] \end{figure}

1. Find the exact coordinates of P and Q .
2. Find the exact gradient of the curve at the point P . Show also that the turning points of the curve occur when \(x \tan 3 x = \frac { 1 } { 3 }\).
3. Find the area of the region enclosed by the curve and the \(x\)-axis between O and P , giving your answer in exact form.

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 \(\frac { 1 } { 8 } \pi \left( \pi ^ { 2 } - 2 \right)\).

4 Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } ( 1 + \sin x ) ^ { 2 } \mathrm {~d} x\).

2 Use the substitution \(x = \tan \theta\) to find the exact value of $$\int _ { 1 } ^ { \sqrt { 3 } } \frac { 1 - x ^ { 2 } } { 1 + x ^ { 2 } } \mathrm {~d} x$$

5 Solve the equation \(2 \sec ^ { 2 } \theta = 5 \tan \theta\), for \(0 \leqslant \theta \leqslant \pi\).

5 It is given that \(I = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \cos \theta } { 1 + \cos \theta } \mathrm { d } \theta\).

1. By using the substitution \(t = \tan \frac { 1 } { 2 } \theta\), show that \(I = \int _ { 0 } ^ { 1 } \left( \frac { 2 } { 1 + t ^ { 2 } } - 1 \right) \mathrm { d } t\).
2. Hence find \(I\) in terms of \(\pi\).

The curve \(C\) has equation \(y = 2 \sec x\), for \(0 \leqslant x \leqslant \frac { 1 } { 4 } \pi\). Show that the arc length \(s\) of \(C\) is given by $$S = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \left( 2 \sec ^ { 2 } x - 1 \right) d x$$ Find the exact value of \(s\). The surface area generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis is denoted by \(S\). Show that

1. \(S = 4 \pi \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \left( 2 \sec ^ { 3 } x - \sec x \right) \mathrm { d } x\),
2. \(\frac { \mathrm { d } } { \mathrm { d } x } ( \sec x \tan x ) = 2 \sec ^ { 3 } x - \sec x\). Hence find the exact value of \(S\).

10

1. Using de Moivre's theorem, show that $$\tan 5 \theta = \frac { 5 \tan \theta - 10 \tan ^ { 3 } \theta + \tan ^ { 5 } \theta } { 1 - 10 \tan ^ { 2 } \theta + 5 \tan ^ { 4 } \theta } .$$
2. Hence show that the equation \(x ^ { 2 } - 10 x + 5 = 0\) has roots \(\tan ^ { 2 } \left( \frac { 1 } { 5 } \pi \right)\) and \(\tan ^ { 2 } \left( \frac { 2 } { 5 } \pi \right)\).
3. Deduce a quadratic equation, with integer coefficients, having roots \(\sec ^ { 2 } \left( \frac { 1 } { 5 } \pi \right)\) and \(\sec ^ { 2 } \left( \frac { 2 } { 5 } \pi \right)\). [3]

8 A particle is projected with speed \(49 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation \(\theta\) from a point \(O\) on a horizontal plane, and moves freely under gravity. The horizontal and upward vertical displacements of the particle from \(O\) at time \(t\) seconds after projection are \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively.

1. Express \(x\) and \(y\) in terms of \(\theta\) and \(t\), and hence show that $$y = x \tan \theta - \frac { x ^ { 2 } \left( 1 + \tan ^ { 2 } \theta \right) } { 490 } .$$
   \includegraphics[max width=\textwidth, alt={}]{35477eb8-59e0-4de6-889c-1f5841f65eec-4_627_1249_1699_447}
   The particle passes through the point where \(x = 70\) and \(y = 30\). The two possible values of \(\theta\) are \(\theta _ { 1 }\) and \(\theta _ { 2 }\), and the corresponding points where the particle returns to the plane are \(A _ { 1 }\) and \(A _ { 2 }\) respectively (see diagram).
2. Find \(\theta _ { 1 }\) and \(\theta _ { 2 }\).
3. Calculate the distance between \(A _ { 1 }\) and \(A _ { 2 }\).

5

1. Show that \(\tan x\) is an integrating factor for the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { \sec ^ { 2 } x } { \tan x } y = \tan x$$ (2 marks)
2. Hence solve this differential equation, given that \(y = 3\) when \(x = \frac { \pi } { 4 }\).
   (6 marks)

6 The function f is defined by $$\mathrm { f } ( x ) = ( 9 + \tan x ) ^ { \frac { 1 } { 2 } }$$

1. 1. Find \(f ^ { \prime \prime } ( x )\).
   2. By using Maclaurin's theorem, show that, for small values of \(x\), $$( 9 + \tan x ) ^ { \frac { 1 } { 2 } } \approx 3 + \frac { x } { 6 } - \frac { x ^ { 2 } } { 216 }$$
2. Find $$\lim _ { x \rightarrow 0 } \left[ \frac { f ( x ) - 3 } { \sin 3 x } \right]$$

4 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \cot x ) y = \sin 2 x , \quad 0 < x < \frac { \pi } { 2 }$$ given that \(y = \frac { 1 } { 2 }\) when \(x = \frac { \pi } { 6 }\).
(10 marks)

5

1. Given that \(y = \ln ( 1 + 2 \tan x )\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
   (You may leave your expression for \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) unsimplified.)
2. Hence, using Maclaurin's theorem, find the first two non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln ( 1 + 2 \tan x )\).
   (2 marks)
3. Find $$\lim _ { x \rightarrow 0 } \left[ \frac { \ln ( 1 + 2 \tan x ) } { \ln ( 1 - x ) } \right]$$ (4 marks)

12 In this question you must show detailed reasoning. \includegraphics[max width=\textwidth, alt={}, center]{1ba9fa5f-310f-4429-9bd1-4004852d5b3e-6_716_479_292_794} The diagram shows the curve \(y = \frac { 4 \cos 2 x } { 3 - \sin 2 x }\), for \(x \geqslant 0\), and the normal to the curve at the point \(\left( \frac { 1 } { 4 } \pi , 0 \right)\). Show that the exact area of the shaded region enclosed by the curve, the normal to the curve and the \(y\)-axis is \(\ln \frac { 9 } { 4 } + \frac { 1 } { 128 } \pi ^ { 2 }\).
[0pt] [10]