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

8 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 \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).

7

1. Using the triangle, show that \(\sin ^ { 2 } x + \cos ^ { 2 } x = 1\).
2. Hence prove that \includegraphics[max width=\textwidth, alt={}]{73d1c02b-1b7b-426d-a171-c762597cfed4-2_255_501_1779_1022} \(1 + \tan ^ { 2 } x = \frac { 1 } { \cos ^ { 2 } x }\).

6.

1. Prove the identity $$2 \cot 2 x + \tan x \equiv \cot x , \quad x \neq \frac { n } { 2 } \pi , \quad n \in \mathbb { Z }$$
2. Solve, for \(0 \leq x < \pi\), the equation $$2 \cot 2 x + \tan x = \operatorname { cosec } ^ { 2 } x - 7$$ giving your answers to 2 decimal places.

1. Giving your answers to 1 decimal place, solve the equation

$$5 \tan ^ { 2 } 2 \theta - 13 \sec 2 \theta = 1 ,$$ for \(\theta\) in the interval \(0 \leq \theta \leq 360 ^ { \circ }\).

2 Solve, for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), the equation \(\sec ^ { 2 } \theta = 4 \tan \theta - 2\).

8 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.

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.

1 Fig. 4 shows the curve \(y = \mathrm { f } ( x )\), where $$f ( x ) = a + \cos b x , 0 \leqslant x \leqslant 2 \pi ,$$ and \(a\) and \(b\) are positive constants. The curve has stationary points at \(( 0,3 )\) and \(( 2 \pi , 1 )\). \begin{figure}[h] \end{figure}

1. Find \(a\) and \(b\).
2. Find \(\mathrm { f } ^ { - 1 } ( x )\), and state its domain and range.

5 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\). The endpoints of the curve are \(\mathrm { P } ( - \pi , 1 )\) and \(\mathrm { Q } ( \pi , 3 )\), and \(\mathrm { f } ( x ) = a + \sin b x\), where \(a\) and \(b\) are constants. \begin{figure}[h] \end{figure}

1. Using Fig. 9, show that \(a = 2\) and \(b = \frac { 1 } { 2 }\).
2. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point ( 0,2 ). Show that there is no point on the curve at which the gradient is greater than this.
3. Find \(\mathrm { f } ^ { - 1 } ( x )\), and state its domain and range. Write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 2,0 )\).
4. Find the area enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \pi\).

4 Fig. 8 shows parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where \(\mathrm { f } ( x ) = \tan x\) and \(\mathrm { g } ( x ) = 1 + \mathrm { f } \left( x - \frac { 1 } { 4 } \pi \right)\). \begin{figure}[h] \end{figure}

1. Describe a sequence of two transformations which maps the curve \(y = \mathrm { f } ( x )\) to the curve \(y = \mathrm { g } ( x )\). [4] It can be shown that \(\mathrm { g } ( x ) = \frac { 2 \sin x } { \sin x + \cos x }\).
2. Show that \(\mathrm { g } ^ { \prime } ( x ) = \frac { 2 } { ( \sin x + \cos x ) ^ { 2 } }\). Hence verify that the gradient of \(y = \mathrm { g } ( x )\) at the point \(\left( \frac { 1 } { 4 } \pi , 1 \right)\) is the same as that of \(y = \mathrm { f } ( x )\) at the origin.
3. By writing \(\tan x = \frac { \sin x } { \cos x }\) and using the substitution \(u = \cos x\), show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \mathrm { f } ( x ) \mathrm { d } x = \int _ { \frac { 1 } { \sqrt { 2 } } } ^ { 1 } \frac { 1 } { u } \mathrm {~d} u\). Evaluate this integral exactly.
4. Hence find the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the lines \(x = \frac { 1 } { 4 } \pi\) and \(x = \frac { 1 } { 2 } \pi\).

1 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 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 \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).

1. Differentiate \(x \cos 2 x\) with respect to \(x\).
2. Integrate \(x \cos 2 x\) with respect to \(x\).

4 Show that \(\frac { 1 + \tan ^ { 2 } \theta } { 1 - \tan ^ { 2 } \theta } = \sec 2 \theta\).
Hence, or otherwise, solve the equation \(\frac { 1 + \tan ^ { 2 } \theta } { 1 - \tan ^ { 2 } \theta } = 2\), for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

3 Solve the equation \(\sec ^ { 2 } \theta = 2 \tan \theta + 4\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).

2 Solve the equation \(3 \operatorname { cosec } ^ { 2 } x = 2 \cot ^ { 2 } x + 3\) for values of \(x\) in the range \(0 ^ { \circ } < x < 360 ^ { \circ }\).

5.

1. Given that $$x = \sec \frac { y } { 2 } , \quad 0 \leq y < \pi$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x \sqrt { x ^ { 2 } - 1 } }$$
2. Find an equation for the tangent to the curve \(y = \sqrt { 3 + 2 \cos x }\) at the point where \(x = \frac { \pi } { 3 }\).

2

1. Show that \(\cos ( \alpha + \beta ) = \frac { 1 - \tan \alpha \tan \beta } { \sec \alpha \sec \beta }\).
2. Hence show that \(\cos 2 \alpha = \frac { 1 - \tan ^ { 2 } \alpha } { 1 + \tan ^ { 2 } \alpha }\).
3. Hence or otherwise solve the equation \(\frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } = \frac { 1 } { 2 }\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

4 Prove that \(\sec ^ { 2 } \theta + \operatorname { cosec } ^ { 2 } \theta = \sec ^ { 2 } \theta \operatorname { cosec } ^ { 2 } \theta\).

5 Solve the equation \(\operatorname { cosec } ^ { 2 } \theta = 1 + 2 \cot \theta\), for \(- 180 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

6 Given that \(\operatorname { cosec } ^ { 2 } \theta - \cot \theta = 3\), show that \(\cot ^ { 2 } \theta - \cot \theta - 2 = 0\).
Hence solve the equation \(\operatorname { cosec } ^ { 2 } \theta - \cot \theta = 3\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

7 Given that \(x = 2 \sec \theta\) and \(y = 3 \tan \theta\), show that \(\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 9 } = 1\).

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