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

8 \includegraphics[max width=\textwidth, alt={}, center]{5b219e1c-e5a0-4f75-910d-fca9761e5088-3_435_895_799_625} The diagram shows the curve \(y = 5 \sin ^ { 3 } x \cos ^ { 2 } x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Using the substitution \(u = \cos x\), find by integration the area of the shaded region bounded by the curve and the \(x\)-axis.

5 \includegraphics[max width=\textwidth, alt={}, center]{4c71f68a-efb9-4408-bf03-874e0d4426d5-2_458_807_1786_667} The diagram shows the curve $$y = 8 \sin \frac { 1 } { 2 } x - \tan \frac { 1 } { 2 } x$$ for \(0 \leqslant x < \pi\). The \(x\)-coordinate of the maximum point is \(\alpha\) and the shaded region is enclosed by the curve and the lines \(x = \alpha\) and \(y = 0\).

1. Show that \(\alpha = \frac { 2 } { 3 } \pi\).
2. Find the exact value of the area of the shaded region.

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1. It is given that \(2 \tan 2 x + 5 \tan ^ { 2 } x = 0\). Denoting \(\tan x\) by \(t\), form an equation in \(t\) and hence show that either \(t = 0\) or \(t = \sqrt [ 3 ] { } ( t + 0.8 )\).
2. It is given that there is exactly one real value of \(t\) satisfying the equation \(t = \sqrt [ 3 ] { } ( t + 0.8 )\). Verify by calculation that this value lies between 1.2 and 1.3 .
3. Use the iterative formula \(t _ { n + 1 } = \sqrt [ 3 ] { } \left( t _ { n } + 0.8 \right)\) to find the value of \(t\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
4. Using the values of \(t\) found in previous parts of the question, solve the equation $$2 \tan 2 x + 5 \tan ^ { 2 } x = 0$$ for \(- \pi \leqslant x \leqslant \pi\).

9 \includegraphics[max width=\textwidth, alt={}, center]{436d891d-92ee-4076-8369-db756d413979-3_307_601_1553_772} The diagram shows the curve \(y = \sin ^ { 2 } 2 x \cos x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Using the substitution \(u = \sin x\), find by integration the area of the shaded region bounded by the curve and the \(x\)-axis.

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1. Find \(\int \left( 4 + \tan ^ { 2 } 2 x \right) \mathrm { d } x\).
2. Find the exact value of \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } \frac { \sin \left( x + \frac { 1 } { 6 } \pi \right) } { \sin x } \mathrm {~d} x\).

4 By first expressing the equation \(\cot \theta - \cot \left( \theta + 45 ^ { \circ } \right) = 3\) as a quadratic equation in \(\tan \theta\), solve the equation for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

3 Showing all necessary working, solve the equation \(\cot 2 \theta = 2 \tan \theta\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

2 Show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x ^ { 2 } \cos 2 x \mathrm {~d} x = \frac { 1 } { 32 } \left( \pi ^ { 2 } - 8 \right)\).

5 Let \(I = \int _ { 0 } ^ { 1 } \frac { 9 } { \left( 3 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x\).

1. Using the substitution \(x = ( \sqrt { } 3 ) \tan \theta\), show that \(I = \sqrt { } 3 \int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos ^ { 2 } \theta \mathrm {~d} \theta\).
2. Hence find the exact value of \(I\).

5 The curve with equation \(y = \mathrm { e } ^ { - a x } \tan x\), where \(a\) is a positive constant, has only one point in the interval \(0 < x < \frac { 1 } { 2 } \pi\) at which the tangent is parallel to the \(x\)-axis. Find the value of \(a\) and state the exact value of the \(x\)-coordinate of this point.

5 The variables \(x\) and \(y\) satisfy the relation \(\sin y = \tan x\), where \(- \frac { 1 } { 2 } \pi < y < \frac { 1 } { 2 } \pi\). Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \cos x \sqrt { } ( \cos 2 x ) } .$$

3 The equation of a curve is \(y = x + \cos 2 x\). Find the \(x\)-coordinates of the stationary points of the curve for which \(0 \leqslant x \leqslant \pi\), and determine the nature of each of these stationary points.

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1. Use the substitution \(x = 2 \tan \theta\) to show that $$\int _ { 0 } ^ { 2 } \frac { 8 } { \left( 4 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos ^ { 2 } \theta \mathrm {~d} \theta$$
2. Hence find the exact value of $$\int _ { 0 } ^ { 2 } \frac { 8 } { \left( 4 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x$$

5 Let \(I = \int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { \sqrt { } \left( 4 - x ^ { 2 } \right) } \mathrm { d } x\).

1. Using the substitution \(x = 2 \sin \theta\), show that $$I = \int _ { 0 } ^ { \frac { 1 } { 6 } \pi } 4 \sin ^ { 2 } \theta \mathrm {~d} \theta$$
2. Hence find the exact value of \(I\).

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1. Given that \(\sec \theta + 2 \operatorname { cosec } \theta = 3 \operatorname { cosec } 2 \theta\), show that \(2 \sin \theta + 4 \cos \theta = 3\).
2. Express \(2 \sin \theta + 4 \cos \theta\) in the form \(R \sin ( \theta + \alpha )\) where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the value of \(\alpha\) correct to 2 decimal places.
3. Hence solve the equation \(\sec \theta + 2 \operatorname { cosec } \theta = 3 \operatorname { cosec } 2 \theta\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).

2 A curve is defined for \(0 < \theta < \frac { 1 } { 2 } \pi\) by the parametric equations $$x = \tan \theta , \quad y = 2 \cos ^ { 2 } \theta \sin \theta$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 \cos ^ { 5 } \theta - 4 \cos ^ { 3 } \theta\).

5 Use the substitution \(u = 4 - 3 \cos x\) to find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 9 \sin 2 x } { \sqrt { ( 4 - 3 \cos x ) } } \mathrm { d } x\).

9 \includegraphics[max width=\textwidth, alt={}, center]{ccadf73b-16f5-463a-8f69-1394839d5325-3_481_483_1434_831} The diagram shows the curves \(y = x \cos x\) and \(y = \frac { k } { x }\), where \(k\) is a constant, for \(0 < x \leqslant \frac { 1 } { 2 } \pi\). The curves touch at the point where \(x = a\).

1. Show that \(a\) satisfies the equation \(\tan a = \frac { 2 } { a }\).
2. Use the iterative formula \(a _ { n + 1 } = \tan ^ { - 1 } \left( \frac { 2 } { a _ { n } } \right)\) to determine \(a\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
3. Hence find the value of \(k\) correct to 2 decimal places.

7 \includegraphics[max width=\textwidth, alt={}, center]{c861e691-66da-4269-9057-4a343be9835e-12_357_565_260_790} The diagram shows the curve \(y = 5 \sin ^ { 2 } x \cos ^ { 3 } x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\). The shaded region \(R\) is bounded by the curve and the \(x\)-axis.

1. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.
2. Using the substitution \(u = \sin x\) and showing all necessary working, find the exact area of \(R\). [4]

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1. By differentiating \(\frac { \cos x } { \sin x }\), show that if \(y = \cot x\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x\).
2. Show that \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } x \operatorname { cosec } ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 4 } ( \pi + \ln 4 )\). \(7 \quad\) Two lines \(l\) and \(m\) have equations \(\mathbf { r } = a \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } )\) and \(\mathbf { r } = 2 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } + \mu ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )\) respectively, where \(a\) is a constant. It is given that the lines intersect.

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1. Show that \(2 \operatorname { cosec } 2 \theta \cot \theta \equiv \operatorname { cosec } ^ { 2 } \theta\).
2. Hence show that \(\operatorname { cosec } ^ { 2 } 15 ^ { \circ } \tan 15 ^ { \circ } = 4\).
3. Solve the equation \(2 \operatorname { cosec } \phi \cot \frac { 1 } { 2 } \phi + \operatorname { cosec } \frac { 1 } { 2 } \phi = 12\) for \(- 360 ^ { \circ } < \phi < 360 ^ { \circ }\). Show all necessary working.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

1 Solve the equation \(\sec ^ { 2 } \theta + \tan ^ { 2 } \theta = 5 \tan \theta + 4\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\). Show all necessary working.

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1. Express \(\cos \theta + \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), giving the exact values of \(R\) and \(\alpha\).
2. Hence show that $$\frac { 1 } { ( \cos \theta + \sin \theta ) ^ { 2 } } = \frac { 1 } { 2 } \sec ^ { 2 } \left( \theta - \frac { 1 } { 4 } \pi \right)$$
3. By differentiating \(\frac { \sin x } { \cos x }\), show that if \(y = \tan x\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec ^ { 2 } x\).
4. Using the results of parts (ii) and (iii), show that $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { ( \cos \theta + \sin \theta ) ^ { 2 } } \mathrm {~d} \theta = 1$$

6 \includegraphics[max width=\textwidth, alt={}, center]{4029c46c-50a1-4d23-bc29-589417a6b7f5-3_501_497_269_826} The diagram shows the part of the curve \(y = \frac { \mathrm { e } ^ { 2 x } } { x }\) for \(x > 0\), and its minimum point \(M\).

1. Find the coordinates of \(M\).
2. Use the trapezium rule with 2 intervals to estimate the value of $$\int _ { 1 } ^ { 2 } \frac { \mathrm { e } ^ { 2 x } } { x } \mathrm {~d} x$$ giving your answer correct to 1 decimal place.
3. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii).
4. Given that \(y = \tan 2 x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
5. Hence, or otherwise, show that $$\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sec ^ { 2 } 2 x \mathrm {~d} x = \frac { 1 } { 2 } \sqrt { } 3$$ and, by using an appropriate trigonometrical identity, find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \tan ^ { 2 } 2 x \mathrm {~d} x\).
6. Use the identity \(\cos 4 x \equiv 2 \cos ^ { 2 } 2 x - 1\) to find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \frac { 1 } { 1 + \cos 4 x } \mathrm {~d} x$$

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1. (a) Prove the identity $$\sec ^ { 2 } x + \sec x \tan x \equiv \frac { 1 + \sin x } { \cos ^ { 2 } x }$$ (b) Hence prove that $$\sec ^ { 2 } x + \sec x \tan x \equiv \frac { 1 } { 1 - \sin x }$$
2. By differentiating \(\frac { 1 } { \cos x }\), show that if \(y = \sec x\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec x \tan x\).
3. Using the results of parts (i) and (ii), find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 } { 1 - \sin x } \mathrm {~d} x$$