4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1

4

1. Starting from the definitions of \(\sinh x\) and \(\cosh x\) in terms of exponentials, prove that $$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$$
2. Solve the equation \(4 \cosh ^ { 2 } x + 9 \sinh x = 13\), giving the answers in exact logarithmic form.
3. Show that there is only one stationary point on the curve $$y = 4 \cosh ^ { 2 } x + 9 \sinh x$$ and find the \(y\)-coordinate of the stationary point.
4. Show that \(\int _ { 0 } ^ { \ln 2 } \left( 4 \cosh ^ { 2 } x + 9 \sinh x \right) \mathrm { d } x = 2 \ln 2 + \frac { 33 } { 8 }\).

4

1. Prove, using exponential functions, that $$\sinh 2 x = 2 \sinh x \cosh x$$ Differentiate this result to obtain a formula for \(\cosh 2 x\).
2. Sketch the curve with equation \(y = \cosh x - 1\). The region bounded by this curve, the \(x\)-axis, and the line \(x = 2\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Find, correct to 3 decimal places, the volume generated. (You must show your working; numerical integration by calculator will receive no credit.)
3. Show that the curve with equation $$y = \cosh 2 x + \sinh x$$ has exactly one stationary point.
   Determine, in exact logarithmic form, the \(x\)-coordinate of the stationary point.

8

1. Define tanh \(y\) in terms of \(\mathrm { e } ^ { y }\) and \(\mathrm { e } ^ { - y }\).
2. Given that \(y = \tanh ^ { - 1 } x\), where \(- 1 < x < 1\), prove that \(y = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\).
3. Find the exact solution of the equation \(3 \cosh x = 4 \sinh x\), giving the answer in terms of a logarithm.
4. Solve the equation $$\tanh ^ { - 1 } x + \ln ( 1 - x ) = \ln \left( \frac { 4 } { 5 } \right)$$

4

1. Using the definition of \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), prove that $$\cosh 2 x = 2 \cosh ^ { 2 } x - 1$$
2. Hence solve the equation $$\cosh 2 x - 7 \cosh x = 3$$ giving your answer in logarithmic form.

7

1. Using the definitions of hyperbolic functions in terms of exponentials, prove that $$\cosh x \cosh y - \sinh x \sinh y = \cosh ( x - y )$$
2. Given that \(\cosh x \cosh y = 9\) and \(\sinh x \sinh y = 8\), show that \(x = y\).
3. Hence find the values of \(x\) and \(y\) which satisfy the equations given in part (ii), giving the answers in logarithmic form.

1

1. By first expanding \(\left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right) ^ { 3 }\), or otherwise, show that \(\cosh 3 x \equiv 4 \cosh ^ { 3 } x - 3 \cosh x\).
2. Solve the equation \(\cosh 3 x = 6 \cosh x\), giving your answers in exact logarithmic form.

1

1. Starting from the definition of \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\), show that \(\cosh 2 x = 2 \cosh ^ { 2 } x - 1\).
2. Given that \(\cosh 2 x = k\), where \(k > 1\), express each of \(\cosh x\) and \(\sinh x\) in terms of \(k\).

4

1. 1. Prove, from definitions involving exponentials, that $$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$$
   2. Given that \(\sinh x = \tan y\), where \(- \frac { 1 } { 2 } \pi < y < \frac { 1 } { 2 } \pi\), show that
      (A) \(\tanh x = \sin y\),
      (B) \(x = \ln ( \tan y + \sec y )\).
2. 1. Given that \(y = \operatorname { artanh } x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\). Hence show that \(\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { 1 - x ^ { 2 } } \mathrm {~d} x = 2 \operatorname { artanh } \frac { 1 } { 2 }\).
   2. Express \(\frac { 1 } { 1 - x ^ { 2 } }\) in partial fractions and hence find an expression for \(\int \frac { 1 } { 1 - x ^ { 2 } } \mathrm {~d} x\) in terms of logarithms.
   3. Use the results in parts (i) and (ii) to show that \(\operatorname { artanh } \frac { 1 } { 2 } = \frac { 1 } { 2 } \ln 3\).

4

1. Prove, using exponential functions, that $$\cosh 2 x = 1 + 2 \sinh ^ { 2 } x$$ Differentiate this result to obtain a formula for \(\sinh 2 x\).
2. Solve the equation $$2 \cosh 2 x + 3 \sinh x = 3$$ expressing your answers in exact logarithmic form.
3. Given that \(\cosh t = \frac { 5 } { 4 }\), show by using exponential functions that \(t = \pm \ln 2\). Find the exact value of the integral $$\int _ { 4 } ^ { 5 } \frac { 1 } { \sqrt { x ^ { 2 } - 16 } } \mathrm {~d} x$$

4

1. Prove, from definitions involving exponential functions, that $$\cosh 2 u = 2 \sinh ^ { 2 } u + 1$$
2. Prove that, if \(y \geqslant 0\) and \(\cosh y = u\), then \(y = \ln \left( u + \sqrt { } \left( u ^ { 2 } - 1 \right) \right)\).
3. Using the substitution \(2 x = \cosh u\), show that $$\int \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x = a x \sqrt { 4 x ^ { 2 } - 1 } - b \operatorname { arcosh } 2 x + c$$ where \(a\) and \(b\) are constants to be determined and \(c\) is an arbitrary constant.
4. Find \(\int _ { \frac { 1 } { 2 } } ^ { 1 } \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x\), expressing your answer in an exact form involving logarithms.

4

1. Prove, using exponential functions, that \(\cosh ^ { 2 } u - \sinh ^ { 2 } u = 1\).
2. Given that \(y = \operatorname { arsinh } x\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }$$ and that $$y = \ln \left( x + \sqrt { 1 + x ^ { 2 } } \right)$$
3. Show that $$\int _ { 0 } ^ { 2 } \frac { 1 } { \sqrt { 4 + 9 x ^ { 2 } } } \mathrm {~d} x = \frac { 1 } { 3 } \ln ( 3 + \sqrt { 10 } )$$
4. Find, in exact logarithmic form, $$\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 + x ^ { 2 } } } \operatorname { arsinh } x \mathrm {~d} x$$

4

1. Use the definitions of hyperbolic functions in terms of exponentials to prove that $$8 \sinh ^ { 4 } x \equiv \cosh 4 x - 4 \cosh 2 x + 3$$
2. Solve the equation $$\cosh 4 x - 3 \cosh 2 x + 1 = 0$$ giving your answer(s) in logarithmic form.

8

1. Using the definitions of \(\sinh x\) and \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that
   1. \(\cosh ( \ln a ) \equiv \frac { a ^ { 2 } + 1 } { 2 a }\), where \(a > 0\),
   2. \(\cosh x \cosh y - \sinh x \sinh y \equiv \cosh ( x - y )\).
   3. Use part (i)(b) to show that \(\cosh ^ { 2 } x - \sinh ^ { 2 } x \equiv 1\).
   4. Given that \(R > 0\) and \(a > 1\), find \(R\) and \(a\) such that $$13 \cosh x - 5 \sinh x \equiv R \cosh ( x - \ln a )$$
   5. Hence write down the coordinates of the minimum point on the curve with equation \(y = 13 \cosh x - 5 \sinh x\).

8

1. Using the definition of \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that $$4 \cosh ^ { 3 } x - 3 \cosh x \equiv \cosh 3 x$$
2. Use the substitution \(u = \cosh x\) to find, in terms of \(5 ^ { \frac { 1 } { 3 } }\), the real root of the equation $$20 u ^ { 3 } - 15 u - 13 = 0 .$$

3

1. By quoting results given in the List of Formulae (MF1), prove that \(\tanh 2 x \equiv \frac { 2 \tanh x } { 1 + \tanh ^ { 2 } x }\).
2. Solve the equation \(5 \tanh 2 x = 1 + 6 \tanh x\), giving your answers in logarithmic form.

2

1. Using the definitions for \(\cosh x\) and \(\sinh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that \(\cosh ^ { 2 } x - \sinh ^ { 2 } x \equiv 1\).
2. Hence solve the equation \(\sinh ^ { 2 } x = 5 \cosh x - 7\), giving your answers in logarithmic form.

7

1. By using the definitions of \(\cosh u\) and \(\sinh u\) in terms of \(\mathrm { e } ^ { u }\) and \(\mathrm { e } ^ { - u }\), show that \(\sinh 2 u \equiv 2 \sinh u \cosh u\). The equation of a curve, \(C\), is \(\mathrm { y } = 16 \cosh \mathrm { x } - \sinh 2 \mathrm { x }\).
2. Show that there is only one solution to the equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } = 0\) You are now given that \(C\) has exactly one point of inflection.
3. Use your answer to part (b) to determine the exact coordinates of this point of inflection. Give your answer in a logarithmic form where appropriate.

8

1. Using exponentials, show that \(\cosh 2 u \equiv 2 \sinh ^ { 2 } u + 1\).
2. By differentiating both sides of the identity in part (a) with respect to \(u\), show that \(\sinh 2 u \equiv 2 \sinh u \cosh u\).
3. Use the substitution \(\mathrm { x } = \sinh ^ { 2 } \mathrm { u }\) to find \(\int \sqrt { \frac { x } { x + 1 } } \mathrm {~d} x\). Give your answer in the form asinh \(^ { - 1 } \mathrm {~b} \sqrt { \mathrm { x } } + \mathrm { f } ( \mathrm { x } )\) where \(a\) and \(b\) are integers and \(\mathrm { f } ( x )\) is a function to be determined.
4. Hence determine the exact area of the region between the curve \(\mathrm { y } = \sqrt { \frac { \mathrm { x } } { \mathrm { x } + 1 } }\), the \(x\)-axis, the line \(x = 1\) and the line \(x = 2\). Give your answer in the form \(\mathrm { p } + \mathrm { q } \mid \mathrm { nr }\) where \(p , q\) and \(r\) are numbers to be determined.

5

1. By using the exponential definitions of \(\sinh x\) and \(\cosh x\), prove the identity \(\cosh 2 x \equiv \cosh ^ { 2 } x + \sinh ^ { 2 } x\).
2. Hence find an expression for \(\cosh 2 x\) in terms of \(\cosh x\).
3. Determine the solutions of the equation \(5 \cosh 2 x = 16 \cosh x + 21\), giving your answers in exact logarithmic form.

5 In this question you must show detailed reasoning.

1. Using the definition of \(\cosh x\) in terms of exponentials, show that \(\cosh 2 x \equiv 2 \cosh ^ { 2 } x - 1\).
2. Solve the equation \(\cosh 2 x = 3 \cosh x + 1\), giving all your answers in exact logarithmic form.

5 A designer intends to manufacture a product using a 3-D printer. The product will take the form of a surface \(S\) which must meet a number of design specifications. The designer chooses to model \(S\) with the equation \(\mathrm { Z } = \mathrm { y } \cosh \mathrm { x }\) for \(- \ln 20 \leqslant x \leqslant \ln 20 , - 2 \leqslant y \leqslant 2\).

1. 1. In the Printed Answer Booklet, on the axes provided, sketch the section of \(S\) given by \(y = 1\).
   2. One of the design specifications of the product is that this section should have a length no greater than 20 units. Determine whether the product meets this requirement according to the model.
2. 1. In the Printed Answer Booklet, on the axes provided, sketch the contour of \(S\) given by \(z = 1\).
   2. When this contour is rotated through \(2 \pi\) radians about the \(x\)-axis, the surface \(T\) is generated. The surface area of \(T\) is denoted by \(A\). Show that \(A\) can be written in the form \(\mathrm { k } \pi \int _ { 0 } ^ { \ln 20 } \frac { 1 } { \cosh ^ { 3 } \mathrm { x } } \sqrt { \cosh ^ { 4 } \mathrm { x } + \cosh ^ { 2 } \mathrm { x } - 1 } \mathrm { dx }\) for some
      integer \(k\) to be determined. integer \(k\) to be determined.
   3. A second design specification is that the surface area of \(T\) must not be greater than 20 square units. Use your calculator to decide whether the product meets this requirement according to the model.

1

1. Use the definitions \(\cosh x = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right)\) and \(\sinh x = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } \right)\) to show that $$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$$
2. 1. Express $$5 \cosh ^ { 2 } x + 3 \sinh ^ { 2 } x$$ in terms of \(\cosh x\).
   2. Sketch the curve \(y = \cosh x\).
   3. Hence solve the equation $$5 \cosh ^ { 2 } x + 3 \sinh ^ { 2 } x = 9.5$$ giving your answers in logarithmic form.

5

1. Use the definition \(\cosh x = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right)\) to show that \(\cosh 2 x = 2 \cosh ^ { 2 } x - 1\).
   (2 marks)
2. 1. The arc of the curve \(y = \cosh x\) between \(x = 0\) and \(x = \ln a\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Show that \(S\), the surface area generated, is given by $$S = 2 \pi \int _ { 0 } ^ { \ln a } \cosh ^ { 2 } x \mathrm {~d} x$$
   2. Hence show that $$S = \pi \left( \ln a + \frac { a ^ { 4 } - 1 } { 4 a ^ { 2 } } \right)$$

5

1. Using the identities $$\cosh ^ { 2 } t - \sinh ^ { 2 } t = 1 , \quad \tanh t = \frac { \sinh t } { \cosh t } \quad \text { and } \quad \operatorname { sech } t = \frac { 1 } { \cosh t }$$ show that:
   1. \(\tanh ^ { 2 } t + \operatorname { sech } ^ { 2 } t = 1\);
   2. \(\frac { \mathrm { d } } { \mathrm { d } t } ( \tanh t ) = \operatorname { sech } ^ { 2 } t\);
   3. \(\frac { \mathrm { d } } { \mathrm { d } t } ( \operatorname { sech } t ) = - \operatorname { sech } t \tanh t\).
2. A curve \(C\) is given parametrically by $$x = \operatorname { sech } t , y = 4 - \tanh t$$
   1. Show that the arc length, \(s\), of \(C\) between the points where \(t = 0\) and \(t = \frac { 1 } { 2 } \ln 3\) is given by $$s = \int _ { 0 } ^ { \frac { 1 } { 2 } \ln 3 } \operatorname { sech } t \mathrm {~d} t$$
   2. Using the substitution \(u = \mathrm { e } ^ { t }\), find the exact value of \(s\).

      REFERENCE
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6

1. Show that $$\frac { 1 } { 4 } ( \cosh 4 x + 2 \cosh 2 x + 1 ) = \cosh ^ { 2 } x \cosh 2 x$$
2. Show that, if \(y = \cosh ^ { 2 } x\), then $$1 + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = \cosh ^ { 2 } 2 x$$
3. The arc of the curve \(y = \cosh ^ { 2 } x\) between the points where \(x = 0\) and \(x = \ln 2\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Show that the area \(S\) of the curved surface formed is given by $$S = \frac { \pi } { 256 } ( a \ln 2 + b )$$ where \(a\) and \(b\) are integers.