Higher Engineering Mathematics, Sixth Edition

340 Higher Engineering Mathematics

(iv) press ) to close the brackets

(v) press=and 1.443635475 appears

Hence,sinh−^12 = 1. 4436 , correct to 4 decimal places.

Problem 10. Show that

tanh−^1

x

a

=

1

2

ln

(

a+x

a−x

)

and evaluate, correct

to 4 decimal places, tanh−^1

3

5

Ify=tanh−^1

x

a

then

x

a

=tanhy.

From Chapter 5,

tanhy=

sinhx

coshx

=

1

2 (e

y−e−y)

1

2 (e

y+e−y)=

e^2 y− 1

e^2 y+ 1

by dividing each term by e−y

Thus,

x

a

=

e^2 y− 1

e^2 y+ 1

from which, x(e^2 y+ 1 )=a(e^2 y− 1 )

Hencex+a=ae^2 y−xe^2 y=e^2 y(a−x)

from which e^2 y=

(

a+x

a−x

)

Taking Napierian logarithms of both sides gives:

2 y=ln

(

a+x

a−x

)

and y=

1

2

ln

(

a+x

a−x

)

Hence,tanh−^1

x

a

=

1

2

ln

(

a+x

a−x

)

Substitutingx=3anda=5gives:

tanh−^1

3

5

=

1

2

ln

(

5 + 3

5 − 3

)

=

1

2

ln4

=0.6931,correct to 4 decimal places.

Problem 11. Prove that

cosh−^1

x

a

=ln

{

x+

√

x^2 −a^2

a

}

and hence evaluate cosh−^1 1.4 correct to

4 decimal places.

Ify=cosh−^1

x

a

then

x

a

=cosy

ey=coshy+sinhy=coshy±

√

cosh^2 y− 1

=

x

a

±

√[

(x

a

) 2

− 1

]

=

x

a

±

√

x^2 −a^2

a

=

x±

√

x^2 −a^2

a

Taking Napierian logarithms of both sides gives:

y=ln

{

x±

√

x^2 −a^2

a

}

Thus, assuming the principal value,

cosh−^1

x

a

=ln

{

x+

√

x^2 −a^2

a

}

cosh−^11. 4 =cosh−^1

14

10

=cosh−^1

7

5

In the equation for cosh−^1

x

a

,letx=7anda= 5

Then cosh−^1

7

5

=ln

{

7 +

√

72 − 52

5

}

=ln2. 3798 =0.8670,

correct to 4 decimal places.

Now try the following exercise

Exercise 136 Further problems on

logarithmic forms of the inverse hyperbolic

functions

In Problems 1 to 3 use logarithmic equivalents of

inverse hyperbolic functions to evaluate correct to

4 decimal places.

1. (a) sinh−^1

1

2

(b) sinh−^1 4(c)sinh−^1 0.9

[(a) 0.4812 (b) 2.0947 (c) 0.8089]

2. (a) cosh−^1

5

4

(b) cosh−^1 3 (c) cosh−^1 4.3

[(a) 0.6931 (b) 1.7627 (c) 2.1380]

3. (a) tanh−^1

1

4

(b) tanh−^1

5

8

(c) tanh−^1 0.7

[(a) 0.2554 (b) 0.7332 (c) 0.8673]