340 Higher Engineering Mathematics
(iv) press ) to close the brackets
(v) press = and 1.443635475 appears
Hence, sinh
−1 2 = 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
If y = tanh −1 x
a
then
x
a
= tanh y.
From Chapter 5,
tanh y =
sinh x
cosh x
=
1
2 (e y − e −y )
1
2 (e y + e −y )
=
e 2y − 1
e 2y + 1
by dividing each term by e
−y
Thus,
x
a
=
e 2y − 1
e 2y + 1
from which, x(e 2y + 1) = a(e 2y − 1)
Hence x + a = ae 2y − xe 2y = e 2y (a − x)
from which e 2y =
a + x
a − x
Taking Napierian logarithms of both sides gives:
2y = 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
Substituting x = 3 and a = 5 gives:
tanh
−1 3
5
=
1
2
ln
5 + 3
5 − 3
=
1
2
ln 4
= 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.
If y = cosh −1 x
a
then
x
a
= cos y
e y = cosh y + sinh y = cosh y ±
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 −1 1.4 = cosh −1 14
10
= cosh −1 7
5
In the equation for cosh −1 x
a
, let x = 7 and a = 5
Then cosh
−1 7
5
= ln
7 +
√
7 2 − 5 2
5
= ln 2.3798 = 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]