50 Higher Engineering Mathematics
In the series expansion for sh x, let x = 2θ, then:
sh 2θ = 2θ +
(2θ) 3
3!
+
(2θ) 5
5!
+ · · ·
= 2θ +
4
3
θ
3
+
4
15
θ
5
+ · · ·
Hence
ch
θ
2
− sh 2θ =
2 +
θ 2
4
+
θ 4
192
+ · · ·
−
2θ +
4
3
θ
3
+
4
15
θ
5
+ · · ·
= 2 −2θ +
θ
2
4
−
4
3
θ
3
+
θ
4
192
−
4
15
θ
5
+ · · · as far the term in θ
5
Now try the following exercise
Exercise 23 Further problems on series
expansions for cosh x and sinh x
1. Use the series expansion for ch x to evaluate,
correct to 4 decimal places: (a) ch 1.5 (b) ch 0.8
[(a) 2.3524 (b) 1.3374]
2. Use the series expansion for sh x to evaluate, correct to 4 decimal places: (a) sh 0.5
(b) sh 2
[(a) 0.5211 (b) 3.6269]
3. Expand the following as a power series as far
as the term in x 5 : (a) sh 3x (b) ch 2x
⎡
⎢
⎣
(a) 3x +
9
2
x 3 +
81
40
x 5
(b) 1 + 2x 2 +
2
3
x 4
⎤
⎥
⎦
In Problems 4 and 5, prove the given identities,
the series being taken as far as the term in θ
5
only.
4. sh 2θ − sh θ ≡ θ +
7
6
θ 3 +
31
120
θ 5
5. 2 sh
θ
2
− ch
θ
2
≡ − 1 + θ −
θ 2
8
+
θ 3
24
−
θ 4
384
+
θ 5
1920