Hyperbolic functions: sine and cosine's cousins from eˣ

The problem

Hang a chain between two poles and it sags into a curve that looks like a parabola, but isn't. Its true shape needs a new function built from eˣ, called cosh. Its partner is sinh.

They are called hyperbolic for a reason. The point (cosh t, sinh t) always lies on the hyperbola x² − y² = 1. In the same way, (cos t, sin t) lies on the circle x² + y² = 1. One sign changes everything.

sinh x = ½(eˣ − e⁻ˣ), with domain and range all real numbers. cosh x = ½(eˣ + e⁻ˣ), with domain all real numbers and range [1, +∞). The other four are built from these two.

Key terms

sinh x

Half the difference of eˣ and e⁻ˣ. It is zero at x = 0, odd like sin x, and takes every real value.

cosh x

Half the sum of eˣ and e⁻ˣ. It equals 1 at x = 0 and is never smaller, so its range is [1, +∞). It is even like cos x.

tanh, coth, sech, csch

tanh x = sinh x/cosh x and sech x = 1/cosh x work for all x. coth x = cosh x/sinh x and csch x = 1/sinh x need x ≠ 0, because sinh 0 = 0.

Inverse hyperbolic

Because sinh and cosh are built from eˣ, their inverses can be written with natural logarithms, e.g. sinh⁻¹x = ln(x + √(x² + 1)). The book says these will be studied in higher classes.

Short questions with model answers

1. Q1. Using e = 2.7183, find sinh 1 and cosh 1.

   • e⁻¹ = 1 ÷ 2.7183 = 0.3679
   • sinh 1 = ½(2.7183 − 0.3679) = 1.1752
   • cosh 1 = ½(2.7183 + 0.3679) = 1.5431

   sinh 1 = 1.1752 · cosh 1 = 1.5431

2. Q2. Prove that cosh²x − sinh²x = 1.

   • cosh²x = (e²ˣ + 2 + e⁻²ˣ) ÷ 4
   • sinh²x = (e²ˣ − 2 + e⁻²ˣ) ÷ 4
   • cosh²x − sinh²x = (2 + 2) ÷ 4 = 1

   cosh²x − sinh²x = 1

3. Q3. Prove that sinh 2x = 2 sinh x cosh x.

   • 2 · ½(eˣ − e⁻ˣ) · ½(eˣ + e⁻ˣ) = ½(e²ˣ − e⁻²ˣ)
   • ½(e²ˣ − e⁻²ˣ) = sinh 2x

   sinh 2x = 2 sinh x cosh x

Common mistakes

• ✗ Writing cosh²x + sinh²x = 1, copied from the trigonometric identity.

  ✓ It is cosh²x − sinh²x = 1. The sum cosh²x + sinh²x equals cosh 2x (book Example 2).

• ✗ Giving the range of cosh x as all real numbers, like sinh x.

  ✓ cosh x = ½(eˣ + e⁻ˣ) is smallest at x = 0, where it equals 1. Its range is [1, +∞).

• ✗ Swapping the definitions: sinh x = ½(eˣ + e⁻ˣ).

  ✓ Check at x = 0: sinh 0 must be 0 (like sin 0), so sinh uses the minus sign; cosh 0 = 1 uses the plus.

MCQs

1. 1. The range of cosh x is:

   1. (a) all real numbers
   2. (b) [0, +∞)
   3. (c) [1, +∞)
   4. (d) [−1, 1]
   Show answer

   (c) Its minimum is cosh 0 = 1, so the range is [1, +∞).

2. 2. sinh 0 equals:

   1. (a) 0
   2. (b) 1
   3. (c) e
   4. (d) ½
   Show answer

   (a) ½(e⁰ − e⁰) = ½(1 − 1) = 0.

3. 3. Which identity is correct?

   1. (a) cosh²x + sinh²x = 1
   2. (b) cosh²x − sinh²x = 1
   3. (c) sinh²x − cosh²x = 1
   4. (d) cosh x − sinh x = 1
   Show answer

   (b) cosh²x − sinh²x = 1, the hyperbolic twin of cos²x + sin²x = 1.

4. 4. tanh x in terms of e is:

   1. (a) (eˣ + e⁻ˣ)/(eˣ − e⁻ˣ)
   2. (b) (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ)
   3. (c) 2/(eˣ + e⁻ˣ)
   4. (d) 2/(eˣ − e⁻ˣ)
   Show answer

   (b) tanh = sinh/cosh; the ½s cancel. The first option is coth, the third sech, the fourth csch.

5. 5. For which x is csch x undefined?

   1. (a) x = 0
   2. (b) x = 1
   3. (c) x < 0
   4. (d) never
   Show answer

   (a) csch x = 1/sinh x and sinh 0 = 0, so x = 0 is excluded.

Quick revision

• sinh x = ½(eˣ − e⁻ˣ) (range ℝ); cosh x = ½(eˣ + e⁻ˣ) (range [1, +∞)); tanh, coth, sech, csch come from them.
• Key identities: cosh²x − sinh²x = 1, cosh²x + sinh²x = cosh 2x, sinh 2x = 2 sinh x cosh x.
• Inverse hyperbolic functions are written with ln, e.g. sinh⁻¹x = ln(x + √(x² + 1)).