Questions on hyperbolic functions and identities

Question 1

Write down the definitions of cosh x and sinh x in terms of exponentials, and state the value of each at x = 0.

Accepted Answer

cosh x = (e^x + e^−x)/2 and sinh x = (e^x − e^−x)/2. At zero, cosh 0 = 1 and sinh 0 = 0, mirroring cos 0 and sin 0. B1 for the two definitions, B1 for both values at zero.

Question 2

Using the definitions, find the exact values of cosh(ln 2) and sinh(ln 2).

Accepted Answer

e^ln 2 = 2 and e^−ln 2 = 1/2, so cosh(ln 2) = (2 + 1/2)/2 = 5/4 and sinh(ln 2) = (2 − 1/2)/2 = 3/4. M1 for both exponential values, A1 for 5/4, A1 for 3/4. As a check, (5/4)² − (3/4)² = 1.

Question 3

Write down sech x, cosech x and coth x in terms of cosh x, sinh x and tanh x, and state which of the three are undefined at x = 0.

Accepted Answer

sech x = 1/cosh x, cosech x = 1/sinh x and coth x = 1/tanh x. B1 for sech, B1 for cosech and coth. cosech x and coth x are undefined at x = 0, since sinh 0 = 0 and tanh 0 = 0, so each would need 1/0; sech 0 = 1/1 = 1. B1 for identifying both with a reason. Only sech is defined at every real x, since cosh x is never zero.

Question 4

Use Osborn's rule to write down the hyperbolic identity corresponding to cos²θ = (1 + cos 2θ)/2.

Accepted Answer

Neither term here carries an isolated sine factor, so nothing changes sign: cosh²x = (1 + cosh 2x)/2. B1 for replacing cos by cosh throughout, B1 for no sign change.

Question 5

Solve sinh x = 2, giving the answer in exact logarithmic form.

Accepted Answer

arsinh x = ln(x + √(x² + 1)), so x = ln(2 + √5) ≈ 1.444. sinh is one-to-one, so this is the only solution. M1 for the logarithmic form of arsinh, A1 for ln(2 + √5), B1 for the uniqueness. No ± appears here, unlike the cosh case in the next question.

Question 6

Solve cosh x = 5/4, giving both answers exactly, and explain why there are two.

Accepted Answer

arcosh(5/4) = ln(5/4 + √(25/16 − 1)) = ln(5/4 + 3/4) = ln 2. cosh is even, so x = ±ln 2. M1 for the logarithmic form, A1 for ln 2, A1 for the second root, B1 for the evenness argument. Every value of cosh above 1 is hit twice, once on each side of the valley.

Question 7

Starting from cosh²x − sinh²x = 1, derive an identity connecting tanh²x and sech²x.

Accepted Answer

Divide the whole identity by cosh²x: 1 − tanh²x = sech²x. M1 for dividing through by cosh²x, A1 for the identity. The trig counterpart is 1 + tan²x = sec²x; the sign flip on the tanh² term is Osborn's rule at work, since tanh² contains a product of two sinh terms.

Question 8

Using cosh²x = 1 + sinh²x, solve 2 cosh²x − sinh x = 2, giving exact answers.

Accepted Answer

Substitute: 2 + 2 sinh²x − sinh x = 2, so sinh x (2 sinh x − 1) = 0. Either sinh x = 0, giving x = 0, or sinh x = 1/2, giving x = ln(1/2 + √(5/4)) = ln((1 + √5)/2). M1 for the substitution, A1 for the factorised quadratic, B1 for x = 0, M1 for the logarithmic form, A1 for ln((1 + √5)/2).

Question 9

Sketch the graphs of sinh x and cosh x on the same axes for −2 ≤ x ≤ 2, marking the point where each crosses an axis, and explain why the two curves never meet.

Accepted Answer

sinh x is increasing through the origin, crossing the x-axis only at (0, 0). cosh x crosses the y-axis at its minimum, (0, 1), and is symmetric about that axis, rising on both sides. B1 for sinh through (0, 0), rising left to right, B1 for cosh with minimum (0, 1), symmetric and never below 1. The two curves never meet, because cosh x − sinh x = e^−x, and e^−x > 0 for every real x, so cosh x is always strictly greater than sinh x. B1 for the difference e^−x, B1 for concluding it is never zero. The gap e^−x shrinks as x increases, so the curves draw closer to the right, but it never reaches 0.

Question 10

Solve tanh x = 3/5, giving the answer in exact logarithmic form.

Accepted Answer

artanh x = ½ln((1 + x)/(1 − x)), so x = ½ln((1 + 3/5)/(1 − 3/5)) = ½ln(8/5 ÷ 2/5) = ½ln 4 = ln 2. M1 for the logarithmic form of artanh, M1 for the fraction (1 + 3/5)/(1 − 3/5), A1 for ln 2. tanh is one-to-one, so this is the only solution; no ± is needed, as for sinh and unlike cosh. B1 for the uniqueness. Checking back, tanh(ln 2) = (2 − 1/2)/(2 + 1/2) = (3/2)/(5/2) = 3/5.

Question 11

Solve cosech x = 3/4, giving the answer in exact logarithmic form.

Accepted Answer

cosech x = 3/4 means sinh x = 4/3. Then arsinh(4/3) = ln(4/3 + √(16/9 + 1)) = ln(4/3 + 5/3) = ln 3. M1 for rewriting as sinh x = 4/3, M1 for the logarithmic form of arsinh, A1 for ln 3. sinh is one-to-one, so cosech is too away from x = 0, and this is the only solution. B1 for the check. Checking back, sinh(ln 3) = (3 − 1/3)/2 = 4/3, so cosech(ln 3) = 3/4.

Question 12

Solve 5 cosh x − 4 sinh x = 3, giving your answer in exact form, and explain why the equation has only one solution.

Accepted Answer

Go back to the definitions. The left side is 5(e^x + e^−x)/2 − 4(e^x − e^−x)/2 = (e^x + 9e^−x)/2, so the equation becomes e^x + 9e^−x = 6. Writing u = e^x and multiplying by u gives u² − 6u + 9 = 0, that is (u − 3)² = 0. The repeated root u = 3 gives x = ln 3 and nothing else. Geometrically the curve y = 5 cosh x − 4 sinh x has minimum value 3, since √(5² − 4²) = 3, so the line y = 3 touches it rather than cutting it. M1 for using the exponential definitions, A1 for e^x + 9e^−x = 6, M1 for the quadratic in u, A1 for the repeated root, A1 for ln 3, B1 for the reason there is only one solution. Substituting u for e^x is quicker here than finding the single-function form 3 cosh(x − ln 3).

Question 13

State the domain and range of sinh x, cosh x, tanh x, sech x, cosech x and coth x, and of arsinh x, arcosh x and artanh x. Explain why cosh must be restricted before its inverse can be defined.

Accepted Answer

sinh x: domain all real x, range all real y. cosh x: domain all real x, range y ≥ 1. tanh x: domain all real x, range −1 < y < 1. sech x: domain all real x, range 0 < y ≤ 1. cosech x: domain x ≠ 0, range y ≠ 0. coth x: domain x ≠ 0, range y < −1 or y > 1. B1 for sinh and cosh, B1 for tanh, B1 for the three reciprocals. arsinh x: domain all real x, range all real y. arcosh x: domain x ≥ 1, range y ≥ 0. artanh x: domain −1 < x < 1, range all real y. B1 for arsinh and artanh, B1 for arcosh. cosh is not one-to-one, since cosh x = cosh(−x) for every x, so without a restriction an inverse would need to map one input to two different outputs, which no function can do; restricting to x ≥ 0 makes it one-to-one, and the domain of that restriction, x ≥ 0, becomes the range of arcosh. B1 for the reasoning. sinh and tanh need no such restriction, since each is already one-to-one on the whole real line.

Question 14

Prove that cosh 3x = 4cosh³x − 3cosh x. You may quote the addition and double angle formulae for cosh and sinh.

Accepted Answer

Write cosh 3x = cosh(2x + x), and by Osborn's rule on cos(2θ + θ) = cos 2θ cos θ − sin 2θ sin θ, with the sin 2θ sin θ term carrying a pair of sine factors, its sign flips: cosh 3x = cosh 2x cosh x + sinh 2x sinh x. M1 for the addition formula by Osborn's rule, A1 for the correctly signed result. With cosh 2x = 2cosh²x − 1 and sinh 2x = 2 sinh x cosh x: cosh 3x = (2cosh²x − 1)cosh x + (2 sinh x cosh x)sinh x = 2cosh³x − cosh x + 2 sinh²x cosh x. M1 for substituting the double angle identities, A1 for the expanded expression. Now replace sinh²x using cosh²x − sinh²x = 1, so sinh²x = cosh²x − 1: 2cosh³x − cosh x + 2(cosh²x − 1)cosh x = 2cosh³x − cosh x + 2cosh³x − 2cosh x = 4cosh³x − 3cosh x, as required. M1 for substituting sinh²x = cosh²x − 1, A1 for the printed result. This is the hyperbolic counterpart of cos 3θ = 4cos³θ − 3cos θ, and here the final sign is unchanged from the trig version, since every sine factor in the working appears in a pair.

Question 15

Using Osborn's rule on cos(x + y) = cos x cos y − sin x sin y and cos(x − y) = cos x cos y + sin x sin y, write down the corresponding identities for cosh(x + y) and cosh(x − y). Hence, given that cosh a cosh b = 2 and sinh a sinh b = 1, find the exact values of a + b and a − b.

Accepted Answer

Each formula has one term with a pair of sine factors, so that term's sign flips: cosh(x + y) = cosh x cosh y + sinh x sinh y and cosh(x − y) = cosh x cosh y − sinh x sinh y. M1 for applying Osborn's rule to each formula, A1 for both identities with the flipped sign. Substituting a and b: cosh(a + b) = 2 + 1 = 3 and cosh(a − b) = 2 − 1 = 1. M1 for the substitution. Then a + b = arcosh 3 = ln(3 + √(9 − 1)) = ±ln(3 + 2√2): cosh is even, and (a, b) and (−a, −b) satisfy the same conditions, so both signs occur. M1 for the logarithmic form, A1 for ±ln(3 + 2√2). cosh(a − b) = 1 has the unique solution a − b = 0, since arcosh 1 = ln 1 = 0 and cosh being even would only offer ±0, which are the same point. B1 for a − b = 0. Together the two results fix a and b as well, since a = b and a + b is known.