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Hyperbolic functions and identities questions
The hyperbolic functions are defined directly from exponentials, so every property can be checked from the definition. This lesson covers their graphs, the identities that parallel the trigonometric ones with some signs changed, and the inverses, which turn out to be logarithms.
7 original questions · 25 marks · the hyperbolic functions and identities notes · Hyperbolic functions
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Write down the definitions of cosh x and sinh x in terms of exponentials, and state the value of each at x = 0.
Worked answer
cosh x = (ex + e−x)/2 and sinh x = (ex − 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.Using the definitions, find the exact values of cosh(ln 2) and sinh(ln 2).
Worked answer
eln 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.Solve sinh x = 2, giving the answer in exact logarithmic form.
Worked 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.Solve cosh x = 5/4, giving both answers exactly, and explain why there are two.
Worked 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.Starting from cosh²x − sinh²x = 1, derive an identity connecting tanh²x and sech²x.
Worked 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.Using cosh²x = 1 + sinh²x, solve 2 cosh²x − sinh x = 2, giving exact answers.
Worked 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). Converting cosh² into sinh² first turns the equation into a quadratic in one variable.Solve 5 cosh x − 4 sinh x = 3, giving your answer in exact form, and explain why the equation has only one solution.
Worked answer
Go back to the definitions. The left side is 5(ex + e−x)/2 − 4(ex − e−x)/2 = (ex + 9e−x)/2, so the equation becomes ex + 9e−x = 6. Writing u = ex 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 ex + 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 ex beats hunting for an identity here, because 5 cosh x − 4 sinh x has no neat single-function form.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise hyperbolic functions and identities one question at a time
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