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Hyperbolic functions and identities questions

Build trig's cousins from exponentials and everything about them becomes checkable. Graphs, identities, and inverses that turn out to be plain logarithms once the definition is unpacked.

7 original questions · 25 marks · the hyperbolic functions and identities notes · Hyperbolic functions

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2 marks · write · warm-up
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 = (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.
3 marks · warm-up
Using the definitions, find the exact values of cosh(ln 2) and sinh(ln 2).
Worked 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.
3 marks
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.
4 marks
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.
2 marks
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.
5 marks
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.
6 marks · stretch
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(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 beats hunting for an identity here, because 5 cosh x − 4 sinh x has no neat single-function form.

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