EAPCET Mathematics · Intermediate first year · Paper IA

Hyperbolic Functions

Hyperbolic Functions defines sinh, cosh and tanh using exponentials, along with their identities and their inverses written as logarithms. It is a short chapter with a small, fixed set of results.

Subject
Mathematics
Taught in
Intermediate first year · Paper IA
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • sinh⁡x=ex−e−x2\sinh x = \dfrac{e^{x} - e^{-x}}{2} and cosh⁡x=ex+e−x2\cosh x = \dfrac{e^{x} + e^{-x}}{2}
  • cosh⁡2x−sinh⁡2x=1\cosh^{2}x - \sinh^{2}x = 1
  • sinh⁡2x=2sinh⁡xcosh⁡x\sinh 2x = 2\sinh x \cosh x; cosh⁡2x=cosh⁡2x+sinh⁡2x\cosh 2x = \cosh^{2}x + \sinh^{2}x
  • sinh⁡−1x=log⁡(x+x2+1)\sinh^{-1}x = \log\left(x + \sqrt{x^{2} + 1}\right)
  • cosh⁡−1x=log⁡(x+x2−1)\cosh^{-1}x = \log\left(x + \sqrt{x^{2} - 1}\right) for x≥1x \ge 1
  • tanh⁡−1x=12log⁡1+x1−x\tanh^{-1}x = \tfrac{1}{2}\log\dfrac{1 + x}{1 - x} for ∣x∣<1|x| < 1
  • sech⁡2x+tanh⁡2x=1\operatorname{sech}^{2}x + \tanh^{2}x = 1 and coth⁡2x−cosech⁡2x=1\coth^{2}x - \operatorname{cosech}^{2}x = 1
  • sinh⁡(x±y)=sinh⁡xcosh⁡y±cosh⁡xsinh⁡y\sinh(x \pm y) = \sinh x \cosh y \pm \cosh x \sinh y; cosh⁡(x±y)=cosh⁡xcosh⁡y±sinh⁡xsinh⁡y\cosh(x \pm y) = \cosh x \cosh y \pm \sinh x \sinh y
  • tanh⁡(x±y)=tanh⁡x±tanh⁡y1±tanh⁡xtanh⁡y\tanh(x \pm y) = \dfrac{\tanh x \pm \tanh y}{1 \pm \tanh x \tanh y}
  • tanh⁡2x=2tanh⁡x1+tanh⁡2x\tanh 2x = \dfrac{2\tanh x}{1 + \tanh^{2}x}; cosh⁡2x=1+2sinh⁡2x=2cosh⁡2x−1\cosh 2x = 1 + 2\sinh^{2}x = 2\cosh^{2}x - 1
  • sinh⁡3x=3sinh⁡x+4sinh⁡3x\sinh 3x = 3\sinh x + 4\sinh^{3}x; cosh⁡3x=4cosh⁡3x−3cosh⁡x\cosh 3x = 4\cosh^{3}x - 3\cosh x
  • coth⁡−1x=12log⁡x+1x−1\coth^{-1}x = \tfrac{1}{2}\log\dfrac{x + 1}{x - 1} for ∣x∣>1|x| > 1
  • sech⁡−1x=log⁡1+1−x2x\operatorname{sech}^{-1}x = \log\dfrac{1 + \sqrt{1 - x^{2}}}{x} for 0<x≤10 < x \le 1
  • Ranges: sinh⁡x\sinh x takes all real values, cosh⁡x∈[1,∞)\cosh x \in [1, \infty), tanh⁡x∈(−1,1)\tanh x \in (-1, 1), sech⁡x∈(0,1]\operatorname{sech} x \in (0, 1]

What this chapter covers

  • Definitions of hyperbolic functions
  • Hyperbolic identities
  • Inverse hyperbolic functions in logarithmic form
  • Domains and ranges

How it is usually asked

  • Find the other hyperbolic values given one of them
  • Evaluate an inverse hyperbolic value in log form
  • Simplify an expression using hyperbolic identities

Understanding Hyperbolic Functions

Hyperbolic functions are built from the exponential function: sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2, with tanh x = sinh x/cosh x and the reciprocals defined as in trigonometry. cosh x is even and never less than 1, sinh x is odd and takes every real value, and tanh x always lies strictly between −1 and 1.

Their identities mirror the trigonometric ones with some sign changes. The basic one is cosh²x − sinh²x = 1, which you can check directly from the definitions. Double-argument results such as sinh 2x = 2 sinh x cosh x and cosh 2x = cosh²x + sinh²x = 1 + 2 sinh²x follow the same pattern, so given one hyperbolic value you can find all the others.

Because the functions are made of exponentials, their inverses are logarithms: sinh⁻¹x = log(x + √(x² + 1)) for all x, cosh⁻¹x = log(x + √(x² − 1)) for x ≥ 1, and tanh⁻¹x = ½ log[(1 + x)/(1 − x)] for |x| < 1. EAPCET questions test these formulas, the identities and the domains directly.

Worked example

EAPCET-style question

If sinh⁡x=34\sinh x = \tfrac{3}{4}, then cosh⁡2x\cosh 2x equals

  1. (1)178\dfrac{17}{8}
  2. (2)2516\dfrac{25}{16}
  3. (3)98\dfrac{9}{8}
  4. (4)1

Solution

  1. Use cosh⁡2x=1+2sinh⁡2x\cosh 2x = 1 + 2\sinh^{2}x.
  2. sinh⁡2x=916\sinh^{2}x = \tfrac{9}{16}, so cosh⁡2x=1+1816=3416=178\cosh 2x = 1 + \tfrac{18}{16} = \tfrac{34}{16} = \tfrac{17}{8}.
  3. Check: x=sinh⁡−134=log⁡(34+54)=log⁡2x = \sinh^{-1}\tfrac{3}{4} = \log\left(\tfrac{3}{4} + \tfrac{5}{4}\right) = \log 2, and cosh⁡(2log⁡2)=4+142=178\cosh(2\log 2) = \dfrac{4 + \frac{1}{4}}{2} = \tfrac{17}{8}.

Answer: 178\dfrac{17}{8}

Shortcuts and results to remember

  • sinh, tanh, coth and cosech are odd functions; cosh and sech are even.
  • sinh⁡0=0\sinh 0 = 0, cosh⁡0=1\cosh 0 = 1, tanh⁡0=0\tanh 0 = 0, and cosh⁡x≥1\cosh x \ge 1 for every real xx.
  • Osborn’s rule: a trig identity turns into the hyperbolic one when you change sin to sinh and cos to cosh and flip the sign of any product of two sines (cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^{2}x becomes cosh⁡2x=1+2sinh⁡2x\cosh 2x = 1 + 2\sinh^{2}x).

More practice questions

Practice question 1

If sinh⁡x=34\sinh x = \tfrac{3}{4}, then cosh⁡2x\cosh 2x equals

  1. (1)178\dfrac{17}{8}
  2. (2)2516\dfrac{25}{16}
  3. (3)98\dfrac{9}{8}
  4. (4)54\dfrac{5}{4}

Solution

  1. Use cosh⁡2x=1+2sinh⁡2x\cosh 2x = 1 + 2\sinh^{2}x
  2. sinh⁡2x=916\sinh^{2}x = \tfrac{9}{16}
  3. cosh⁡2x=1+2×916=1+98\cosh 2x = 1 + 2 \times \tfrac{9}{16} = 1 + \tfrac{9}{8}
  4. =178= \tfrac{17}{8}

Answer: 178\dfrac{17}{8}

Practice question 2

The value of sinh⁡−134\sinh^{-1}\tfrac{3}{4} is

  1. (1)log⁡2\log 2
  2. (2)log⁡3\log 3
  3. (3)log⁡32\log\tfrac{3}{2}
  4. (4)log⁡4\log 4

Solution

  1. sinh⁡−1x=log⁡(x+x2+1)\sinh^{-1}x = \log\left(x + \sqrt{x^{2} + 1}\right)
  2. 916+1=2516=54\sqrt{\tfrac{9}{16} + 1} = \sqrt{\tfrac{25}{16}} = \tfrac{5}{4}
  3. x+x2+1=34+54=2x + \sqrt{x^{2} + 1} = \tfrac{3}{4} + \tfrac{5}{4} = 2
  4. So sinh⁡−134=log⁡2\sinh^{-1}\tfrac{3}{4} = \log 2

Answer: log⁡2\log 2

Common mistakes to avoid

  • Writing cosh²x + sinh²x = 1 out of trigonometric habit; the identity is cosh²x − sinh²x = 1.
  • Taking cosh x as negative when finding it from sinh x; cosh x ≥ 1 always, so take the positive root.
  • Using the cosh⁻¹ formula for x < 1 or the tanh⁻¹ formula for |x| ≥ 1, where they are not defined.
  • Mixing the signs inside the inverse formulas: sinh⁻¹ uses √(x² + 1), cosh⁻¹ uses √(x² − 1).

How to prepare it

The whole chapter fits on one page of formulas — it is one of the cheapest chapters to make reliable.

Most identities mirror trigonometric ones with a sign change, which helps you recall them.

Once the formulas feel familiar, test them the way the exam does: practise Hyperbolic Functions questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Hyperbolic Functions chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Hyperbolic Functions part of the EAPCET Mathematics syllabus?
Yes. Hyperbolic Functions is taught in Intermediate first year · Paper IA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Hyperbolic Functions for EAPCET?
The whole chapter fits on one page of formulas — it is one of the cheapest chapters to make reliable. Most identities mirror trigonometric ones with a sign change, which helps you recall them.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Hyperbolic Functions?
Writing cosh²x + sinh²x = 1 out of trigonometric habit; the identity is cosh²x − sinh²x = 1. Taking cosh x as negative when finding it from sinh x; cosh x ≥ 1 always, so take the positive root. Using the cosh⁻¹ formula for x < 1 or the tanh⁻¹ formula for |x| ≥ 1, where they are not defined. Mixing the signs inside the inverse formulas: sinh⁻¹ uses √(x² + 1), cosh⁻¹ uses √(x² − 1).