CAIE A-Level Further Math A2 2.1.3 Hyperbolic Functions QuestionsPractise proving and applying hyperbolic identities from exponential definitions with Further Mathematics Paper 2 questions.Syllabus2028–2030CourseFurther Mathematics 9231LevelA2
Exam pointsProve and apply identities linking hyperbolic functions using their exponential definitions.
CAIE A-Level Further Math A2 2.1.3 Hyperbolic Functions Questions question 1[Maximum number: 3]Starting from the definitions of coth and cosech in terms of exponentials, prove thatcoth2x−cosech2x=1.\operatorname{coth}^{2} x-\operatorname{cosech}^{2} x=1 .coth2x−cosech2x=1.The curve C has equation y=lncoth(12x)y=\ln \operatorname{coth}\left(\frac{1}{2} x\right)y=lncoth(21x) for x>0.Show Answercothx=ex+e−xex−e−xcosechx=2ex−e−x\operatorname{coth} x=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{\mathrm{e}^{x}-\mathrm{e}^{-x}} \quad \operatorname{cosech} x=\frac{2}{\mathrm{e}^{x}-\mathrm{e}^{-x}}cothx=ex−e−xex+e−xcosechx=ex−e−x2(ex+e−xex−e−x)2−4(ex−e−x)2=e2x+e−2x−2(ex−e−x)2=1\left(\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{\mathrm{e}^{x}-\mathrm{e}^{-x}}\right)^{2}-\frac{4}{\left(\mathrm{e}^{x}-\mathrm{e}^{-x}\right)^{2}}=\frac{\mathrm{e}^{2 x}+\mathrm{e}^{-2 x}-2}{\left(\mathrm{e}^{x}-\mathrm{e}^{-x}\right)^{2}}=1(ex−e−xex+e−x)2−(ex−e−x)24=(ex−e−x)2e2x+e−2x−2=1Writes over common denominator, AG.3Add to Test