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Hyperbolic function

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NUMBERS We already know that π’†π’Šπœ½ = 𝐜𝐨𝐬 𝜽 + π’Š π’”π’Šπ’ 𝜽 π’†βˆ’π’Šπœ½ = 𝒄𝒐𝒔 𝜽 βˆ’ π’Š π’”π’Šπ’ 𝜽 From the above two equations we have 𝐜𝐨𝐬 𝜽 = π’†π’Šπœ½ + π’†βˆ’π’Šπœ½ 𝟐 & 𝐬𝐒𝐧 𝜽 = π’†π’Šπœ½ βˆ’ π’†βˆ’π’Šπœ½ πŸπ’Š These forms are known as Euler’s exponential forms of circular functions. Further, if z is a complex number we define 𝒄𝒐𝒔 𝒛 = π’†π’Šπ’› + π’†βˆ’π’Šπ’› 𝟐 & π’”π’Šπ’ 𝒛 = π’†π’Šπ’› βˆ’ π’†βˆ’π’Šπ’› πŸπ’Š
HYPERBOLIC FUNCTIONS If x is real or complex 𝒆𝒙+π’†βˆ’π’™ 𝟐 is called hyperbolic cosine of x and is denoted by cosh x and π’†π’™βˆ’π’†βˆ’π’™ 𝟐 is called hyperbolic sine of x and is denoted by sinh x. Thus, π’”π’Šπ’π’‰ 𝒙 = π’†π’™βˆ’π’†βˆ’π’™ 𝟐 , 𝒄𝒐𝒔𝒉 𝒙 = 𝒆𝒙+π’†βˆ’π’™ 𝟐 Relationship between hyperbolic and circular functions: (i) sin ix = i sinh x and sinh x = -i sin ix (ii) cos ix = cosh x (iii) tan ix = i tanh x and tanh x = -i tan ix (iv) sinh ix = i sin x and sin x = -i sinh ix (v) cosh ix = cos x (vi) tanh ix = i tan x and tan x = -i tanh ix
IDENTITIES (i) π’”π’Šπ’π’‰ (βˆ’π’™) = βˆ’π’”π’Šπ’π’‰ 𝒙 (ii) 𝒄𝒐𝒔𝒉(βˆ’π’™) = 𝒄𝒐𝒔𝒉 𝒙 (iii) π’„π’π’”π’‰πŸ 𝒙 βˆ’ π’”π’Šπ’π’‰πŸ 𝒙 = 𝟏 (iv) π’”π’†π’„π’‰πŸπ’™ + π’•π’‚π’π’‰πŸπ’™ = 𝟏 (v) π’„π’π’•π’‰πŸ 𝒙 βˆ’ π’„π’π’”π’†π’„π’‰πŸ 𝒙 = 𝟏 (vi) π’”π’Šπ’π’‰ 𝒙 Β± π’š = π’”π’Šπ’π’‰ 𝒙 𝒄𝒐𝒔𝒉 π’š Β± 𝒄𝒐𝒔𝒉 𝒙 π’”π’Šπ’π’‰ π’š (vii) πœπ¨π¬π’‰ 𝒙 Β± π’š = 𝒄𝒐𝒔𝒉 𝒙 𝒄𝒐𝒔𝒉 π’š Β± π’”π’Šπ’π’‰ 𝒙 π’”π’Šπ’π’‰ π’š (viii) 𝒕𝒂𝒏𝒉 𝒙 Β± π’š = 𝒕𝒂𝒏𝒉 𝒙±𝒕𝒂𝒏𝒉 π’š πŸΒ±π’•π’‚π’π’‰ 𝒙 βˆ™ 𝒕𝒂𝒏𝒉 π’š Formula for 2x and 3x (i) π’”π’Šπ’π’‰ πŸπ’™ = 𝟐 π’”π’Šπ’π’‰ 𝒙 𝒄𝒐𝒔𝒉 𝒙 (ii) 𝒄𝒐𝒔𝒉 πŸπ’™ = π’„π’π’”π’‰πŸ 𝒙 + π’”π’Šπ’π’‰πŸ 𝒙 = πŸπ’„π’π’”π’‰πŸ 𝒙 βˆ’ 𝟏 = 𝟏 + πŸπ’”π’Šπ’π’‰πŸ 𝒙 (iii) 𝐭𝒂𝒏𝒉 πŸπ’™ = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟏+π’•π’‚π’π’‰πŸπ’™ (iv) π’”π’Šπ’π’‰ πŸ‘π’™ = πŸ‘ π’”π’Šπ’π’‰ 𝒙 + πŸ’π’”π’Šπ’π’‰πŸ‘ 𝒙 (v) πœπ¨π’”π’‰ πŸ‘π’™ = πŸ’π’„π’π’”π’‰πŸ‘ 𝒙 βˆ’ πŸ‘π’„π’π’”π’‰ 𝒙 (vi) 𝐭𝒂𝒏𝒉 πŸ‘π’™ = πŸ‘ 𝒕𝒂𝒏𝒉 𝒙+π’•π’‚π’π’‰πŸ‘π’™ 𝟏+πŸ‘ π’•π’‚π’π’‰πŸπ’™ (vii) π’”π’Šπ’π’‰ 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 πŸβˆ’π’•π’‚π’π’‰πŸ 𝒙 𝟐 (viii) πœπ¨π’”π’‰ 𝒙 = 𝟏+π’•π’‚π’π’‰πŸ 𝒙 𝟐 πŸβˆ’π’•π’‚π’π’‰πŸ 𝒙 𝟐 (ix) π’”π’Šπ’π’‰ 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 𝟏+π’•π’‚π’π’‰πŸ 𝒙 𝟐
HYPERBOLIC IDENTITIES(CONTD) FORMULAE FOR 2X AND 3X (I) π’”π’Šπ’π’‰ πŸπ’™ = 𝟐 π’”π’Šπ’π’‰ 𝒙 𝒄𝒐𝒔𝒉 𝒙 (II) 𝒄𝒐𝒔𝒉 πŸπ’™ = π’„π’π’”π’‰πŸπ’™ + π’”π’Šπ’π’‰πŸπ’™ = πŸπ’„π’π’”π’‰πŸπ’™ βˆ’ 𝟏 = 𝟏 + πŸπ’”π’Šπ’π’‰πŸ 𝒙 (III) 𝒕𝒂𝒏𝒉 πŸπ’™ = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟏+π’•π’‚π’π’‰πŸπ’™ (IV) π’”π’Šπ’π’‰ πŸ‘π’™ = πŸ‘ π’”π’Šπ’π’‰ 𝒙 + πŸ’π’”π’Šπ’π’‰πŸ‘π’™ (V) πœπ¨π’”π’‰ πŸ‘π’™ = πŸ’π’„π’π’”π’‰πŸ‘ 𝒙 βˆ’ πŸ‘π’„π’π’”π’‰ 𝒙 (VI) 𝐭𝒂𝒏𝒉 πŸ‘π’™ = πŸ‘ 𝒕𝒂𝒏𝒉 𝒙+π’•π’‚π’π’‰πŸ‘π’™ 𝟏+πŸ‘ π’•π’‚π’π’‰πŸπ’™ (vii) π’”π’Šπ’π’‰ 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 πŸβˆ’π’•π’‚π’π’‰πŸ 𝒙 𝟐 (viii) πœπ¨π’”π’‰ 𝒙 = 𝟏+π’•π’‚π’π’‰πŸ 𝒙 𝟐 πŸβˆ’π’•π’‚π’π’‰πŸ 𝒙 𝟐 (ix) 𝒕𝒂𝒏𝒉 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 𝟏+π’•π’‚π’π’‰πŸ 𝒙 𝟐
HYPERBOLIC IDENTITIES(CONTD) DIFFERENTIATION AND INTEGRATION S.No Function Differentiation Integration 1 π’š = π’”π’Šπ’π’‰ 𝒙 π’…π’š 𝒅𝒙 = 𝒄𝒐𝒔𝒉 𝒙 𝐬𝐒𝐧𝐑 𝒙 𝒅𝒙 = 𝐜𝐨𝐬𝐑 𝒙 2 π’š = 𝒄𝒐𝒔𝒉 𝒙 π’…π’š 𝒅𝒙 = π’”π’Šπ’π’‰ 𝒙 𝐜𝐨𝐬𝐑 𝒙 𝒅𝒙 = 𝐬𝐒𝐧𝐑 𝒙 3 π’š = 𝒕𝒂𝒏𝒉 𝒙 π’…π’š 𝒅𝒙 = π’”π’†π’„π’‰πŸ 𝒙 π’”π’†π’„π’‰πŸ 𝒙 𝒅𝒙 = 𝐭𝐚𝐧𝐑 𝒙
HYPERBOLIC IDENTITIES(CONTD) PRODUCT FORMULAE (I) π’”π’Šπ’π’‰ 𝒙 + π’š + π’”π’Šπ’π’‰ 𝒙 βˆ’ π’š = 𝟐 π’”π’Šπ’π’‰ 𝒙 𝒄𝒐𝒔𝒉 π’š (II) π’”π’Šπ’π’‰ 𝒙 + π’š βˆ’ π’”π’Šπ’π’‰ 𝒙 βˆ’ π’š = 𝟐 𝒄𝒐𝒔𝒉 𝒙 π’”π’Šπ’π’‰ π’š (III) πœπ¨π’”π’‰ 𝒙 + π’š + 𝒄𝒐𝒔𝒉 𝒙 βˆ’ π’š = 𝟐 𝒄𝒐𝒔𝒉 𝒙 𝒄𝒐𝒔𝒉 π’š (IV) πœπ¨π¬π’‰ 𝒙 + π’š βˆ’ 𝒄𝒐𝒔𝒉 𝒙 βˆ’ π’š = 𝟐 π’”π’Šπ’π’‰ 𝒙 π’”π’Šπ’π’‰ π’š (V) ) π’”π’Šπ’π’‰ 𝒙 + π’”π’Šπ’π’‰ π’š = 𝟐 π’”π’Šπ’π’‰ 𝒙+π’š 𝟐 𝒄𝒐𝒔𝒉 π’™βˆ’π’š 𝟐 (VI) π’”π’Šπ’π’‰ 𝒙 βˆ’ π’”π’Šπ’π’‰ π’š = 𝟐 𝒄𝒐𝒔𝒉 𝒙+π’š 𝟐 π’”π’Šπ’π’‰ π’™βˆ’π’š 𝟐 (VII) πœπ¨π’”π’‰ 𝒙 + 𝒄𝒐𝒔𝒉 π’š = πŸπ’„π’π’”π’‰ 𝒙+π’š 𝟐 𝒄𝒐𝒔𝒉 π’™βˆ’π’š 𝟐 (VIII) πœπ¨π’”π’‰ 𝒙 βˆ’ 𝒄𝒐𝒔𝒉 π’š = πŸπ’”π’Šπ’π’‰ 𝒙+π’š 𝟐 π’”π’Šπ’π’‰ π’™βˆ’π’š 𝟐
TABLE OF VALUES OF HYPERBOLIC FUNCTIONS x 0 βˆ’βˆž +∞ π’š = π’”π’Šπ’π’‰ 𝒙 0 βˆ’βˆž ∞ π’š = 𝒄𝒐𝒔𝒉 𝒙 1 ∞ ∞ π’š = 𝒕𝒂𝒏𝒉 𝒙 0 βˆ’1 1
INVERSE HYPERBOLIC FUNCTIONS Definition: if sinh u = z then u is called inverse hyperbolic sine of z and is denoted by 𝒖 = π’”π’Šπ’π’‰βˆ’πŸπ’›. Similarly we can define inverse hyperbolic cosine and inverse hyperbolic tangent and other functions which we denote as π’„π’π’”π’‰βˆ’πŸπ’›, π’•π’‚π’π’‰βˆ’πŸπ’›, 𝒆𝒕𝒄. The inverse hyperbolic functions are many valued but we will consider their principal values only. If Z is real we can show that 1. π’”π’Šπ’π’‰βˆ’πŸπ’› = π’π’π’ˆ 𝒛 + π’›πŸ + 𝟏 2. π’„π’π’”π’‰βˆ’πŸ 𝒛 = π’π’π’ˆ 𝒛 + π’›πŸ βˆ’ 𝟏 3. π’•π’‚π’π’‰βˆ’πŸ 𝒛 = 𝟏 𝟐 π’π’π’ˆ 𝟏+𝒛 πŸβˆ’π’› .
INVERSE HYPERBOLIC FUNCTIONS INTEGRATION FORMULAE 𝒅𝒙 π’™πŸ + π’‚πŸ = π’”π’Šπ’π’‰βˆ’πŸ 𝒙 𝒂 𝒅𝒙 π’™πŸ βˆ’ π’‚πŸ = π’„π’π’”π’‰βˆ’πŸ 𝒙 𝒂 𝒅𝒙 π’‚πŸ βˆ’ π’™πŸ = 𝟏 𝒂 π’•π’‚π’π’‰βˆ’πŸ 𝒙 𝒂
PROBLEMS BASED ON HYPERBOLIC FUNCTIONS 1.Prove that 𝒄𝒐𝒔𝒉 𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒙 𝒏 = 𝒄𝒐𝒔𝒉 𝒏𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒏𝒙 2.Express cot (x +i y) in a + i b form. 3. Solve the equation π’™πŸ• + π’™πŸ’ + π’™πŸ‘ + 𝟏 = 𝟎 4. Prove that 𝟏 + 𝒄𝒐𝒔 𝒙 + π’Š π’”π’Šπ’ 𝒙 𝒏 = πŸπ’π’„π’π’”π’ 𝒙 𝟐 𝒄𝒐𝒔 𝒏𝒙 𝟐 + π’Šπ’”π’Šπ’ 𝒏𝒙 𝟐 5. Prove that π’”π’Šπ’π’‰βˆ’πŸ 𝒙 = π’π’π’ˆ 𝒙 + π’™πŸ + 𝟏 6.Show that π’”π’†π’„π’‰βˆ’πŸ π’”π’Šπ’ 𝜽 = π’π’π’ˆ 𝒄𝒐𝒕 𝜽 𝟐 7.Express sec(x+iy) in a+ib form.
PROBLEMS BASED ON HYPERBOLIC FUNCTIONS 1. Prove that 𝒄𝒐𝒔𝒉 𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒙 𝒏 = 𝒄𝒐𝒔𝒉 𝒏𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒏𝒙 Solution: LHS = 𝒄𝒐𝒔𝒉 𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒙 𝒏 = 𝒆𝒙+π’†βˆ’π’™ 𝟐 βˆ’ π’†π’™βˆ’π’†βˆ’π’™ 𝟐 𝒏 = 𝒆𝒙 + π’†βˆ’π’™ βˆ’ 𝒆𝒙 + π’†βˆ’π’™ 𝟐 𝒏 = π’†βˆ’π’™ 𝒏 = π’†βˆ’π’π’™ RHS = 𝒄𝒐𝒔𝒉 𝒏𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒏𝒙 = 𝒆𝒏𝒙+π’†βˆ’π’π’™ 𝟐 βˆ’ π’†π’π’™βˆ’π’†βˆ’π’π’™ 𝟐 = 𝒆𝒏𝒙+π’†βˆ’π’π’™βˆ’π’†π’π’™+π’†βˆ’π’π’™ 𝟐 = π’†βˆ’π’π’™ ∴ LHS=RHS
PROBLEMS BASED ON HYPERBOLIC FUNCTIONS 2.Express cot(x +i y) in a + i b form. Solution: 𝒄𝒐𝒕 𝒙 + π’Šπ’š = 𝒄𝒐𝒔(𝒙+π’Šπ’š) π’”π’Šπ’(𝒙+π’Šπ’š) = 𝒄𝒐𝒔(𝒙+π’Šπ’š) π’”π’Šπ’(𝒙+π’Šπ’š) Γ— π’”π’Šπ’(π’™βˆ’π’Šπ’š) π’”π’Šπ’(π’™βˆ’π’Šπ’š) = πŸπ’„π’π’”(𝒙+π’Šπ’š) πŸπ’”π’Šπ’(𝒙+π’Šπ’š) Γ— π’”π’Šπ’(π’™βˆ’π’Šπ’š) π’”π’Šπ’(π’™βˆ’π’Šπ’š) = π’”π’Šπ’ 𝒙+π’Šπ’š+π’™βˆ’π’Šπ’š βˆ’π’”π’Šπ’ 𝒙+π’Šπ’šβˆ’π’™βˆ’π’Šπ’š 𝒄𝒐𝒔 𝒙+π’Šπ’šβˆ’(π’™βˆ’π’Šπ’š) βˆ’π’„π’π’” 𝒙+π’Šπ’š+π’™βˆ’π’Šπ’š π’”π’Šπ’ 𝑨 + 𝑩 βˆ’ π’”π’Šπ’ 𝑨 βˆ’ 𝑩 = 𝟐 𝒄𝒐𝒔𝑨 π’”π’Šπ’π‘© 𝒄𝒐𝒔 𝑨 βˆ’ 𝑩 βˆ’ 𝒄𝒐𝒔 𝑨 + 𝑩 = 𝟐 π’”π’Šπ’π‘¨ π’”π’Šπ’π‘© = π’”π’Šπ’ πŸπ’™ βˆ’π’”π’Šπ’ πŸπ’Šπ’š 𝒄𝒐𝒔 πŸπ’Šπ’š βˆ’π’„π’π’” πŸπ’™ = π’”π’Šπ’ πŸπ’™ βˆ’π’Šπ’”π’Šπ’π’‰ πŸπ’š 𝒄𝒐𝒔𝒉 πŸπ’š βˆ’π’„π’π’” πŸπ’™ = π’”π’Šπ’ πŸπ’™ 𝒄𝒐𝒔𝒉 πŸπ’š βˆ’ 𝒄𝒐𝒔 πŸπ’™ + π’Š βˆ’π’”π’Šπ’π’‰ πŸπ’š 𝒄𝒐𝒔𝒉 πŸπ’š βˆ’ 𝒄𝒐𝒔 πŸπ’™ = 𝒂 + π’Šπ’ƒ Where a= π’”π’Šπ’ πŸπ’™ 𝒄𝒐𝒔𝒉 πŸπ’š βˆ’π’„π’π’” πŸπ’™ & b= βˆ’π’”π’Šπ’π’‰ πŸπ’š 𝒄𝒐𝒔𝒉 πŸπ’š βˆ’π’„π’π’” πŸπ’™
PROBLEMS BASED ON HYPERBOLIC FUNCTIONS 3. Solve the equation π’™πŸ• + π’™πŸ’ + π’™πŸ‘ + 𝟏 = 𝟎 Roots of a Complex Number: De Moivre’s Theorem can be used to find all n- roots of a complex number. Since, 𝒄𝒐𝒔 𝜽 = 𝒄𝒐𝒔 πŸπ’Œπ… + 𝜽 & π’”π’Šπ’ 𝜽 = π’”π’Šπ’ (πŸπ’Œπ… + 𝜽) where k is an integer. We have by DeMoivre’s Theorem 𝒄𝒐𝒔 𝜽 + π’Šπ’”π’Šπ’ 𝜽 𝟏 𝒏 = 𝒄𝒐𝒔 πŸπ’Œπ… + 𝜽 + π’”π’Šπ’ (πŸπ’Œπ… + 𝜽) 𝟏 𝒏 𝒄𝒐𝒔 𝜽 + π’Šπ’”π’Šπ’ 𝜽 𝟏 𝒏 = 𝒄𝒐𝒔 πŸπ…π’Œ + 𝜽 𝒏 + π’Šπ’”π’Šπ’ πŸπ…π’Œ + 𝜽 𝒏 By putting π’Œ = 𝟎, 𝟏, 𝟐, … … . . 𝒏 βˆ’ 𝟏 we get n roots of the complex number.
PROBLEMS BASED ON HYPERBOLIC FUNCTIONS 3. Solve the equation π’™πŸ• + π’™πŸ’ + π’™πŸ‘ + 𝟏 = 𝟎 Solution: π’™πŸ• + π’™πŸ’ + π’™πŸ‘ + 𝟏 = 𝟎 ⟹ π’™πŸ’ π’™πŸ‘ + 𝟏 + 𝟏 π’™πŸ‘ + 𝟏 = 𝟎 ⟹ π’™πŸ‘ + 𝟏 π’™πŸ’ + 𝟏 = 𝟎 ⟹ π’™πŸ‘ + 𝟏 = 𝟎, π’™πŸ’ + 𝟏 = 𝟎 βŸΉπ’™πŸ‘ = βˆ’πŸ, π’™πŸ’ = βˆ’πŸ π’™πŸ‘ = βˆ’πŸ βŸΉπ’™πŸ‘ = (𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅) βŸΉπ’™ = 𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅 𝟏 πŸ‘ βŸΉπ’™ = 𝒄𝒐𝒔 πŸπ’Œπ… + 𝝅 + π’Šπ’”π’Šπ’(πŸπ’Œπ… + 𝝅 𝟏 πŸ‘ ⟹ 𝒙 = 𝒄𝒐𝒔 πŸπ’Œ + 𝟏)𝝅 + π’Šπ’”π’Šπ’(πŸπ’Œ + 𝟏)𝝅 𝟏 πŸ‘ ⟹ 𝒙 = 𝒄𝒐𝒔 (πŸπ’Œ+𝟏)𝝅 πŸ‘ + π’Š π’”π’Šπ’ (πŸπ’Œ+𝟏)𝝅 𝒏 where π’Œ = 𝟎, 𝟏, 𝟐 when π’Œ = 𝟎, 𝒙 = 𝒄𝒐𝒔 𝝅 πŸ‘ + π’Šπ’”π’Šπ’ 𝝅 πŸ‘ when k = 1, 𝐱 = 𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅 When k = 2, 𝒙 = 𝒄𝒐𝒔 πŸ“π… πŸ‘ + π’Šπ’”π’Šπ’ πŸ“π… πŸ‘
PROBLEMS BASED ON HYPERBOLIC FUNCTIONS π’™πŸ’ = βˆ’πŸ βŸΉπ’™πŸ’ = (𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅) βŸΉπ’™ = 𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅 𝟏 πŸ’ βŸΉπ’™ = 𝒄𝒐𝒔 πŸπ’Œπ… + 𝝅 + π’Šπ’”π’Šπ’(πŸπ’Œπ… + 𝝅 𝟏 πŸ’ ⟹ 𝒙 = 𝒄𝒐𝒔 πŸπ’Œ + 𝟏)𝝅 + π’Šπ’”π’Šπ’(πŸπ’Œ + 𝟏)𝝅 𝟏 πŸ’ ⟹ 𝒙 = 𝒄𝒐𝒔 (πŸπ’Œ+𝟏)𝝅 πŸ’ + π’Š π’”π’Šπ’ (πŸπ’Œ+𝟏)𝝅 πŸ’ where π’Œ = 𝟎, 𝟏, 𝟐, πŸ‘ when π’Œ = 𝟎, 𝒙 = 𝒄𝒐𝒔 𝝅 πŸ’ + π’Šπ’”π’Šπ’ 𝝅 πŸ’ when k = 1, 𝒙 = 𝒄𝒐𝒔 πŸ‘π… πŸ’ + π’Šπ’”π’Šπ’ πŸ‘π… πŸ’ When k = 2, 𝒙 = 𝒄𝒐𝒔 πŸ“π… πŸ’ + π’Šπ’”π’Šπ’ πŸ“π… πŸ’ when k = 3, 𝒙 = 𝒄𝒐𝒔 πŸ•π… πŸ’ + π’Šπ’”π’Šπ’ πŸ•π… πŸ’ ∴ π‘‡β„Žπ‘’ 𝑠𝑒𝑣𝑒𝑛 π‘Ÿπ‘œπ‘œπ‘‘π‘  π‘œπ‘“ π‘‘β„Žπ‘’ 𝑔𝑖𝑣𝑒𝑛 π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› are given by 𝒙 = 𝒄𝒐𝒔 𝝅 πŸ‘ + π’Šπ’”π’Šπ’ 𝝅 πŸ‘ 𝐱 = 𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅 𝒙 = 𝒄𝒐𝒔 πŸ“π… πŸ‘ + π’Šπ’”π’Šπ’ πŸ“π… πŸ‘ 𝒙 = 𝒄𝒐𝒔 𝝅 πŸ’ + π’Šπ’”π’Šπ’ 𝝅 πŸ’ 𝒙 = 𝒄𝒐𝒔 πŸ‘π… πŸ’ + π’Šπ’”π’Šπ’ πŸ‘π… πŸ’ 𝒙 = 𝒄𝒐𝒔 πŸ“π… πŸ’ + π’Šπ’”π’Šπ’ πŸ“π… πŸ’ 𝒙 = 𝒄𝒐𝒔 πŸ•π… πŸ’ + π’Šπ’”π’Šπ’ πŸ•π… πŸ’
4. Prove that 𝟏 + 𝒄𝒐𝒔 𝒙 + π’Š π’”π’Šπ’ 𝒙 𝒏 = πŸπ’π’„π’π’”π’ 𝒙 𝟐 𝒄𝒐𝒔 𝒏𝒙 𝟐 + π’Šπ’”π’Šπ’ 𝒏𝒙 𝟐 Solution: we use the following formulas 𝐬𝐒𝐧 πŸπ’™ = 𝟐 𝐬𝐒𝐧 𝒙 𝐜𝐨𝐬 𝒙 π’”π’Šπ’ 𝒙 = πŸπ’”π’Šπ’ 𝒙 𝟐 𝒄𝒐𝒔 𝒙 𝟐 𝟏 + 𝒄𝒐𝒔 𝒙 = πŸπ’„π’π’”πŸ 𝒙 𝟐 LHS= 𝟏 + 𝒄𝒐𝒔 𝒙 + π’Š π’”π’Šπ’ 𝒙 𝒏 = πŸπ’„π’π’”πŸ 𝒙 𝟐 + π’ŠπŸπ’”π’Šπ’ 𝒙 𝟐 𝒄𝒐𝒔 𝒙 𝟐 𝒏 = πŸπ’„π’π’” 𝒙 𝟐 𝒄𝒐𝒔 𝒙 𝟐 + π’Šπ’”π’Šπ’ 𝒙 𝟐 𝒏 = πŸπ’„π’π’” 𝒙 𝟐 𝒏 𝒄𝒐𝒔 𝒙 𝟐 + π’Šπ’”π’Šπ’ 𝒙 𝟐 𝒏 = πŸπ’ 𝒄𝒐𝒔𝒏 𝒙 𝟐 𝒄𝒐𝒔 𝒏𝒙 𝟐 + π’Šπ’”π’Šπ’ 𝒏𝒙 𝟐 (by De’Moivre’s Theorem) = RHS
5. Prove that π’”π’Šπ’π’‰βˆ’πŸ 𝒙 = π’π’π’ˆ 𝒙 + π’™πŸ + 𝟏 Solution: Let π’”π’Šπ’π’‰βˆ’πŸ 𝒙 = π’šβŸΉ π’”π’Šπ’π’‰ π’š = 𝒙 ∴ 𝒙 = π’”π’Šπ’π’‰π’š = π’†π’š βˆ’ π’†βˆ’π’š 𝟐 = π’†π’š βˆ’ 𝟏 π’†π’š 𝟐 = π’†πŸπ’š βˆ’ 𝟏 πŸπ’†π’š 𝒙 = π’†πŸπ’šβˆ’πŸ πŸπ’†π’š βŸΉπŸπ’†π’š 𝐱 = π’†πŸπ’š βˆ’ 𝟏 ⟹ πŸπ’†π’š 𝐱 βˆ’ π’†πŸπ’š βˆ’ 𝟏 = 𝟎 ⟹ π’†πŸπ’š βˆ’ πŸπ’™π’†π’š +1=0 ⟹ π’†π’š 𝟐 βˆ’ πŸπ’™ π’†π’š + 𝟏 = 𝟎 This is quadratic in π’†π’š . π’†π’š = βˆ’(βˆ’πŸπ’™) Β± (βˆ’πŸπ’™)𝟐 βˆ’ πŸ’(𝟏)(𝟏) 𝟐(𝟏) = πŸπ’™ Β± πŸ’π’™πŸ βˆ’ πŸ’ 𝟐 = πŸπ’™ Β± 𝟐 π’™πŸ βˆ’ 𝟏 𝟐 = 𝒙 Β± π’™πŸ βˆ’ 𝟏 Conventionally we take positive sign, π’†π’š= 𝒙 Β± π’™πŸ βˆ’ 𝟏. Taking log on both sides log(π’†π’š )=log 𝐱 Β± π’™πŸ βˆ’ 𝟏⟹ 𝐲 = π’π’π’ˆ 𝒙 + π’™πŸ + 𝟏 ⟹ π’”π’Šπ’π’‰βˆ’πŸ 𝒙 = π’π’π’ˆ 𝒙 + π’™πŸ + 𝟏
6. Show that π’”π’†π’„π’‰βˆ’πŸ π’”π’Šπ’ 𝜽 = π’π’π’ˆ 𝒄𝒐𝒕 𝜽 𝟐 ⟹ Solution: Let π’”π’†π’„π’‰βˆ’πŸ π’”π’Šπ’ 𝜽 = π’™βŸΉ 𝒔𝒆𝒄𝒉 𝒙 = π’”π’Šπ’πœ½ ∴ π’”π’Šπ’πœ½ = 𝟏 𝒄𝒐𝒔𝒉 𝒙 = 𝟏 𝒆𝒙 + π’†βˆ’π’™ 𝟐 = 𝟐 𝒆𝒙 + π’†βˆ’π’™ = 𝟐 𝒆𝒙 + 𝟏 𝒆𝒙 = πŸπ’†π’™ π’†πŸπ’™ + 𝟏 π’”π’Šπ’πœ½ = πŸπ’†π’™ π’†πŸπ’™ + 𝟏 π’†πŸπ’™ sin 𝜽 + π’”π’Šπ’πœ½ = πŸπ’†π’™ ⟹ 𝒆𝒙 𝟐 π’”π’Šπ’πœ½ βˆ’ πŸπ’†π’™ + π’”π’Šπ’πœ½ = 𝟎 This is quadratic in 𝒆𝒙 . 𝒆𝒙 = βˆ’(βˆ’πŸ) Β± (βˆ’πŸ)𝟐 βˆ’ πŸ’(π’”π’Šπ’πœ½)(π’”π’Šπ’πœ½) πŸπ’”π’Šπ’πœ½ = 𝟐 Β± πŸ’ βˆ’ πŸ’π’”π’Šπ’πŸπœ½ πŸπ’”π’Šπ’πœ½ = 𝟐 Β± 𝟐 𝟏 βˆ’ π’”π’Šπ’πŸπœ½ πŸπ’”π’Šπ’πœ½ = 𝟏 Β± π’„π’π’”πŸπœ½ πŸπ’”π’Šπ’πœ½ Conventionally we take positive sign, 𝒆𝒙= 𝟏 + 𝒄𝒐𝒔 𝜽 π’”π’Šπ’πœ½ = πŸπ’„π’π’”πŸ 𝜽 𝟐 πŸπ’”π’Šπ’ 𝜽 𝟐 𝒄𝒐𝒔 𝜽 𝟐 = 𝒄𝒐𝒕 𝜽 𝟐 log(𝒆𝒙 )=log 𝒄𝒐𝒕 𝜽 𝟐 ⟹𝐱 = π’π’π’ˆ 𝒄𝒐𝒕 𝜽 𝟐 ⟹ π’”π’†π’„π’‰βˆ’πŸ π’”π’Šπ’ 𝜽 = π’π’π’ˆ 𝒄𝒐𝒕 𝜽 𝟐
7.Express sec(x+iy) in a+ib form. Solution: 𝑠𝑒𝑐 π‘₯ + 𝑖𝑦 = 𝟏 𝒄𝒐𝒔(𝒙+π’Šπ’š) = 𝟏 𝒄𝒐𝒔(𝒙+π’Šπ’š) Γ— 𝒄𝒐𝒔(π’™βˆ’π’Šπ’š) 𝒄𝒐𝒔(π’™βˆ’π’Šπ’š) = πŸπ’„π’π’”(π’™βˆ’π’Šπ’š) 𝟐 𝒄𝒐𝒔(𝒙+π’Šπ’š)𝒄𝒐𝒔(𝒙+π’Šπ’š) = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔 π’Šπ’š +π’”π’Šπ’ 𝒙 π’”π’Šπ’ (π’Šπ’š) 𝒄𝒐𝒔 𝒙+π’Šπ’š+π’™βˆ’π’Šπ’š +𝒄𝒐𝒔(𝒙+π’Šπ’šβˆ’ π’™βˆ’π’Šπ’š ) [Since 𝒄𝒐𝒔 𝑨 βˆ’ 𝑩 = 𝒄𝒐𝒔𝑨 𝒄𝒐𝒔𝑩 + π’”π’Šπ’π‘¨ π’”π’Šπ’π‘© & πŸπ’„π’π’”π‘¨ 𝒄𝒐𝒔𝑩 = 𝐜𝐨𝐬 𝑨 + 𝑩 + 𝐜𝐨𝐬(𝑨 βˆ’ 𝑩)] = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 π’š+π’Š π’”π’Šπ’ 𝒙 π’”π’Šπ’π’‰ π’š 𝒄𝒐𝒔 πŸπ’™ +𝒄𝒐𝒔(πŸπ’Šπ’š) = πŸπ’„π’π’” 𝒙 𝒄𝒐𝒔𝒉 π’š+π’Š 𝟐 π’”π’Šπ’ 𝒙 π’”π’Šπ’π’‰ π’š 𝒄𝒐𝒔 πŸπ’™+𝒄𝒐𝒔𝒉 πŸπ’š = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 π’š 𝒄𝒐𝒔 πŸπ’™+𝒄𝒐𝒔𝒉 πŸπ’š + π’Š 𝟐 π’”π’Šπ’ 𝒙 π’”π’Šπ’π’‰ π’š 𝒄𝒐𝒔 πŸπ’™+𝒄𝒐𝒔𝒉 πŸπ’š = 𝒂 + π’Šπ’ƒ 𝒔𝒆𝒄 𝒙 + π’Šπ’š = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 π’š 𝒄𝒐𝒔 πŸπ’™ + 𝒄𝒐𝒔𝒉 πŸπ’š + π’Š 𝟐 π’”π’Šπ’ 𝒙 π’”π’Šπ’π’‰ π’š 𝒄𝒐𝒔 πŸπ’™ + 𝒄𝒐𝒔𝒉 πŸπ’š

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  • 1. NUMBERS We already know that π’†π’Šπœ½ = 𝐜𝐨𝐬 𝜽 + π’Š π’”π’Šπ’ 𝜽 π’†βˆ’π’Šπœ½ = 𝒄𝒐𝒔 𝜽 βˆ’ π’Š π’”π’Šπ’ 𝜽 From the above two equations we have 𝐜𝐨𝐬 𝜽 = π’†π’Šπœ½ + π’†βˆ’π’Šπœ½ 𝟐 & 𝐬𝐒𝐧 𝜽 = π’†π’Šπœ½ βˆ’ π’†βˆ’π’Šπœ½ πŸπ’Š These forms are known as Euler’s exponential forms of circular functions. Further, if z is a complex number we define 𝒄𝒐𝒔 𝒛 = π’†π’Šπ’› + π’†βˆ’π’Šπ’› 𝟐 & π’”π’Šπ’ 𝒛 = π’†π’Šπ’› βˆ’ π’†βˆ’π’Šπ’› πŸπ’Š
  • 2. HYPERBOLIC FUNCTIONS If x is real or complex 𝒆𝒙+π’†βˆ’π’™ 𝟐 is called hyperbolic cosine of x and is denoted by cosh x and π’†π’™βˆ’π’†βˆ’π’™ 𝟐 is called hyperbolic sine of x and is denoted by sinh x. Thus, π’”π’Šπ’π’‰ 𝒙 = π’†π’™βˆ’π’†βˆ’π’™ 𝟐 , 𝒄𝒐𝒔𝒉 𝒙 = 𝒆𝒙+π’†βˆ’π’™ 𝟐 Relationship between hyperbolic and circular functions: (i) sin ix = i sinh x and sinh x = -i sin ix (ii) cos ix = cosh x (iii) tan ix = i tanh x and tanh x = -i tan ix (iv) sinh ix = i sin x and sin x = -i sinh ix (v) cosh ix = cos x (vi) tanh ix = i tan x and tan x = -i tanh ix
  • 3. IDENTITIES (i) π’”π’Šπ’π’‰ (βˆ’π’™) = βˆ’π’”π’Šπ’π’‰ 𝒙 (ii) 𝒄𝒐𝒔𝒉(βˆ’π’™) = 𝒄𝒐𝒔𝒉 𝒙 (iii) π’„π’π’”π’‰πŸ 𝒙 βˆ’ π’”π’Šπ’π’‰πŸ 𝒙 = 𝟏 (iv) π’”π’†π’„π’‰πŸπ’™ + π’•π’‚π’π’‰πŸπ’™ = 𝟏 (v) π’„π’π’•π’‰πŸ 𝒙 βˆ’ π’„π’π’”π’†π’„π’‰πŸ 𝒙 = 𝟏 (vi) π’”π’Šπ’π’‰ 𝒙 Β± π’š = π’”π’Šπ’π’‰ 𝒙 𝒄𝒐𝒔𝒉 π’š Β± 𝒄𝒐𝒔𝒉 𝒙 π’”π’Šπ’π’‰ π’š (vii) πœπ¨π¬π’‰ 𝒙 Β± π’š = 𝒄𝒐𝒔𝒉 𝒙 𝒄𝒐𝒔𝒉 π’š Β± π’”π’Šπ’π’‰ 𝒙 π’”π’Šπ’π’‰ π’š (viii) 𝒕𝒂𝒏𝒉 𝒙 Β± π’š = 𝒕𝒂𝒏𝒉 𝒙±𝒕𝒂𝒏𝒉 π’š πŸΒ±π’•π’‚π’π’‰ 𝒙 βˆ™ 𝒕𝒂𝒏𝒉 π’š Formula for 2x and 3x (i) π’”π’Šπ’π’‰ πŸπ’™ = 𝟐 π’”π’Šπ’π’‰ 𝒙 𝒄𝒐𝒔𝒉 𝒙 (ii) 𝒄𝒐𝒔𝒉 πŸπ’™ = π’„π’π’”π’‰πŸ 𝒙 + π’”π’Šπ’π’‰πŸ 𝒙 = πŸπ’„π’π’”π’‰πŸ 𝒙 βˆ’ 𝟏 = 𝟏 + πŸπ’”π’Šπ’π’‰πŸ 𝒙 (iii) 𝐭𝒂𝒏𝒉 πŸπ’™ = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟏+π’•π’‚π’π’‰πŸπ’™ (iv) π’”π’Šπ’π’‰ πŸ‘π’™ = πŸ‘ π’”π’Šπ’π’‰ 𝒙 + πŸ’π’”π’Šπ’π’‰πŸ‘ 𝒙 (v) πœπ¨π’”π’‰ πŸ‘π’™ = πŸ’π’„π’π’”π’‰πŸ‘ 𝒙 βˆ’ πŸ‘π’„π’π’”π’‰ 𝒙 (vi) 𝐭𝒂𝒏𝒉 πŸ‘π’™ = πŸ‘ 𝒕𝒂𝒏𝒉 𝒙+π’•π’‚π’π’‰πŸ‘π’™ 𝟏+πŸ‘ π’•π’‚π’π’‰πŸπ’™ (vii) π’”π’Šπ’π’‰ 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 πŸβˆ’π’•π’‚π’π’‰πŸ 𝒙 𝟐 (viii) πœπ¨π’”π’‰ 𝒙 = 𝟏+π’•π’‚π’π’‰πŸ 𝒙 𝟐 πŸβˆ’π’•π’‚π’π’‰πŸ 𝒙 𝟐 (ix) π’”π’Šπ’π’‰ 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 𝟏+π’•π’‚π’π’‰πŸ 𝒙 𝟐
  • 4. HYPERBOLIC IDENTITIES(CONTD) FORMULAE FOR 2X AND 3X (I) π’”π’Šπ’π’‰ πŸπ’™ = 𝟐 π’”π’Šπ’π’‰ 𝒙 𝒄𝒐𝒔𝒉 𝒙 (II) 𝒄𝒐𝒔𝒉 πŸπ’™ = π’„π’π’”π’‰πŸπ’™ + π’”π’Šπ’π’‰πŸπ’™ = πŸπ’„π’π’”π’‰πŸπ’™ βˆ’ 𝟏 = 𝟏 + πŸπ’”π’Šπ’π’‰πŸ 𝒙 (III) 𝒕𝒂𝒏𝒉 πŸπ’™ = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟏+π’•π’‚π’π’‰πŸπ’™ (IV) π’”π’Šπ’π’‰ πŸ‘π’™ = πŸ‘ π’”π’Šπ’π’‰ 𝒙 + πŸ’π’”π’Šπ’π’‰πŸ‘π’™ (V) πœπ¨π’”π’‰ πŸ‘π’™ = πŸ’π’„π’π’”π’‰πŸ‘ 𝒙 βˆ’ πŸ‘π’„π’π’”π’‰ 𝒙 (VI) 𝐭𝒂𝒏𝒉 πŸ‘π’™ = πŸ‘ 𝒕𝒂𝒏𝒉 𝒙+π’•π’‚π’π’‰πŸ‘π’™ 𝟏+πŸ‘ π’•π’‚π’π’‰πŸπ’™ (vii) π’”π’Šπ’π’‰ 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 πŸβˆ’π’•π’‚π’π’‰πŸ 𝒙 𝟐 (viii) πœπ¨π’”π’‰ 𝒙 = 𝟏+π’•π’‚π’π’‰πŸ 𝒙 𝟐 πŸβˆ’π’•π’‚π’π’‰πŸ 𝒙 𝟐 (ix) 𝒕𝒂𝒏𝒉 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 𝟏+π’•π’‚π’π’‰πŸ 𝒙 𝟐
  • 5. HYPERBOLIC IDENTITIES(CONTD) DIFFERENTIATION AND INTEGRATION S.No Function Differentiation Integration 1 π’š = π’”π’Šπ’π’‰ 𝒙 π’…π’š 𝒅𝒙 = 𝒄𝒐𝒔𝒉 𝒙 𝐬𝐒𝐧𝐑 𝒙 𝒅𝒙 = 𝐜𝐨𝐬𝐑 𝒙 2 π’š = 𝒄𝒐𝒔𝒉 𝒙 π’…π’š 𝒅𝒙 = π’”π’Šπ’π’‰ 𝒙 𝐜𝐨𝐬𝐑 𝒙 𝒅𝒙 = 𝐬𝐒𝐧𝐑 𝒙 3 π’š = 𝒕𝒂𝒏𝒉 𝒙 π’…π’š 𝒅𝒙 = π’”π’†π’„π’‰πŸ 𝒙 π’”π’†π’„π’‰πŸ 𝒙 𝒅𝒙 = 𝐭𝐚𝐧𝐑 𝒙
  • 6. HYPERBOLIC IDENTITIES(CONTD) PRODUCT FORMULAE (I) π’”π’Šπ’π’‰ 𝒙 + π’š + π’”π’Šπ’π’‰ 𝒙 βˆ’ π’š = 𝟐 π’”π’Šπ’π’‰ 𝒙 𝒄𝒐𝒔𝒉 π’š (II) π’”π’Šπ’π’‰ 𝒙 + π’š βˆ’ π’”π’Šπ’π’‰ 𝒙 βˆ’ π’š = 𝟐 𝒄𝒐𝒔𝒉 𝒙 π’”π’Šπ’π’‰ π’š (III) πœπ¨π’”π’‰ 𝒙 + π’š + 𝒄𝒐𝒔𝒉 𝒙 βˆ’ π’š = 𝟐 𝒄𝒐𝒔𝒉 𝒙 𝒄𝒐𝒔𝒉 π’š (IV) πœπ¨π¬π’‰ 𝒙 + π’š βˆ’ 𝒄𝒐𝒔𝒉 𝒙 βˆ’ π’š = 𝟐 π’”π’Šπ’π’‰ 𝒙 π’”π’Šπ’π’‰ π’š (V) ) π’”π’Šπ’π’‰ 𝒙 + π’”π’Šπ’π’‰ π’š = 𝟐 π’”π’Šπ’π’‰ 𝒙+π’š 𝟐 𝒄𝒐𝒔𝒉 π’™βˆ’π’š 𝟐 (VI) π’”π’Šπ’π’‰ 𝒙 βˆ’ π’”π’Šπ’π’‰ π’š = 𝟐 𝒄𝒐𝒔𝒉 𝒙+π’š 𝟐 π’”π’Šπ’π’‰ π’™βˆ’π’š 𝟐 (VII) πœπ¨π’”π’‰ 𝒙 + 𝒄𝒐𝒔𝒉 π’š = πŸπ’„π’π’”π’‰ 𝒙+π’š 𝟐 𝒄𝒐𝒔𝒉 π’™βˆ’π’š 𝟐 (VIII) πœπ¨π’”π’‰ 𝒙 βˆ’ 𝒄𝒐𝒔𝒉 π’š = πŸπ’”π’Šπ’π’‰ 𝒙+π’š 𝟐 π’”π’Šπ’π’‰ π’™βˆ’π’š 𝟐
  • 7. TABLE OF VALUES OF HYPERBOLIC FUNCTIONS x 0 βˆ’βˆž +∞ π’š = π’”π’Šπ’π’‰ 𝒙 0 βˆ’βˆž ∞ π’š = 𝒄𝒐𝒔𝒉 𝒙 1 ∞ ∞ π’š = 𝒕𝒂𝒏𝒉 𝒙 0 βˆ’1 1
  • 8. INVERSE HYPERBOLIC FUNCTIONS Definition: if sinh u = z then u is called inverse hyperbolic sine of z and is denoted by 𝒖 = π’”π’Šπ’π’‰βˆ’πŸπ’›. Similarly we can define inverse hyperbolic cosine and inverse hyperbolic tangent and other functions which we denote as π’„π’π’”π’‰βˆ’πŸπ’›, π’•π’‚π’π’‰βˆ’πŸπ’›, 𝒆𝒕𝒄. The inverse hyperbolic functions are many valued but we will consider their principal values only. If Z is real we can show that 1. π’”π’Šπ’π’‰βˆ’πŸπ’› = π’π’π’ˆ 𝒛 + π’›πŸ + 𝟏 2. π’„π’π’”π’‰βˆ’πŸ 𝒛 = π’π’π’ˆ 𝒛 + π’›πŸ βˆ’ 𝟏 3. π’•π’‚π’π’‰βˆ’πŸ 𝒛 = 𝟏 𝟐 π’π’π’ˆ 𝟏+𝒛 πŸβˆ’π’› .
  • 9. INVERSE HYPERBOLIC FUNCTIONS INTEGRATION FORMULAE 𝒅𝒙 π’™πŸ + π’‚πŸ = π’”π’Šπ’π’‰βˆ’πŸ 𝒙 𝒂 𝒅𝒙 π’™πŸ βˆ’ π’‚πŸ = π’„π’π’”π’‰βˆ’πŸ 𝒙 𝒂 𝒅𝒙 π’‚πŸ βˆ’ π’™πŸ = 𝟏 𝒂 π’•π’‚π’π’‰βˆ’πŸ 𝒙 𝒂
  • 10. PROBLEMS BASED ON HYPERBOLIC FUNCTIONS 1.Prove that 𝒄𝒐𝒔𝒉 𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒙 𝒏 = 𝒄𝒐𝒔𝒉 𝒏𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒏𝒙 2.Express cot (x +i y) in a + i b form. 3. Solve the equation π’™πŸ• + π’™πŸ’ + π’™πŸ‘ + 𝟏 = 𝟎 4. Prove that 𝟏 + 𝒄𝒐𝒔 𝒙 + π’Š π’”π’Šπ’ 𝒙 𝒏 = πŸπ’π’„π’π’”π’ 𝒙 𝟐 𝒄𝒐𝒔 𝒏𝒙 𝟐 + π’Šπ’”π’Šπ’ 𝒏𝒙 𝟐 5. Prove that π’”π’Šπ’π’‰βˆ’πŸ 𝒙 = π’π’π’ˆ 𝒙 + π’™πŸ + 𝟏 6.Show that π’”π’†π’„π’‰βˆ’πŸ π’”π’Šπ’ 𝜽 = π’π’π’ˆ 𝒄𝒐𝒕 𝜽 𝟐 7.Express sec(x+iy) in a+ib form.
  • 11. PROBLEMS BASED ON HYPERBOLIC FUNCTIONS 1. Prove that 𝒄𝒐𝒔𝒉 𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒙 𝒏 = 𝒄𝒐𝒔𝒉 𝒏𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒏𝒙 Solution: LHS = 𝒄𝒐𝒔𝒉 𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒙 𝒏 = 𝒆𝒙+π’†βˆ’π’™ 𝟐 βˆ’ π’†π’™βˆ’π’†βˆ’π’™ 𝟐 𝒏 = 𝒆𝒙 + π’†βˆ’π’™ βˆ’ 𝒆𝒙 + π’†βˆ’π’™ 𝟐 𝒏 = π’†βˆ’π’™ 𝒏 = π’†βˆ’π’π’™ RHS = 𝒄𝒐𝒔𝒉 𝒏𝒙 βˆ’ π’”π’Šπ’π’‰ 𝒏𝒙 = 𝒆𝒏𝒙+π’†βˆ’π’π’™ 𝟐 βˆ’ π’†π’π’™βˆ’π’†βˆ’π’π’™ 𝟐 = 𝒆𝒏𝒙+π’†βˆ’π’π’™βˆ’π’†π’π’™+π’†βˆ’π’π’™ 𝟐 = π’†βˆ’π’π’™ ∴ LHS=RHS
  • 12. PROBLEMS BASED ON HYPERBOLIC FUNCTIONS 2.Express cot(x +i y) in a + i b form. Solution: 𝒄𝒐𝒕 𝒙 + π’Šπ’š = 𝒄𝒐𝒔(𝒙+π’Šπ’š) π’”π’Šπ’(𝒙+π’Šπ’š) = 𝒄𝒐𝒔(𝒙+π’Šπ’š) π’”π’Šπ’(𝒙+π’Šπ’š) Γ— π’”π’Šπ’(π’™βˆ’π’Šπ’š) π’”π’Šπ’(π’™βˆ’π’Šπ’š) = πŸπ’„π’π’”(𝒙+π’Šπ’š) πŸπ’”π’Šπ’(𝒙+π’Šπ’š) Γ— π’”π’Šπ’(π’™βˆ’π’Šπ’š) π’”π’Šπ’(π’™βˆ’π’Šπ’š) = π’”π’Šπ’ 𝒙+π’Šπ’š+π’™βˆ’π’Šπ’š βˆ’π’”π’Šπ’ 𝒙+π’Šπ’šβˆ’π’™βˆ’π’Šπ’š 𝒄𝒐𝒔 𝒙+π’Šπ’šβˆ’(π’™βˆ’π’Šπ’š) βˆ’π’„π’π’” 𝒙+π’Šπ’š+π’™βˆ’π’Šπ’š π’”π’Šπ’ 𝑨 + 𝑩 βˆ’ π’”π’Šπ’ 𝑨 βˆ’ 𝑩 = 𝟐 𝒄𝒐𝒔𝑨 π’”π’Šπ’π‘© 𝒄𝒐𝒔 𝑨 βˆ’ 𝑩 βˆ’ 𝒄𝒐𝒔 𝑨 + 𝑩 = 𝟐 π’”π’Šπ’π‘¨ π’”π’Šπ’π‘© = π’”π’Šπ’ πŸπ’™ βˆ’π’”π’Šπ’ πŸπ’Šπ’š 𝒄𝒐𝒔 πŸπ’Šπ’š βˆ’π’„π’π’” πŸπ’™ = π’”π’Šπ’ πŸπ’™ βˆ’π’Šπ’”π’Šπ’π’‰ πŸπ’š 𝒄𝒐𝒔𝒉 πŸπ’š βˆ’π’„π’π’” πŸπ’™ = π’”π’Šπ’ πŸπ’™ 𝒄𝒐𝒔𝒉 πŸπ’š βˆ’ 𝒄𝒐𝒔 πŸπ’™ + π’Š βˆ’π’”π’Šπ’π’‰ πŸπ’š 𝒄𝒐𝒔𝒉 πŸπ’š βˆ’ 𝒄𝒐𝒔 πŸπ’™ = 𝒂 + π’Šπ’ƒ Where a= π’”π’Šπ’ πŸπ’™ 𝒄𝒐𝒔𝒉 πŸπ’š βˆ’π’„π’π’” πŸπ’™ & b= βˆ’π’”π’Šπ’π’‰ πŸπ’š 𝒄𝒐𝒔𝒉 πŸπ’š βˆ’π’„π’π’” πŸπ’™
  • 13. PROBLEMS BASED ON HYPERBOLIC FUNCTIONS 3. Solve the equation π’™πŸ• + π’™πŸ’ + π’™πŸ‘ + 𝟏 = 𝟎 Roots of a Complex Number: De Moivre’s Theorem can be used to find all n- roots of a complex number. Since, 𝒄𝒐𝒔 𝜽 = 𝒄𝒐𝒔 πŸπ’Œπ… + 𝜽 & π’”π’Šπ’ 𝜽 = π’”π’Šπ’ (πŸπ’Œπ… + 𝜽) where k is an integer. We have by DeMoivre’s Theorem 𝒄𝒐𝒔 𝜽 + π’Šπ’”π’Šπ’ 𝜽 𝟏 𝒏 = 𝒄𝒐𝒔 πŸπ’Œπ… + 𝜽 + π’”π’Šπ’ (πŸπ’Œπ… + 𝜽) 𝟏 𝒏 𝒄𝒐𝒔 𝜽 + π’Šπ’”π’Šπ’ 𝜽 𝟏 𝒏 = 𝒄𝒐𝒔 πŸπ…π’Œ + 𝜽 𝒏 + π’Šπ’”π’Šπ’ πŸπ…π’Œ + 𝜽 𝒏 By putting π’Œ = 𝟎, 𝟏, 𝟐, … … . . 𝒏 βˆ’ 𝟏 we get n roots of the complex number.
  • 14. PROBLEMS BASED ON HYPERBOLIC FUNCTIONS 3. Solve the equation π’™πŸ• + π’™πŸ’ + π’™πŸ‘ + 𝟏 = 𝟎 Solution: π’™πŸ• + π’™πŸ’ + π’™πŸ‘ + 𝟏 = 𝟎 ⟹ π’™πŸ’ π’™πŸ‘ + 𝟏 + 𝟏 π’™πŸ‘ + 𝟏 = 𝟎 ⟹ π’™πŸ‘ + 𝟏 π’™πŸ’ + 𝟏 = 𝟎 ⟹ π’™πŸ‘ + 𝟏 = 𝟎, π’™πŸ’ + 𝟏 = 𝟎 βŸΉπ’™πŸ‘ = βˆ’πŸ, π’™πŸ’ = βˆ’πŸ π’™πŸ‘ = βˆ’πŸ βŸΉπ’™πŸ‘ = (𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅) βŸΉπ’™ = 𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅 𝟏 πŸ‘ βŸΉπ’™ = 𝒄𝒐𝒔 πŸπ’Œπ… + 𝝅 + π’Šπ’”π’Šπ’(πŸπ’Œπ… + 𝝅 𝟏 πŸ‘ ⟹ 𝒙 = 𝒄𝒐𝒔 πŸπ’Œ + 𝟏)𝝅 + π’Šπ’”π’Šπ’(πŸπ’Œ + 𝟏)𝝅 𝟏 πŸ‘ ⟹ 𝒙 = 𝒄𝒐𝒔 (πŸπ’Œ+𝟏)𝝅 πŸ‘ + π’Š π’”π’Šπ’ (πŸπ’Œ+𝟏)𝝅 𝒏 where π’Œ = 𝟎, 𝟏, 𝟐 when π’Œ = 𝟎, 𝒙 = 𝒄𝒐𝒔 𝝅 πŸ‘ + π’Šπ’”π’Šπ’ 𝝅 πŸ‘ when k = 1, 𝐱 = 𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅 When k = 2, 𝒙 = 𝒄𝒐𝒔 πŸ“π… πŸ‘ + π’Šπ’”π’Šπ’ πŸ“π… πŸ‘
  • 15. PROBLEMS BASED ON HYPERBOLIC FUNCTIONS π’™πŸ’ = βˆ’πŸ βŸΉπ’™πŸ’ = (𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅) βŸΉπ’™ = 𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅 𝟏 πŸ’ βŸΉπ’™ = 𝒄𝒐𝒔 πŸπ’Œπ… + 𝝅 + π’Šπ’”π’Šπ’(πŸπ’Œπ… + 𝝅 𝟏 πŸ’ ⟹ 𝒙 = 𝒄𝒐𝒔 πŸπ’Œ + 𝟏)𝝅 + π’Šπ’”π’Šπ’(πŸπ’Œ + 𝟏)𝝅 𝟏 πŸ’ ⟹ 𝒙 = 𝒄𝒐𝒔 (πŸπ’Œ+𝟏)𝝅 πŸ’ + π’Š π’”π’Šπ’ (πŸπ’Œ+𝟏)𝝅 πŸ’ where π’Œ = 𝟎, 𝟏, 𝟐, πŸ‘ when π’Œ = 𝟎, 𝒙 = 𝒄𝒐𝒔 𝝅 πŸ’ + π’Šπ’”π’Šπ’ 𝝅 πŸ’ when k = 1, 𝒙 = 𝒄𝒐𝒔 πŸ‘π… πŸ’ + π’Šπ’”π’Šπ’ πŸ‘π… πŸ’ When k = 2, 𝒙 = 𝒄𝒐𝒔 πŸ“π… πŸ’ + π’Šπ’”π’Šπ’ πŸ“π… πŸ’ when k = 3, 𝒙 = 𝒄𝒐𝒔 πŸ•π… πŸ’ + π’Šπ’”π’Šπ’ πŸ•π… πŸ’ ∴ π‘‡β„Žπ‘’ 𝑠𝑒𝑣𝑒𝑛 π‘Ÿπ‘œπ‘œπ‘‘π‘  π‘œπ‘“ π‘‘β„Žπ‘’ 𝑔𝑖𝑣𝑒𝑛 π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› are given by 𝒙 = 𝒄𝒐𝒔 𝝅 πŸ‘ + π’Šπ’”π’Šπ’ 𝝅 πŸ‘ 𝐱 = 𝒄𝒐𝒔 𝝅 + π’Šπ’”π’Šπ’ 𝝅 𝒙 = 𝒄𝒐𝒔 πŸ“π… πŸ‘ + π’Šπ’”π’Šπ’ πŸ“π… πŸ‘ 𝒙 = 𝒄𝒐𝒔 𝝅 πŸ’ + π’Šπ’”π’Šπ’ 𝝅 πŸ’ 𝒙 = 𝒄𝒐𝒔 πŸ‘π… πŸ’ + π’Šπ’”π’Šπ’ πŸ‘π… πŸ’ 𝒙 = 𝒄𝒐𝒔 πŸ“π… πŸ’ + π’Šπ’”π’Šπ’ πŸ“π… πŸ’ 𝒙 = 𝒄𝒐𝒔 πŸ•π… πŸ’ + π’Šπ’”π’Šπ’ πŸ•π… πŸ’
  • 16. 4. Prove that 𝟏 + 𝒄𝒐𝒔 𝒙 + π’Š π’”π’Šπ’ 𝒙 𝒏 = πŸπ’π’„π’π’”π’ 𝒙 𝟐 𝒄𝒐𝒔 𝒏𝒙 𝟐 + π’Šπ’”π’Šπ’ 𝒏𝒙 𝟐 Solution: we use the following formulas 𝐬𝐒𝐧 πŸπ’™ = 𝟐 𝐬𝐒𝐧 𝒙 𝐜𝐨𝐬 𝒙 π’”π’Šπ’ 𝒙 = πŸπ’”π’Šπ’ 𝒙 𝟐 𝒄𝒐𝒔 𝒙 𝟐 𝟏 + 𝒄𝒐𝒔 𝒙 = πŸπ’„π’π’”πŸ 𝒙 𝟐 LHS= 𝟏 + 𝒄𝒐𝒔 𝒙 + π’Š π’”π’Šπ’ 𝒙 𝒏 = πŸπ’„π’π’”πŸ 𝒙 𝟐 + π’ŠπŸπ’”π’Šπ’ 𝒙 𝟐 𝒄𝒐𝒔 𝒙 𝟐 𝒏 = πŸπ’„π’π’” 𝒙 𝟐 𝒄𝒐𝒔 𝒙 𝟐 + π’Šπ’”π’Šπ’ 𝒙 𝟐 𝒏 = πŸπ’„π’π’” 𝒙 𝟐 𝒏 𝒄𝒐𝒔 𝒙 𝟐 + π’Šπ’”π’Šπ’ 𝒙 𝟐 𝒏 = πŸπ’ 𝒄𝒐𝒔𝒏 𝒙 𝟐 𝒄𝒐𝒔 𝒏𝒙 𝟐 + π’Šπ’”π’Šπ’ 𝒏𝒙 𝟐 (by De’Moivre’s Theorem) = RHS
  • 17. 5. Prove that π’”π’Šπ’π’‰βˆ’πŸ 𝒙 = π’π’π’ˆ 𝒙 + π’™πŸ + 𝟏 Solution: Let π’”π’Šπ’π’‰βˆ’πŸ 𝒙 = π’šβŸΉ π’”π’Šπ’π’‰ π’š = 𝒙 ∴ 𝒙 = π’”π’Šπ’π’‰π’š = π’†π’š βˆ’ π’†βˆ’π’š 𝟐 = π’†π’š βˆ’ 𝟏 π’†π’š 𝟐 = π’†πŸπ’š βˆ’ 𝟏 πŸπ’†π’š 𝒙 = π’†πŸπ’šβˆ’πŸ πŸπ’†π’š βŸΉπŸπ’†π’š 𝐱 = π’†πŸπ’š βˆ’ 𝟏 ⟹ πŸπ’†π’š 𝐱 βˆ’ π’†πŸπ’š βˆ’ 𝟏 = 𝟎 ⟹ π’†πŸπ’š βˆ’ πŸπ’™π’†π’š +1=0 ⟹ π’†π’š 𝟐 βˆ’ πŸπ’™ π’†π’š + 𝟏 = 𝟎 This is quadratic in π’†π’š . π’†π’š = βˆ’(βˆ’πŸπ’™) Β± (βˆ’πŸπ’™)𝟐 βˆ’ πŸ’(𝟏)(𝟏) 𝟐(𝟏) = πŸπ’™ Β± πŸ’π’™πŸ βˆ’ πŸ’ 𝟐 = πŸπ’™ Β± 𝟐 π’™πŸ βˆ’ 𝟏 𝟐 = 𝒙 Β± π’™πŸ βˆ’ 𝟏 Conventionally we take positive sign, π’†π’š= 𝒙 Β± π’™πŸ βˆ’ 𝟏. Taking log on both sides log(π’†π’š )=log 𝐱 Β± π’™πŸ βˆ’ 𝟏⟹ 𝐲 = π’π’π’ˆ 𝒙 + π’™πŸ + 𝟏 ⟹ π’”π’Šπ’π’‰βˆ’πŸ 𝒙 = π’π’π’ˆ 𝒙 + π’™πŸ + 𝟏
  • 18. 6. Show that π’”π’†π’„π’‰βˆ’πŸ π’”π’Šπ’ 𝜽 = π’π’π’ˆ 𝒄𝒐𝒕 𝜽 𝟐 ⟹ Solution: Let π’”π’†π’„π’‰βˆ’πŸ π’”π’Šπ’ 𝜽 = π’™βŸΉ 𝒔𝒆𝒄𝒉 𝒙 = π’”π’Šπ’πœ½ ∴ π’”π’Šπ’πœ½ = 𝟏 𝒄𝒐𝒔𝒉 𝒙 = 𝟏 𝒆𝒙 + π’†βˆ’π’™ 𝟐 = 𝟐 𝒆𝒙 + π’†βˆ’π’™ = 𝟐 𝒆𝒙 + 𝟏 𝒆𝒙 = πŸπ’†π’™ π’†πŸπ’™ + 𝟏 π’”π’Šπ’πœ½ = πŸπ’†π’™ π’†πŸπ’™ + 𝟏 π’†πŸπ’™ sin 𝜽 + π’”π’Šπ’πœ½ = πŸπ’†π’™ ⟹ 𝒆𝒙 𝟐 π’”π’Šπ’πœ½ βˆ’ πŸπ’†π’™ + π’”π’Šπ’πœ½ = 𝟎 This is quadratic in 𝒆𝒙 . 𝒆𝒙 = βˆ’(βˆ’πŸ) Β± (βˆ’πŸ)𝟐 βˆ’ πŸ’(π’”π’Šπ’πœ½)(π’”π’Šπ’πœ½) πŸπ’”π’Šπ’πœ½ = 𝟐 Β± πŸ’ βˆ’ πŸ’π’”π’Šπ’πŸπœ½ πŸπ’”π’Šπ’πœ½ = 𝟐 Β± 𝟐 𝟏 βˆ’ π’”π’Šπ’πŸπœ½ πŸπ’”π’Šπ’πœ½ = 𝟏 Β± π’„π’π’”πŸπœ½ πŸπ’”π’Šπ’πœ½ Conventionally we take positive sign, 𝒆𝒙= 𝟏 + 𝒄𝒐𝒔 𝜽 π’”π’Šπ’πœ½ = πŸπ’„π’π’”πŸ 𝜽 𝟐 πŸπ’”π’Šπ’ 𝜽 𝟐 𝒄𝒐𝒔 𝜽 𝟐 = 𝒄𝒐𝒕 𝜽 𝟐 log(𝒆𝒙 )=log 𝒄𝒐𝒕 𝜽 𝟐 ⟹𝐱 = π’π’π’ˆ 𝒄𝒐𝒕 𝜽 𝟐 ⟹ π’”π’†π’„π’‰βˆ’πŸ π’”π’Šπ’ 𝜽 = π’π’π’ˆ 𝒄𝒐𝒕 𝜽 𝟐
  • 19. 7.Express sec(x+iy) in a+ib form. Solution: 𝑠𝑒𝑐 π‘₯ + 𝑖𝑦 = 𝟏 𝒄𝒐𝒔(𝒙+π’Šπ’š) = 𝟏 𝒄𝒐𝒔(𝒙+π’Šπ’š) Γ— 𝒄𝒐𝒔(π’™βˆ’π’Šπ’š) 𝒄𝒐𝒔(π’™βˆ’π’Šπ’š) = πŸπ’„π’π’”(π’™βˆ’π’Šπ’š) 𝟐 𝒄𝒐𝒔(𝒙+π’Šπ’š)𝒄𝒐𝒔(𝒙+π’Šπ’š) = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔 π’Šπ’š +π’”π’Šπ’ 𝒙 π’”π’Šπ’ (π’Šπ’š) 𝒄𝒐𝒔 𝒙+π’Šπ’š+π’™βˆ’π’Šπ’š +𝒄𝒐𝒔(𝒙+π’Šπ’šβˆ’ π’™βˆ’π’Šπ’š ) [Since 𝒄𝒐𝒔 𝑨 βˆ’ 𝑩 = 𝒄𝒐𝒔𝑨 𝒄𝒐𝒔𝑩 + π’”π’Šπ’π‘¨ π’”π’Šπ’π‘© & πŸπ’„π’π’”π‘¨ 𝒄𝒐𝒔𝑩 = 𝐜𝐨𝐬 𝑨 + 𝑩 + 𝐜𝐨𝐬(𝑨 βˆ’ 𝑩)] = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 π’š+π’Š π’”π’Šπ’ 𝒙 π’”π’Šπ’π’‰ π’š 𝒄𝒐𝒔 πŸπ’™ +𝒄𝒐𝒔(πŸπ’Šπ’š) = πŸπ’„π’π’” 𝒙 𝒄𝒐𝒔𝒉 π’š+π’Š 𝟐 π’”π’Šπ’ 𝒙 π’”π’Šπ’π’‰ π’š 𝒄𝒐𝒔 πŸπ’™+𝒄𝒐𝒔𝒉 πŸπ’š = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 π’š 𝒄𝒐𝒔 πŸπ’™+𝒄𝒐𝒔𝒉 πŸπ’š + π’Š 𝟐 π’”π’Šπ’ 𝒙 π’”π’Šπ’π’‰ π’š 𝒄𝒐𝒔 πŸπ’™+𝒄𝒐𝒔𝒉 πŸπ’š = 𝒂 + π’Šπ’ƒ 𝒔𝒆𝒄 𝒙 + π’Šπ’š = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 π’š 𝒄𝒐𝒔 πŸπ’™ + 𝒄𝒐𝒔𝒉 πŸπ’š + π’Š 𝟐 π’”π’Šπ’ 𝒙 π’”π’Šπ’π’‰ π’š 𝒄𝒐𝒔 πŸπ’™ + 𝒄𝒐𝒔𝒉 πŸπ’š