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![7.Express sec(x+iy) in a+ib form.
Solution: 𝑠𝑒𝑐 𝑥 + 𝑖𝑦 =
𝟏
𝒄𝒐𝒔(𝒙+𝒊𝒚)
=
𝟏
𝒄𝒐𝒔(𝒙+𝒊𝒚)
×
𝒄𝒐𝒔(𝒙−𝒊𝒚)
𝒄𝒐𝒔(𝒙−𝒊𝒚)
=
𝟐𝒄𝒐𝒔(𝒙−𝒊𝒚)
𝟐 𝒄𝒐𝒔(𝒙+𝒊𝒚)𝒄𝒐𝒔(𝒙+𝒊𝒚)
=
𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔 𝒊𝒚 +𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏 (𝒊𝒚)
𝒄𝒐𝒔 𝒙+𝒊𝒚+𝒙−𝒊𝒚 +𝒄𝒐𝒔(𝒙+𝒊𝒚− 𝒙−𝒊𝒚 )
[Since 𝒄𝒐𝒔 𝑨 − 𝑩 = 𝒄𝒐𝒔𝑨 𝒄𝒐𝒔𝑩 + 𝒔𝒊𝒏𝑨 𝒔𝒊𝒏𝑩
& 𝟐𝒄𝒐𝒔𝑨 𝒄𝒐𝒔𝑩 = 𝐜𝐨𝐬 𝑨 + 𝑩 + 𝐜𝐨𝐬(𝑨 − 𝑩)]
=
𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 𝒚+𝒊 𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒚
𝒄𝒐𝒔 𝟐𝒙 +𝒄𝒐𝒔(𝟐𝒊𝒚)
=
𝟐𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 𝒚+𝒊 𝟐 𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒚
𝒄𝒐𝒔 𝟐𝒙+𝒄𝒐𝒔𝒉 𝟐𝒚
=
𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 𝒚
𝒄𝒐𝒔 𝟐𝒙+𝒄𝒐𝒔𝒉 𝟐𝒚
+ 𝒊
𝟐 𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒚
𝒄𝒐𝒔 𝟐𝒙+𝒄𝒐𝒔𝒉 𝟐𝒚
= 𝒂 + 𝒊𝒃
𝒔𝒆𝒄 𝒙 + 𝒊𝒚 =
𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 𝒚
𝒄𝒐𝒔 𝟐𝒙 + 𝒄𝒐𝒔𝒉 𝟐𝒚
+ 𝒊
𝟐 𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒚
𝒄𝒐𝒔 𝟐𝒙 + 𝒄𝒐𝒔𝒉 𝟐𝒚](https://image.slidesharecdn.com/06complexnumbershyperbolicfunctions-230610202305-4e4cceb7/75/06_Complex-Numbers_Hyperbolic-Functions-pptx-19-2048.jpg)
06_Complex Numbers_Hyperbolic Functions.pptx
- 1. NUMBERS We already knowthat 𝒆𝒊𝜽 = 𝐜𝐨𝐬 𝜽 + 𝒊 𝒔𝒊𝒏 𝜽 𝒆−𝒊𝜽 = 𝒄𝒐𝒔 𝜽 − 𝒊 𝒔𝒊𝒏 𝜽 From the above two equations we have 𝐜𝐨𝐬 𝜽 = 𝒆𝒊𝜽 + 𝒆−𝒊𝜽 𝟐 & 𝐬𝐢𝐧 𝜽 = 𝒆𝒊𝜽 − 𝒆−𝒊𝜽 𝟐𝒊 These forms are known as Euler’s exponential forms of circular functions. Further, if z is a complex number we define 𝒄𝒐𝒔 𝒛 = 𝒆𝒊𝒛 + 𝒆−𝒊𝒛 𝟐 & 𝒔𝒊𝒏 𝒛 = 𝒆𝒊𝒛 − 𝒆−𝒊𝒛 𝟐𝒊
- 2. HYPERBOLIC FUNCTIONS If xis real or complex 𝒆𝒙+𝒆−𝒙 𝟐 is called hyperbolic cosine of x and is denoted by cosh x and 𝒆𝒙−𝒆−𝒙 𝟐 is called hyperbolic sine of x and is denoted by sinh x. Thus, 𝒔𝒊𝒏𝒉 𝒙 = 𝒆𝒙−𝒆−𝒙 𝟐 , 𝒄𝒐𝒔𝒉 𝒙 = 𝒆𝒙+𝒆−𝒙 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) 𝒔𝒊𝒏𝒉 (−𝒙)= −𝒔𝒊𝒏𝒉 𝒙 (ii) 𝒄𝒐𝒔𝒉(−𝒙) = 𝒄𝒐𝒔𝒉 𝒙 (iii) 𝒄𝒐𝒔𝒉𝟐 𝒙 − 𝒔𝒊𝒏𝒉𝟐 𝒙 = 𝟏 (iv) 𝒔𝒆𝒄𝒉𝟐𝒙 + 𝒕𝒂𝒏𝒉𝟐𝒙 = 𝟏 (v) 𝒄𝒐𝒕𝒉𝟐 𝒙 − 𝒄𝒐𝒔𝒆𝒄𝒉𝟐 𝒙 = 𝟏 (vi) 𝒔𝒊𝒏𝒉 𝒙 ± 𝒚 = 𝒔𝒊𝒏𝒉 𝒙 𝒄𝒐𝒔𝒉 𝒚 ± 𝒄𝒐𝒔𝒉 𝒙 𝒔𝒊𝒏𝒉 𝒚 (vii) 𝐜𝐨𝐬𝒉 𝒙 ± 𝒚 = 𝒄𝒐𝒔𝒉 𝒙 𝒄𝒐𝒔𝒉 𝒚 ± 𝒔𝒊𝒏𝒉 𝒙 𝒔𝒊𝒏𝒉 𝒚 (viii) 𝒕𝒂𝒏𝒉 𝒙 ± 𝒚 = 𝒕𝒂𝒏𝒉 𝒙±𝒕𝒂𝒏𝒉 𝒚 𝟏±𝒕𝒂𝒏𝒉 𝒙 ∙ 𝒕𝒂𝒏𝒉 𝒚 Formula for 2x and 3x (i) 𝒔𝒊𝒏𝒉 𝟐𝒙 = 𝟐 𝒔𝒊𝒏𝒉 𝒙 𝒄𝒐𝒔𝒉 𝒙 (ii) 𝒄𝒐𝒔𝒉 𝟐𝒙 = 𝒄𝒐𝒔𝒉𝟐 𝒙 + 𝒔𝒊𝒏𝒉𝟐 𝒙 = 𝟐𝒄𝒐𝒔𝒉𝟐 𝒙 − 𝟏 = 𝟏 + 𝟐𝒔𝒊𝒏𝒉𝟐 𝒙 (iii) 𝐭𝒂𝒏𝒉 𝟐𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟏+𝒕𝒂𝒏𝒉𝟐𝒙 (iv) 𝒔𝒊𝒏𝒉 𝟑𝒙 = 𝟑 𝒔𝒊𝒏𝒉 𝒙 + 𝟒𝒔𝒊𝒏𝒉𝟑 𝒙 (v) 𝐜𝐨𝒔𝒉 𝟑𝒙 = 𝟒𝒄𝒐𝒔𝒉𝟑 𝒙 − 𝟑𝒄𝒐𝒔𝒉 𝒙 (vi) 𝐭𝒂𝒏𝒉 𝟑𝒙 = 𝟑 𝒕𝒂𝒏𝒉 𝒙+𝒕𝒂𝒏𝒉𝟑𝒙 𝟏+𝟑 𝒕𝒂𝒏𝒉𝟐𝒙 (vii) 𝒔𝒊𝒏𝒉 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 𝟏−𝒕𝒂𝒏𝒉𝟐 𝒙 𝟐 (viii) 𝐜𝐨𝒔𝒉 𝒙 = 𝟏+𝒕𝒂𝒏𝒉𝟐 𝒙 𝟐 𝟏−𝒕𝒂𝒏𝒉𝟐 𝒙 𝟐 (ix) 𝒔𝒊𝒏𝒉 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 𝟏+𝒕𝒂𝒏𝒉𝟐 𝒙 𝟐
- 4. HYPERBOLIC IDENTITIES(CONTD) FORMULAE FOR2X AND 3X (I) 𝒔𝒊𝒏𝒉 𝟐𝒙 = 𝟐 𝒔𝒊𝒏𝒉 𝒙 𝒄𝒐𝒔𝒉 𝒙 (II) 𝒄𝒐𝒔𝒉 𝟐𝒙 = 𝒄𝒐𝒔𝒉𝟐𝒙 + 𝒔𝒊𝒏𝒉𝟐𝒙 = 𝟐𝒄𝒐𝒔𝒉𝟐𝒙 − 𝟏 = 𝟏 + 𝟐𝒔𝒊𝒏𝒉𝟐 𝒙 (III) 𝒕𝒂𝒏𝒉 𝟐𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟏+𝒕𝒂𝒏𝒉𝟐𝒙 (IV) 𝒔𝒊𝒏𝒉 𝟑𝒙 = 𝟑 𝒔𝒊𝒏𝒉 𝒙 + 𝟒𝒔𝒊𝒏𝒉𝟑𝒙 (V) 𝐜𝐨𝒔𝒉 𝟑𝒙 = 𝟒𝒄𝒐𝒔𝒉𝟑 𝒙 − 𝟑𝒄𝒐𝒔𝒉 𝒙 (VI) 𝐭𝒂𝒏𝒉 𝟑𝒙 = 𝟑 𝒕𝒂𝒏𝒉 𝒙+𝒕𝒂𝒏𝒉𝟑𝒙 𝟏+𝟑 𝒕𝒂𝒏𝒉𝟐𝒙 (vii) 𝒔𝒊𝒏𝒉 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 𝟏−𝒕𝒂𝒏𝒉𝟐 𝒙 𝟐 (viii) 𝐜𝐨𝒔𝒉 𝒙 = 𝟏+𝒕𝒂𝒏𝒉𝟐 𝒙 𝟐 𝟏−𝒕𝒂𝒏𝒉𝟐 𝒙 𝟐 (ix) 𝒕𝒂𝒏𝒉 𝒙 = 𝟐 𝒕𝒂𝒏𝒉 𝒙 𝟐 𝟏+𝒕𝒂𝒏𝒉𝟐 𝒙 𝟐
- 5. HYPERBOLIC IDENTITIES(CONTD) DIFFERENTIATION ANDINTEGRATION S.No Function Differentiation Integration 1 𝒚 = 𝒔𝒊𝒏𝒉 𝒙 𝒅𝒚 𝒅𝒙 = 𝒄𝒐𝒔𝒉 𝒙 𝐬𝐢𝐧𝐡 𝒙 𝒅𝒙 = 𝐜𝐨𝐬𝐡 𝒙 2 𝒚 = 𝒄𝒐𝒔𝒉 𝒙 𝒅𝒚 𝒅𝒙 = 𝒔𝒊𝒏𝒉 𝒙 𝐜𝐨𝐬𝐡 𝒙 𝒅𝒙 = 𝐬𝐢𝐧𝐡 𝒙 3 𝒚 = 𝒕𝒂𝒏𝒉 𝒙 𝒅𝒚 𝒅𝒙 = 𝒔𝒆𝒄𝒉𝟐 𝒙 𝒔𝒆𝒄𝒉𝟐 𝒙 𝒅𝒙 = 𝐭𝐚𝐧𝐡 𝒙
- 6. HYPERBOLIC IDENTITIES(CONTD) PRODUCT FORMULAE (I)𝒔𝒊𝒏𝒉 𝒙 + 𝒚 + 𝒔𝒊𝒏𝒉 𝒙 − 𝒚 = 𝟐 𝒔𝒊𝒏𝒉 𝒙 𝒄𝒐𝒔𝒉 𝒚 (II) 𝒔𝒊𝒏𝒉 𝒙 + 𝒚 − 𝒔𝒊𝒏𝒉 𝒙 − 𝒚 = 𝟐 𝒄𝒐𝒔𝒉 𝒙 𝒔𝒊𝒏𝒉 𝒚 (III) 𝐜𝐨𝒔𝒉 𝒙 + 𝒚 + 𝒄𝒐𝒔𝒉 𝒙 − 𝒚 = 𝟐 𝒄𝒐𝒔𝒉 𝒙 𝒄𝒐𝒔𝒉 𝒚 (IV) 𝐜𝐨𝐬𝒉 𝒙 + 𝒚 − 𝒄𝒐𝒔𝒉 𝒙 − 𝒚 = 𝟐 𝒔𝒊𝒏𝒉 𝒙 𝒔𝒊𝒏𝒉 𝒚 (V) ) 𝒔𝒊𝒏𝒉 𝒙 + 𝒔𝒊𝒏𝒉 𝒚 = 𝟐 𝒔𝒊𝒏𝒉 𝒙+𝒚 𝟐 𝒄𝒐𝒔𝒉 𝒙−𝒚 𝟐 (VI) 𝒔𝒊𝒏𝒉 𝒙 − 𝒔𝒊𝒏𝒉 𝒚 = 𝟐 𝒄𝒐𝒔𝒉 𝒙+𝒚 𝟐 𝒔𝒊𝒏𝒉 𝒙−𝒚 𝟐 (VII) 𝐜𝐨𝒔𝒉 𝒙 + 𝒄𝒐𝒔𝒉 𝒚 = 𝟐𝒄𝒐𝒔𝒉 𝒙+𝒚 𝟐 𝒄𝒐𝒔𝒉 𝒙−𝒚 𝟐 (VIII) 𝐜𝐨𝒔𝒉 𝒙 − 𝒄𝒐𝒔𝒉 𝒚 = 𝟐𝒔𝒊𝒏𝒉 𝒙+𝒚 𝟐 𝒔𝒊𝒏𝒉 𝒙−𝒚 𝟐
- 7. TABLE OF VALUESOF HYPERBOLIC FUNCTIONS x 0 −∞ +∞ 𝒚 = 𝒔𝒊𝒏𝒉 𝒙 0 −∞ ∞ 𝒚 = 𝒄𝒐𝒔𝒉 𝒙 1 ∞ ∞ 𝒚 = 𝒕𝒂𝒏𝒉 𝒙 0 −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 = 𝒔𝒊𝒏𝒉−𝟏𝒛. Similarly we can define inverse hyperbolic cosine and inverse hyperbolic tangent and other functions which we denote as 𝒄𝒐𝒔𝒉−𝟏𝒛, 𝒕𝒂𝒏𝒉−𝟏𝒛, 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. 𝒔𝒊𝒏𝒉−𝟏𝒛 = 𝒍𝒐𝒈 𝒛 + 𝒛𝟐 + 𝟏 2. 𝒄𝒐𝒔𝒉−𝟏 𝒛 = 𝒍𝒐𝒈 𝒛 + 𝒛𝟐 − 𝟏 3. 𝒕𝒂𝒏𝒉−𝟏 𝒛 = 𝟏 𝟐 𝒍𝒐𝒈 𝟏+𝒛 𝟏−𝒛 .
- 9. INVERSE HYPERBOLIC FUNCTIONS INTEGRATIONFORMULAE 𝒅𝒙 𝒙𝟐 + 𝒂𝟐 = 𝒔𝒊𝒏𝒉−𝟏 𝒙 𝒂 𝒅𝒙 𝒙𝟐 − 𝒂𝟐 = 𝒄𝒐𝒔𝒉−𝟏 𝒙 𝒂 𝒅𝒙 𝒂𝟐 − 𝒙𝟐 = 𝟏 𝒂 𝒕𝒂𝒏𝒉−𝟏 𝒙 𝒂
- 10. PROBLEMS BASED ONHYPERBOLIC FUNCTIONS 1.Prove that 𝒄𝒐𝒔𝒉 𝒙 − 𝒔𝒊𝒏𝒉 𝒙 𝒏 = 𝒄𝒐𝒔𝒉 𝒏𝒙 − 𝒔𝒊𝒏𝒉 𝒏𝒙 2.Express cot (x +i y) in a + i b form. 3. Solve the equation 𝒙𝟕 + 𝒙𝟒 + 𝒙𝟑 + 𝟏 = 𝟎 4. Prove that 𝟏 + 𝒄𝒐𝒔 𝒙 + 𝒊 𝒔𝒊𝒏 𝒙 𝒏 = 𝟐𝒏𝒄𝒐𝒔𝒏 𝒙 𝟐 𝒄𝒐𝒔 𝒏𝒙 𝟐 + 𝒊𝒔𝒊𝒏 𝒏𝒙 𝟐 5. Prove that 𝒔𝒊𝒏𝒉−𝟏 𝒙 = 𝒍𝒐𝒈 𝒙 + 𝒙𝟐 + 𝟏 6.Show that 𝒔𝒆𝒄𝒉−𝟏 𝒔𝒊𝒏 𝜽 = 𝒍𝒐𝒈 𝒄𝒐𝒕 𝜽 𝟐 7.Express sec(x+iy) in a+ib form.
- 11. PROBLEMS BASED ONHYPERBOLIC FUNCTIONS 1. Prove that 𝒄𝒐𝒔𝒉 𝒙 − 𝒔𝒊𝒏𝒉 𝒙 𝒏 = 𝒄𝒐𝒔𝒉 𝒏𝒙 − 𝒔𝒊𝒏𝒉 𝒏𝒙 Solution: LHS = 𝒄𝒐𝒔𝒉 𝒙 − 𝒔𝒊𝒏𝒉 𝒙 𝒏 = 𝒆𝒙+𝒆−𝒙 𝟐 − 𝒆𝒙−𝒆−𝒙 𝟐 𝒏 = 𝒆𝒙 + 𝒆−𝒙 − 𝒆𝒙 + 𝒆−𝒙 𝟐 𝒏 = 𝒆−𝒙 𝒏 = 𝒆−𝒏𝒙 RHS = 𝒄𝒐𝒔𝒉 𝒏𝒙 − 𝒔𝒊𝒏𝒉 𝒏𝒙 = 𝒆𝒏𝒙+𝒆−𝒏𝒙 𝟐 − 𝒆𝒏𝒙−𝒆−𝒏𝒙 𝟐 = 𝒆𝒏𝒙+𝒆−𝒏𝒙−𝒆𝒏𝒙+𝒆−𝒏𝒙 𝟐 = 𝒆−𝒏𝒙 ∴ LHS=RHS
- 12. PROBLEMS BASED ONHYPERBOLIC FUNCTIONS 2.Express cot(x +i y) in a + i b form. Solution: 𝒄𝒐𝒕 𝒙 + 𝒊𝒚 = 𝒄𝒐𝒔(𝒙+𝒊𝒚) 𝒔𝒊𝒏(𝒙+𝒊𝒚) = 𝒄𝒐𝒔(𝒙+𝒊𝒚) 𝒔𝒊𝒏(𝒙+𝒊𝒚) × 𝒔𝒊𝒏(𝒙−𝒊𝒚) 𝒔𝒊𝒏(𝒙−𝒊𝒚) = 𝟐𝒄𝒐𝒔(𝒙+𝒊𝒚) 𝟐𝒔𝒊𝒏(𝒙+𝒊𝒚) × 𝒔𝒊𝒏(𝒙−𝒊𝒚) 𝒔𝒊𝒏(𝒙−𝒊𝒚) = 𝒔𝒊𝒏 𝒙+𝒊𝒚+𝒙−𝒊𝒚 −𝒔𝒊𝒏 𝒙+𝒊𝒚−𝒙−𝒊𝒚 𝒄𝒐𝒔 𝒙+𝒊𝒚−(𝒙−𝒊𝒚) −𝒄𝒐𝒔 𝒙+𝒊𝒚+𝒙−𝒊𝒚 𝒔𝒊𝒏 𝑨 + 𝑩 − 𝒔𝒊𝒏 𝑨 − 𝑩 = 𝟐 𝒄𝒐𝒔𝑨 𝒔𝒊𝒏𝑩 𝒄𝒐𝒔 𝑨 − 𝑩 − 𝒄𝒐𝒔 𝑨 + 𝑩 = 𝟐 𝒔𝒊𝒏𝑨 𝒔𝒊𝒏𝑩 = 𝒔𝒊𝒏 𝟐𝒙 −𝒔𝒊𝒏 𝟐𝒊𝒚 𝒄𝒐𝒔 𝟐𝒊𝒚 −𝒄𝒐𝒔 𝟐𝒙 = 𝒔𝒊𝒏 𝟐𝒙 −𝒊𝒔𝒊𝒏𝒉 𝟐𝒚 𝒄𝒐𝒔𝒉 𝟐𝒚 −𝒄𝒐𝒔 𝟐𝒙 = 𝒔𝒊𝒏 𝟐𝒙 𝒄𝒐𝒔𝒉 𝟐𝒚 − 𝒄𝒐𝒔 𝟐𝒙 + 𝒊 −𝒔𝒊𝒏𝒉 𝟐𝒚 𝒄𝒐𝒔𝒉 𝟐𝒚 − 𝒄𝒐𝒔 𝟐𝒙 = 𝒂 + 𝒊𝒃 Where a= 𝒔𝒊𝒏 𝟐𝒙 𝒄𝒐𝒔𝒉 𝟐𝒚 −𝒄𝒐𝒔 𝟐𝒙 & b= −𝒔𝒊𝒏𝒉 𝟐𝒚 𝒄𝒐𝒔𝒉 𝟐𝒚 −𝒄𝒐𝒔 𝟐𝒙
- 13. PROBLEMS BASED ONHYPERBOLIC FUNCTIONS 3. Solve the equation 𝒙𝟕 + 𝒙𝟒 + 𝒙𝟑 + 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 𝟐𝒌𝝅 + 𝜽 & 𝒔𝒊𝒏 𝜽 = 𝒔𝒊𝒏 (𝟐𝒌𝝅 + 𝜽) where k is an integer. We have by DeMoivre’s Theorem 𝒄𝒐𝒔 𝜽 + 𝒊𝒔𝒊𝒏 𝜽 𝟏 𝒏 = 𝒄𝒐𝒔 𝟐𝒌𝝅 + 𝜽 + 𝒔𝒊𝒏 (𝟐𝒌𝝅 + 𝜽) 𝟏 𝒏 𝒄𝒐𝒔 𝜽 + 𝒊𝒔𝒊𝒏 𝜽 𝟏 𝒏 = 𝒄𝒐𝒔 𝟐𝝅𝒌 + 𝜽 𝒏 + 𝒊𝒔𝒊𝒏 𝟐𝝅𝒌 + 𝜽 𝒏 By putting 𝒌 = 𝟎, 𝟏, 𝟐, … … . . 𝒏 − 𝟏 we get n roots of the complex number.
- 14. PROBLEMS BASED ONHYPERBOLIC FUNCTIONS 3. Solve the equation 𝒙𝟕 + 𝒙𝟒 + 𝒙𝟑 + 𝟏 = 𝟎 Solution: 𝒙𝟕 + 𝒙𝟒 + 𝒙𝟑 + 𝟏 = 𝟎 ⟹ 𝒙𝟒 𝒙𝟑 + 𝟏 + 𝟏 𝒙𝟑 + 𝟏 = 𝟎 ⟹ 𝒙𝟑 + 𝟏 𝒙𝟒 + 𝟏 = 𝟎 ⟹ 𝒙𝟑 + 𝟏 = 𝟎, 𝒙𝟒 + 𝟏 = 𝟎 ⟹𝒙𝟑 = −𝟏, 𝒙𝟒 = −𝟏 𝒙𝟑 = −𝟏 ⟹𝒙𝟑 = (𝒄𝒐𝒔 𝝅 + 𝒊𝒔𝒊𝒏 𝝅) ⟹𝒙 = 𝒄𝒐𝒔 𝝅 + 𝒊𝒔𝒊𝒏 𝝅 𝟏 𝟑 ⟹𝒙 = 𝒄𝒐𝒔 𝟐𝒌𝝅 + 𝝅 + 𝒊𝒔𝒊𝒏(𝟐𝒌𝝅 + 𝝅 𝟏 𝟑 ⟹ 𝒙 = 𝒄𝒐𝒔 𝟐𝒌 + 𝟏)𝝅 + 𝒊𝒔𝒊𝒏(𝟐𝒌 + 𝟏)𝝅 𝟏 𝟑 ⟹ 𝒙 = 𝒄𝒐𝒔 (𝟐𝒌+𝟏)𝝅 𝟑 + 𝒊 𝒔𝒊𝒏 (𝟐𝒌+𝟏)𝝅 𝒏 where 𝒌 = 𝟎, 𝟏, 𝟐 when 𝒌 = 𝟎, 𝒙 = 𝒄𝒐𝒔 𝝅 𝟑 + 𝒊𝒔𝒊𝒏 𝝅 𝟑 when k = 1, 𝐱 = 𝒄𝒐𝒔 𝝅 + 𝒊𝒔𝒊𝒏 𝝅 When k = 2, 𝒙 = 𝒄𝒐𝒔 𝟓𝝅 𝟑 + 𝒊𝒔𝒊𝒏 𝟓𝝅 𝟑
- 15. PROBLEMS BASED ONHYPERBOLIC FUNCTIONS 𝒙𝟒 = −𝟏 ⟹𝒙𝟒 = (𝒄𝒐𝒔 𝝅 + 𝒊𝒔𝒊𝒏 𝝅) ⟹𝒙 = 𝒄𝒐𝒔 𝝅 + 𝒊𝒔𝒊𝒏 𝝅 𝟏 𝟒 ⟹𝒙 = 𝒄𝒐𝒔 𝟐𝒌𝝅 + 𝝅 + 𝒊𝒔𝒊𝒏(𝟐𝒌𝝅 + 𝝅 𝟏 𝟒 ⟹ 𝒙 = 𝒄𝒐𝒔 𝟐𝒌 + 𝟏)𝝅 + 𝒊𝒔𝒊𝒏(𝟐𝒌 + 𝟏)𝝅 𝟏 𝟒 ⟹ 𝒙 = 𝒄𝒐𝒔 (𝟐𝒌+𝟏)𝝅 𝟒 + 𝒊 𝒔𝒊𝒏 (𝟐𝒌+𝟏)𝝅 𝟒 where 𝒌 = 𝟎, 𝟏, 𝟐, 𝟑 when 𝒌 = 𝟎, 𝒙 = 𝒄𝒐𝒔 𝝅 𝟒 + 𝒊𝒔𝒊𝒏 𝝅 𝟒 when k = 1, 𝒙 = 𝒄𝒐𝒔 𝟑𝝅 𝟒 + 𝒊𝒔𝒊𝒏 𝟑𝝅 𝟒 When k = 2, 𝒙 = 𝒄𝒐𝒔 𝟓𝝅 𝟒 + 𝒊𝒔𝒊𝒏 𝟓𝝅 𝟒 when k = 3, 𝒙 = 𝒄𝒐𝒔 𝟕𝝅 𝟒 + 𝒊𝒔𝒊𝒏 𝟕𝝅 𝟒 ∴ 𝑇ℎ𝑒 𝑠𝑒𝑣𝑒𝑛 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 are given by 𝒙 = 𝒄𝒐𝒔 𝝅 𝟑 + 𝒊𝒔𝒊𝒏 𝝅 𝟑 𝐱 = 𝒄𝒐𝒔 𝝅 + 𝒊𝒔𝒊𝒏 𝝅 𝒙 = 𝒄𝒐𝒔 𝟓𝝅 𝟑 + 𝒊𝒔𝒊𝒏 𝟓𝝅 𝟑 𝒙 = 𝒄𝒐𝒔 𝝅 𝟒 + 𝒊𝒔𝒊𝒏 𝝅 𝟒 𝒙 = 𝒄𝒐𝒔 𝟑𝝅 𝟒 + 𝒊𝒔𝒊𝒏 𝟑𝝅 𝟒 𝒙 = 𝒄𝒐𝒔 𝟓𝝅 𝟒 + 𝒊𝒔𝒊𝒏 𝟓𝝅 𝟒 𝒙 = 𝒄𝒐𝒔 𝟕𝝅 𝟒 + 𝒊𝒔𝒊𝒏 𝟕𝝅 𝟒
- 16. 4. Prove that𝟏 + 𝒄𝒐𝒔 𝒙 + 𝒊 𝒔𝒊𝒏 𝒙 𝒏 = 𝟐𝒏𝒄𝒐𝒔𝒏 𝒙 𝟐 𝒄𝒐𝒔 𝒏𝒙 𝟐 + 𝒊𝒔𝒊𝒏 𝒏𝒙 𝟐 Solution: we use the following formulas 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 𝒔𝒊𝒏 𝒙 = 𝟐𝒔𝒊𝒏 𝒙 𝟐 𝒄𝒐𝒔 𝒙 𝟐 𝟏 + 𝒄𝒐𝒔 𝒙 = 𝟐𝒄𝒐𝒔𝟐 𝒙 𝟐 LHS= 𝟏 + 𝒄𝒐𝒔 𝒙 + 𝒊 𝒔𝒊𝒏 𝒙 𝒏 = 𝟐𝒄𝒐𝒔𝟐 𝒙 𝟐 + 𝒊𝟐𝒔𝒊𝒏 𝒙 𝟐 𝒄𝒐𝒔 𝒙 𝟐 𝒏 = 𝟐𝒄𝒐𝒔 𝒙 𝟐 𝒄𝒐𝒔 𝒙 𝟐 + 𝒊𝒔𝒊𝒏 𝒙 𝟐 𝒏 = 𝟐𝒄𝒐𝒔 𝒙 𝟐 𝒏 𝒄𝒐𝒔 𝒙 𝟐 + 𝒊𝒔𝒊𝒏 𝒙 𝟐 𝒏 = 𝟐𝒏 𝒄𝒐𝒔𝒏 𝒙 𝟐 𝒄𝒐𝒔 𝒏𝒙 𝟐 + 𝒊𝒔𝒊𝒏 𝒏𝒙 𝟐 (by De’Moivre’s Theorem) = RHS
- 17. 5. Prove that𝒔𝒊𝒏𝒉−𝟏 𝒙 = 𝒍𝒐𝒈 𝒙 + 𝒙𝟐 + 𝟏 Solution: Let 𝒔𝒊𝒏𝒉−𝟏 𝒙 = 𝒚⟹ 𝒔𝒊𝒏𝒉 𝒚 = 𝒙 ∴ 𝒙 = 𝒔𝒊𝒏𝒉𝒚 = 𝒆𝒚 − 𝒆−𝒚 𝟐 = 𝒆𝒚 − 𝟏 𝒆𝒚 𝟐 = 𝒆𝟐𝒚 − 𝟏 𝟐𝒆𝒚 𝒙 = 𝒆𝟐𝒚−𝟏 𝟐𝒆𝒚 ⟹𝟐𝒆𝒚 𝐱 = 𝒆𝟐𝒚 − 𝟏 ⟹ 𝟐𝒆𝒚 𝐱 − 𝒆𝟐𝒚 − 𝟏 = 𝟎 ⟹ 𝒆𝟐𝒚 − 𝟐𝒙𝒆𝒚 +1=0 ⟹ 𝒆𝒚 𝟐 − 𝟐𝒙 𝒆𝒚 + 𝟏 = 𝟎 This is quadratic in 𝒆𝒚 . 𝒆𝒚 = −(−𝟐𝒙) ± (−𝟐𝒙)𝟐 − 𝟒(𝟏)(𝟏) 𝟐(𝟏) = 𝟐𝒙 ± 𝟒𝒙𝟐 − 𝟒 𝟐 = 𝟐𝒙 ± 𝟐 𝒙𝟐 − 𝟏 𝟐 = 𝒙 ± 𝒙𝟐 − 𝟏 Conventionally we take positive sign, 𝒆𝒚= 𝒙 ± 𝒙𝟐 − 𝟏. Taking log on both sides log(𝒆𝒚 )=log 𝐱 ± 𝒙𝟐 − 𝟏⟹ 𝐲 = 𝒍𝒐𝒈 𝒙 + 𝒙𝟐 + 𝟏 ⟹ 𝒔𝒊𝒏𝒉−𝟏 𝒙 = 𝒍𝒐𝒈 𝒙 + 𝒙𝟐 + 𝟏
- 18. 6. Show that𝒔𝒆𝒄𝒉−𝟏 𝒔𝒊𝒏 𝜽 = 𝒍𝒐𝒈 𝒄𝒐𝒕 𝜽 𝟐 ⟹ Solution: Let 𝒔𝒆𝒄𝒉−𝟏 𝒔𝒊𝒏 𝜽 = 𝒙⟹ 𝒔𝒆𝒄𝒉 𝒙 = 𝒔𝒊𝒏𝜽 ∴ 𝒔𝒊𝒏𝜽 = 𝟏 𝒄𝒐𝒔𝒉 𝒙 = 𝟏 𝒆𝒙 + 𝒆−𝒙 𝟐 = 𝟐 𝒆𝒙 + 𝒆−𝒙 = 𝟐 𝒆𝒙 + 𝟏 𝒆𝒙 = 𝟐𝒆𝒙 𝒆𝟐𝒙 + 𝟏 𝒔𝒊𝒏𝜽 = 𝟐𝒆𝒙 𝒆𝟐𝒙 + 𝟏 𝒆𝟐𝒙 sin 𝜽 + 𝒔𝒊𝒏𝜽 = 𝟐𝒆𝒙 ⟹ 𝒆𝒙 𝟐 𝒔𝒊𝒏𝜽 − 𝟐𝒆𝒙 + 𝒔𝒊𝒏𝜽 = 𝟎 This is quadratic in 𝒆𝒙 . 𝒆𝒙 = −(−𝟐) ± (−𝟐)𝟐 − 𝟒(𝒔𝒊𝒏𝜽)(𝒔𝒊𝒏𝜽) 𝟐𝒔𝒊𝒏𝜽 = 𝟐 ± 𝟒 − 𝟒𝒔𝒊𝒏𝟐𝜽 𝟐𝒔𝒊𝒏𝜽 = 𝟐 ± 𝟐 𝟏 − 𝒔𝒊𝒏𝟐𝜽 𝟐𝒔𝒊𝒏𝜽 = 𝟏 ± 𝒄𝒐𝒔𝟐𝜽 𝟐𝒔𝒊𝒏𝜽 Conventionally we take positive sign, 𝒆𝒙= 𝟏 + 𝒄𝒐𝒔 𝜽 𝒔𝒊𝒏𝜽 = 𝟐𝒄𝒐𝒔𝟐 𝜽 𝟐 𝟐𝒔𝒊𝒏 𝜽 𝟐 𝒄𝒐𝒔 𝜽 𝟐 = 𝒄𝒐𝒕 𝜽 𝟐 log(𝒆𝒙 )=log 𝒄𝒐𝒕 𝜽 𝟐 ⟹𝐱 = 𝒍𝒐𝒈 𝒄𝒐𝒕 𝜽 𝟐 ⟹ 𝒔𝒆𝒄𝒉−𝟏 𝒔𝒊𝒏 𝜽 = 𝒍𝒐𝒈 𝒄𝒐𝒕 𝜽 𝟐
- 19. 7.Express sec(x+iy) ina+ib form. Solution: 𝑠𝑒𝑐 𝑥 + 𝑖𝑦 = 𝟏 𝒄𝒐𝒔(𝒙+𝒊𝒚) = 𝟏 𝒄𝒐𝒔(𝒙+𝒊𝒚) × 𝒄𝒐𝒔(𝒙−𝒊𝒚) 𝒄𝒐𝒔(𝒙−𝒊𝒚) = 𝟐𝒄𝒐𝒔(𝒙−𝒊𝒚) 𝟐 𝒄𝒐𝒔(𝒙+𝒊𝒚)𝒄𝒐𝒔(𝒙+𝒊𝒚) = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔 𝒊𝒚 +𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏 (𝒊𝒚) 𝒄𝒐𝒔 𝒙+𝒊𝒚+𝒙−𝒊𝒚 +𝒄𝒐𝒔(𝒙+𝒊𝒚− 𝒙−𝒊𝒚 ) [Since 𝒄𝒐𝒔 𝑨 − 𝑩 = 𝒄𝒐𝒔𝑨 𝒄𝒐𝒔𝑩 + 𝒔𝒊𝒏𝑨 𝒔𝒊𝒏𝑩 & 𝟐𝒄𝒐𝒔𝑨 𝒄𝒐𝒔𝑩 = 𝐜𝐨𝐬 𝑨 + 𝑩 + 𝐜𝐨𝐬(𝑨 − 𝑩)] = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 𝒚+𝒊 𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒚 𝒄𝒐𝒔 𝟐𝒙 +𝒄𝒐𝒔(𝟐𝒊𝒚) = 𝟐𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 𝒚+𝒊 𝟐 𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒚 𝒄𝒐𝒔 𝟐𝒙+𝒄𝒐𝒔𝒉 𝟐𝒚 = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 𝒚 𝒄𝒐𝒔 𝟐𝒙+𝒄𝒐𝒔𝒉 𝟐𝒚 + 𝒊 𝟐 𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒚 𝒄𝒐𝒔 𝟐𝒙+𝒄𝒐𝒔𝒉 𝟐𝒚 = 𝒂 + 𝒊𝒃 𝒔𝒆𝒄 𝒙 + 𝒊𝒚 = 𝟐 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝒉 𝒚 𝒄𝒐𝒔 𝟐𝒙 + 𝒄𝒐𝒔𝒉 𝟐𝒚 + 𝒊 𝟐 𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝒉 𝒚 𝒄𝒐𝒔 𝟐𝒙 + 𝒄𝒐𝒔𝒉 𝟐𝒚