Hyperbolic Functions PDF

60 152 Hyperbolic Functions 6.1 Definition. We know that parametric co-ordinates of any point on the unit circle x 2  y 2  1 is (cos  , sin  ) ;  e   e  e   e  hyperbola x 2  y 2  1 is  , 2 2  E3 so that these functions are called circular functions and co-ordinates of any point on...

60 152 Hyperbolic Functions 6.1 Definition. We know that parametric co-ordinates of any point on the unit circle x 2  y 2  1 is (cos  , sin  ) ;  e   e  e   e  hyperbola x 2  y 2  1 is  , 2 2  E3 so that these functions are called circular functions and co-ordinates of any point on unit   i.e., (cosh  , sinh  ). It means that the relation which   ID exists amongst cos  , sin  and unit circle, that relation also exist amongst cosh  , sinh  and unit hyperbola. Because of this reason these functions are called as Hyperbolic functions. For any (real or complex) variable quantity x, e x  e x [Read as 'hyperbolic sine x'] (2) 2 cosine x'] sinh x e x  e  x  cosh x e x  e  x (5) cosech x  D YG (3) tanh x  1 2  x sinh x e  e  x Note :  sinh 0  0, cosh 0  1, cosh x  U (1) sinh x  e x  e x 2 [Read as 'hyperbolic (4) coth x  cosh x e x  e  x  x sinh x e  e x (6) sec h x  1 2  x cosh x e  e  x tanh 0  0 U 6.2 Domain and Range of Hyperbolic Functions. ST Let x is any real number Function Domain Range sinh x R R cosh x R [1, ) tanh x R (1, 1) coth x R0 R  [1, 1] sec h x R (0, 1] cosech x R0 R0 Hyperbolic Functions 153 6.3 Graph of Real Hyperbolic Functions. (1) sinh x (2) cosh x Y 60 Y (0, 1) X E3 Y Y=1 Y = –1 Y = X O y = – 1 (6) sec h x Y D YG O y 1 U X O (5) cosech x Y (4) coth x ID (3) tanh x X O O (0, 1) X O X U 6.4 Formulae for Hyperbolic Functions. The following formulae can easily be established directly from above definitions ST (1) Reciprocal formulae (i) cosech x  (iv) 1 sinh x tanh x  (ii) sech x  sinh x cosh x (v) 1 cosh x coth x  (iii) coth x  1 tanh x cosh x sinh x (2) Square formulae (i) cosh 2 x  sinh 2 x  1 (ii) sec h 2 x  tanh 2 x  1 (iii) coth 2 x  cosech 2 x  1 (iv) cosh 2 x  sinh 2 x  cosh 2 x (3) Expansion or Sum and difference formulae (i) sinh( x  y )  sinh x cosh y  cosh x sinh y (ii) cosh (x  y )  cosh x cosh y  sinh x sinh y 154 Hyperbolic Functions tanh (x  y)  (iii) tanh x  tanh y 1  tanh x tanh y (xi) (cosh x  sinh x )n  cosh nx  sinh nx E3 (ix) cosh x  sinh x  e x (x) cosh x  sinh x  e  x (5) Trigonometric ratio of multiple of an angle 2 tanh x (i) sinh 2 x  2 sinh x cosh x = 1  tanh 2 x 60 (4) Formulae to transform the product into sum or difference xy x y xy x y (i) sinh x  sinh y  2 sinh (ii) sinh x  sinh y  2 cos h cosh sin h 2 2 2 2 xy xy x y x y (iii) cosh x  cosh y  2 cosh (iv) cosh x  cosh y  2 sinh cosh sinh 2 2 2 2 (v) 2 sinh x cosh y  sinh (x  y )  sinh (x  y ) (vi) 2 cosh x sinh y  sinh (x  y )  sinh (x  y ) (vii) 2 cosh x cosh y  cosh (x  y )  cosh (x  y ) (viii) 2 sinh x sinh y  cosh (x  y )  cosh (x  y ) 1  tanh 2 x 1  tanh 2 x ID (ii) cosh 2 x  cosh 2 x  sinh 2 x = 2 cosh 2 x  1 = 1  2 sinh 2 x = 2 tanh x 1  tanh 2 x 3 tanh x  tanh 3 x (vi) sinh 3 x  3 sinh x  4 sinh 3 x (vii) cosh 3 x  4 cosh 3 x  3 cosh x (viii) tanh 3 x  1  3 tanh 2 x x x (6) (i) cosh x  sinh x  e (ii) cosh x  sinh x  e (iii) (iv) 2 sinh 2 x  cosh 2 x  1 D YG (cosh x  sinh x )n  cosh nx  sinh nx Example: 1 sinh x  sinh y is equal to cosh x  cosh y x y (b) tanh    2  (a) 2 coth (x  y) Example: 2 xy (c) coth    2  x y (d) coth    2  x y x y 2 cosh sinh sinh x  sinh y 2 2  coth  x  y .  x y x y cosh x  cosh y  2  2 sinh sinh 2 2 U Solution: (c) (v) tanh 2 x  U (iii) 2 cosh 2 x  cosh 2 x  1 If tanh 2 x  tan 2  , then cosh 2 x is equal to (a)  sin 2 (b) sec 2 [EAMCET 1998] (c) cos 3 (d) cos 2 ST 1  tanh x 1  tan  1 1  =   sec 2. 2 2 2 cos 2 1  tan  1  tanh x 1  tan  1  tan 2  2 2 Solution: (b) cosh 2 x  Example: 3  x  If u  log tan    , then cosh u is equal to 4 2 [EAMCET 1991; Rajasthan PET 1999] (a) sec x Solution: (a) (b) cosec x x 1  tan eu  x  2 u  log tan       x 1 4 2 1  tan 2 (c) tan x (d) sin x Hyperbolic Functions 155 2 cosh 2  sinh 2  If cosh   sec x, then tan 2  x  2 1  cos x  1 x 2 x 1  tan 2 2 1  tan 2 x 2 x 1  tan 2 2 1  tan 2 (d) e 2 (c) cot 2 2 D YG cosh   1 2 [Rajasthan PET 1998] (b) sin 2 2 1 e2 ID Example: 5 cosh   sec x  (c) e 2  e 2 e 2  e 2   e 2. 2 2 cosh 2  sinh 2  Solution: (d) 2  E3 (b) e Solution: (d) (a) cos 2 1  sec x. cos x  [EAMCET 2000] 1 e (a)   2 (d) tanh 2  2 U Example: 4 e u  e u 2 60 cosh u  x   1  tan  2  1  x  x    2 1  tan 2  1  tan  1  tan 2 1 e 2u  1  2 2   2 = =    x  x x   2  2e u 1  tan 2 2 1  tan   1  tan  1  tan  1  tan  2 2 2 2     2.  x   1  tan 2 1  tan   2  cosh   1   cosh   1 x 2  tan 2 x  2 2 2 tan 2 2 sinh 2  2  tan 2 x  tanh 2   tan 2 x. 2 2 2 2 cosh 2 2  6.5 Transformation of a Hyperbolic Functions. U Since, cosh 2 x  sinh 2 x  1 ST  sinh x  cosh 2 x  1  sinh x  tanh x 1  tanh 2 x Also, sinh x   sinh x   sinh x  1  sech 2 x sec h x 1 coth 2 x  1 1 cosech x In a similar manner we can express cosh x , tanh x , coth x ,................ in terms of other hyperbolic functions. 6.6 Expansion of Hyperbolic Functions. (1) sinh x  e x  e x x3 x5 x7 x   .... 2 3! 5! 7! (2) cosh x  e x  e x x2 x4 x6 1    .... 2 2! 4 ! 6! (3) tanh x  e x  e x x3 17 7  x   2x5  x ..... x x 3 315 e e 60 156 Hyperbolic Functions E3 The expansion of coth x , cosech x does not exist because coth (0)  , cosech (0)  . 6.7 Relation between Hyperbolic and Circular Functions. We have from Euler formulae, Adding (i) and (ii)  cos x  and e ix  cos x  i sin x........(ii) e ix  e  ix 2 Subtracting (ii) from (i)  sin x  e ix  e ix 2i ID........(i) U e ix  cos x  i sin x D YG Replacing x by ix in these values, we get cos(ix )  e x  e x  cosh x 2  cos(ix )  cosh x sin( ix)   e x  e x e x  e x  i  2i 2       sin( ix )  i sinh x sin( ix ) i sinh x  cos(ix ) cosh x U Also tan( ix)  tan( ix)  i tanh x replacing x by ix in the definitions of sinh x and cosh x , we get ST Similarly cosh (ix )  e ix  e ix  cos x 2 Also, tanh (ix)  sinh (ix) i sin x   i tan x cosh (ix) cos x Thus, we obtain the following relations between hyperbolic and trigonometrical functions. (1) sin( ix)  i sinh x (2) cos(ix )  cosh x sinh (ix )  i sin x cosh( ix)  cos x sinh x  i sin( ix) cosh x  cos (ix) Hyperbolic Functions 157 sin x  i sin h (ix) (3) tan( ix)  i tanh x (4) cot(ix)  i coth x tanh( ix)  i tan x coth( ix)  i cot x tanh x  i tan( ix) coth x  i cot(ix) tan x  i tanh (ix) cot x  i coth( ix) (6) cosec (ix)  i cosech x (5) sec(ix)  sech x cosech (ix )  i cosec x sec hx  sec(ix) cosech x  i cosec (ix) sec x  sech(ix) co sec x  i cosech (ix) E3 sec h(ix)  sec x 60 cos x  cos h (ix) Important Tips sin x  i sinh x cos x  cosh x sin 2 x   sinh 2 x cos 2 x  cosh 2 x For example, tan x  i tanh x tan 2 x   tanh 2 x in terms of tanh x , replace tan x by i tanh x and tan 2 x by tan 2 x by D YG For finding out the formula for cosh 2 x ID For obtaining any formula given in (5)th article, use the following substitutions in the corresponding formula for trigonometric functions. U   tanh 2 x in the following formula of trigonometric function of cos 2 x : cos 2 x  1  tan 2 x 1  tan x 2 we get, cosh 2 x  1  tanh 2 x 1  tanh 2 x 6.8 Period of Hyperbolic Functions. If for any function f (x ), f (x  T )  f (x ), then f (x ) is called the Periodic function and least positive U value of T is called the Period of the function.  sinh x  sinh( 2i  x ) ST cosh x  cosh( 2i  x ) and tanh x  tanh( i  x ) Therefore the period of these functions are respectively 2i, 2i and i. Also period of cosech x, sech x and coth x are respectively 2i , 2i and i. Note :  Remember that if the period of T  f (x ) is T, then period of f (nx ) will be  . n  Hyperbolic function are neither periodic functions nor their curves are periodic but they show the algebraic properties of periodic functions and having imaginary period. 158 Hyperbolic Functions Example: 6 If cos( x  iy)  A  i B , then A equals (a) cos x cosh y Solution: (a) [Rajasthan PET 1994] (b) sin x sinh y (c)  sin x sinh y (d) cos x sinh y cos( x  iy)  A  iB  cos x cos(iy)  sin x sin(iy)  A  iB  cos x cosh y  i sin x sinh y  A  iB If cos(u  iv)  x  iy, then x 2  y 2  1 is equal to (b) sin 2 u  cosh 2 v (a) cos 2 u  sinh 2 v Solution: (c) [Rajasthan PET 1999] (c) cos 2 u  cosh 2 v cos(u  iv)  x  iy  cos u cos(iv)  sin u sin(iv)  x  iy  cos u cosh v  i sin u sinh v  x  iy (d) sin 2 u  sinh 2 v E3 Example: 7 60  A  cos x cosh y  x  cos u cosh v y   sin u sinh v x 2  y 2  cos 2 u. cosh 2 v  sin 2 u. sinh 2 v ID = (1  sin 2 u) cosh 2 v  sin 2 v[cosh 2 v  1] = cosh 2 v  sin 2 v  x 2  y 2  1  cosh 2 v  1  sin 2 u = cosh 2 v  cos 2 u. Example: 8 The value of sec h(i ) is Example: 9 Solution: (d) = cosh x  i.i sinh x = cosh x  sinh x  Example: 10 (c) 0  sin 6 e x  e x  e x  e x = e x. 2  i  is equal to  i i (b) 2 2  1 i    sinh  i   i. sin  i. . 6 6 2 2     sec h(i)  co sech  i  equals 2  (a)  U Solution: (b) Example: 11 (b) 1  i    sec h(i)  co sech  i  = sec   icosec  1  i. 2 2  (a) 1  i ST Solution: (c) Example: 12 The period of cosh (a) 6i Solution: (a) Example: 13  3 (c) i 3 2 (c) 1  i (d)  i (b) 2i (b) 2 (c) 4i x Since period of sinh x is 2i , therefore period of sinh   will be 4i. 2 6.9 Inverse Hyperbolic Functions. 3 2 (d) 1  i is (c) i  Since the period of cosh  is 2i, so the period of cosh is 3.2i  6i. 3 x The period of sinh   is 2 (a) 2i Solution: (c) (d) 1 (d) e  x (c) e x D YG Solution: (a) (b) i 2 sec h(i )  i = sec   1. e  e i cos ix  i sin ix equals (a) e ix (b) e  ix cos ix  i sin ix U (a) – 1 [Rajasthan PET 1999] (d) 9i (d) 4 Hyperbolic Functions 159 If sinh y  x , then y is called the inverse hyperbolic sine of x and it is written as y  sinh 1 x. Similarly cosech 1 x , cosh 1 x , tanh 1 x etc. can be defined. (1) Domain and range of Inverse hyperbolic function Domain Range sinh 1 x R cosh 1 x [1, ) tanh 1 x (1, 1) coth 1 x R – [–1, 1] R0 sech 1 x (0, 1] R cosech 1 x R0 60 Function R R E3 R ID R0 Method : Let sinh 1 x  y U (2) Relation between inverse hyperbolic function and inverse circular function  x  sinh y =  i sin(iy)  ix  sin(iy)  iy  sin 1 (ix) D YG  y  i sin 1 (ix)  sinh 1 x  i sin 1(ix) Therefore we get the following relations (ii) cosh 1 x  i cos 1 x (i) sinh 1 x  i sin 1(ix) (iv) sec h 1 x  i sec 1 x (iii) tanh 1 x  i tan 1 (ix) (v) cosech –1 x  i cosec 1 (ix) (3) To express any one inverse hyperbolic function in terms of the other inverse hyperbolic functions U To express sinh 1 x in terms of the others ST (i) Let sinh 1 x  y  x  sinh y  cosech y  1 1  y  cosec 1   x x  cosh y  1  sinh 2 y  1  x 2 (ii)  y  cosh 1 1  x 2  sinh 1 x  cosh 1 1  x 2 (iii)  tanh y   y  tanh 1 (iv) sinh y sinh y =  cosh y 1  sinh 2 y x 1  x2  coth y   sinh 1 x  tanh 1 1  sinh 2 y 1 x2  sinh y x x 1  x2 x 1  x2 160 Hyperbolic Functions 1 x2 1 x2  sinh 1 x  coth 1 x x 1 1 1 (v)  sec hy    2 cosh y 1  sinh y 1 x2  y  coth 1 1 1 x 2  sinh 1 x  sec h 1 1 60 y  sec h 1 1 x2 1 Also, sinh 1 x  cosech 1   x From the above, it is clear that E3 (vi) 1 coth 1 x  tanh 1   x ID 1 sec h 1 x  cosh 1   x 1 cosech 1  sinh 1   x Note :  If x U is real then all the above six inverse functions are single valued. (4) Relation between inverse hyperbolic functions and logarithmic functions D YG Method : Let sinh 1 x  y  x  sinh y  e y  e y 2x  4 x 2  4  x  x2 1  e 2 y  2 xe y  1  0  e y  2 2 But e y  0, y and x  x 2  1  ey  x  x 2  1  y  log( x  x 2  1 ) U  sinh 1 x  log( x  x 2  1) By the above method we can obtain the following relations between inverse hyperbolic functions and principal values of logarithmic functions. ST (i) sinh 1 x  log( x  x 2  1) (iii) tanh 1 x  (  x  ) 1 1  x  log   2 1  x  1  1  x2 (v) sec h 1 x  log   x      Note :  Formulae for values of by | x| 1 0  x 1 (ii) cosh 1 x  log( x  x 2  1) (iv) coth 1 x  (x  1) 1  x 1 log   2  x 1  1  1  x2 (vi) cosech 1 x  log   x      | x|1 ( x  0) cosech –1 x , sec h –1 x and coth 1 x may be obtained by replacing x 1 in the values of sinh 1 x , cosh 1 x and tanh 1 x respectively. x Hyperbolic Functions 161 6.10 Separation of Inverse Trigonometric and Inverse Hyperbolic Functions. If sin(  i ) = x  iy then (  i ) , is called the inverse sine of (x  iy). We can write it as, sin 1 (x  iy)    i  (2) sin 1 (x  iy)   2  2  cos 1 (x  iy) i 1 cos 1 (x 2  y 2 )  (1  x 2  y 2 )2  4 x 2 y 2   cosh 1 (x 2  y 2 )  (1  x 2  y 2 )2  4 x 2 y 2       2  2 (3) tan 1 (x  iy)   1 2x tan 1  2 2 2 1  x  y  i  2y   tanh 1  2 2  2 1  x  y  1  2x  = tan 1  2 2  2 1  x  y ID = 1 i cos 1 (x 2  y 2 )  (1  x 2  y 2 )2  4 x 2 y 2   cosh 1 (x 2  y 2 )  (1  x 2  y 2 )2  4 x 2 y 2    2   2 E3 (1) cos 1 (x  iy)  60 Here the following results for inverse functions may be easily established.  x 2  (1  y )2   i   log  2 2  4  x  (1  y )  (4) sin 1 (cos   i sin  )  cos 1 ( sin  )  i sinh 1 ( sin  ) or cos 1 ( sin  )  i log( sin   1  sin  )  4  i  1  sin   log   , (cos  )  0 4  1  sin   D YG (6) tan 1 (cos   i sin  )  U (5) cos 1 (cos   i sin  )  sin 1 ( sin  )  i sinh 1 ( sin  ) or sin 1 ( sin  )  i log( sin   1  sin  )   1  1  sin   and tan 1 (cos   i sin  )      log   , (cos  )  0  4 4  1  sin   Since each inverse hyperbolic function can be expressed in terms of logarithmic function, therefore for separation into real and imaginary parts of inverse hyperbolic function of complex quantities use the appropriate method. U Note :  Both inverse circular and inverse hyperbolic functions are many valued. Example: 14 If x  log( y  y 2  1 ), then y  ST (a) tanh x Solution: (c) x  sinh 1 y  y  sinh x. Example: 15 If cosh 1 x  log( 2  3 ), then x  (a) 2 Solution: (a) cosh 1 [EAMCET 1995] (b) coth x (c) sinh x (d) cosh x [EAMCET 2000] (b) 1 (c) 3 (d) 5 x  log( x  x  1 ) = log( 2  3 ) 2  x 2. Example: 16 log( 3  2 2 )  (a) sinh Solution: (b) 1 3 [Rajasthan PET 1990] (b) cosh 1 3 (c) tanh 1 3 (d) cosh 1 3 log( 3  2 2 )  log( 3  8 ) = log( 3  9  1 )  log( 3  3 2  1 ) log( 3  2 2 )  cosh 1 3. Example: 17 1 sec h -1   is 2 [Rajasthan PET 1998] 162 Hyperbolic Functions (c) log( 2  3 ) (b) log( 3  1) (a) log( 3  2 ) Solution: (c) 1 sec h 1    cosh 1 (2) = log( 2  2 2  2 ) = log( 2  3 ). 2 Example: 18 sinh 1 (2 3 / 2 ) is [EAMCET 2003] sinh 1 (2 3 / 2 )  log( 2 3 / 2  (2 3 / 2 )2  1 ) = log( 3  8 ). Example: 19 If tan 1 (  i )  x  iy, then x   1 2 tan 1  2 2 2  1    Solution: (a)      1 2 tan 1  2 2 2 1    (b)     tan 1 (  i )  x  iy tan 1 (  i )  x  iy 2 x  x  iy  x  iy = tan 1 (  i )  tan 1 (  i ) If   2  1 2   i    i 1 1 2 tan 1 = tan 1 = tan 1  2 2 2 2 2 2 2 1  (  i )(  i ) 1     1    x  2     (d) None of these  .   , then the value of log sec x is x   (a) 2 coth 1  cosec 2  1  2   Let log sec x  y ;  x   (b) 2 coth 1  cosec 2  1  2   1 ey / 2  y / 2 cos x e D YG Solution: (a) ID Example: 20  2 (c) tan 1  2 2  1    x x     (c) 2 cosech -1  cot 2  1  (d) 2 cosech -1  cot 2  1  2 2     U  x 60 Solution: (b) (d) log( 8  27 ) (c) log( 3  8 ) E3 (b) log( 3  8 ) (a) log( 2  18 ) (a) (d) None of these By componendo and Dividendo rule, 1  cos x e y / 2  e  y / 2 x y  cot 2    coth    1  cos x e y / 2  e  y / 2 2 2 x    y  2 coth 1  cosec 2  1 . 2   Example: 21 The value of cosh 1 (sec x ) is  1  sin x  (a) log    cos x   1  sin x  (b) log    cos x   1  cos x  (c) log    sin x   1  sin x  Here cosh 1 (sec x )  log(sec x  sec 2 x  1 ) = log(sec x  tan x ) = log  .  cos x  Example: 22 2 sinh 1 ( ) is equal to ST U Solution: (a)  1  cos x  (d) log    sin x  (a) sinh 1 (2 1   2 ) Solution: (a) (b) sinh 1 (2 1   2 ) (c) sinh 1 ( 1   2 ) (d) None of these We know that, 2 sin 1 x  sin 1 (2 x 1  x 2 ) Putting the value of x  i 2 sin 1 (i )  sin 1 (2i 1  i 2 2 ) 2i sinh 1 ( )  sin 1 (2i 1   2 ) or 2i sinh 1 ( )  i sinh 1 (2 1   2 )  2 sinh 1 ( )  sinh 1 (2 1   2 ). Example: 23  x The value of sinh 1   1  x2    is   (  sin 1 (ix )  i sinh 1 x ) Hyperbolic Functions 163 (b) coth 1 x (a) tanh 1 x Let x  tanh y , then  x  sinh 1   2  1 x Example: 24 x 1 x  2 (d) cosh 1 (2 x ) tanh y  sinh y sec hy    sinh 1 (sinh y )  y  tanh 1 (x ).   60 Solution: (a) (c) sinh 1 (2 x ) If cos  cosh   1, then  is equal to   (a) log sec   2   (d) log sin   2 (c) log(sec   tan  ) (b) log tan  cos . cosh   1  cosh   sec     cosh 1 (sec  ) = log(sec   sec 2   1 ) = log(sec   tan  ) Example: 25 sinh 1 (sinh 1  ) is equal to (a) i 1 1 1 (d)   i (c) i (b)  1 E3 Solution: (c) 1 Solution: (b) sinh (sinh  ) =  i sin(i sinh  ) =  i sin[i(i sin (i )]   sin[sin (i )] =  i.i  i   . Example: 26 If sinh 1 x  cosech –1 y ,then the correct statement is (b) xy  1 ID (a) x  y 2 (c) xy  1 (d) x  y  0   1  1 1 Given that, sinh 1 x  cosech –1y or sinh 1 x  sinh 1   or x  sinh sinh 1   or x   xy  1. y y   y       Example: 27 Find real part of tan 1 (1  i) 1 tan 1 (2) 2 (b) 1 tan 1 (2) 2 D YG (a)  U Solution: (c) 1 1 tan 1   2 2 (d) 0 1 2(1) 1 tan 1   tan 1 (2). 2 1 1 1 2 Solution: (a) Real part = Example: 28 Find real part of cosh 1 (1) (a) – 1 (b) 1 Solution: (c) (c)  (c) 0 (d) None of these (c) 0 (d) None of these  3 1   (c) log   2    (d) None of these We know that cosh 1 x  log  x  x 2  1    U  cosh 1 (1)  log  1  1 2  1  = log 1  0.    5 7  9i   Find imaginary part of sin 1   16    (a) log 2 (b)  log 2 ST Example: 29 Solution: (b)   5 7 9 i       log  9  1  9  =  log( 2). sin 1     16 16 16 16      Example: 30  3 i   Find real part of cos 1   2 2   (a) Solution: (b)  3 (b)  4  Expression cos 1 (cos   i sin  )  sin 1 sin   i log( sin   1  sin  )  Where   6 164 Hyperbolic Functions  3 i    sin 1  cos 1   2 2   Real part =  3 1   . , Imaginary part = log   2  4   Find imaginary part of sin 1 (cosec  )   (a) log  cot  2  Solution: (a) (b)  2 (c) 1   log  cot  2 2  Let sin 1 (cosec  )  x  iy  cosec   sin( x  iy) = sin x cosh y  i cos x sinh y  sin 1 (cosec  ) = Real part =  2    i log  cot  2     , Imaginary part = log  cot . 2 2  U 164.........(ii) ID    y = log[ cosec   cot  ] = log  cot  2  (d) None of these E3 By comparing we get, sin x cosh y  cosec ......(i) and cos x sinh y  0  From (ii), cos x  0  x  2   from (i) sin. cosh y  cosec  or y  cosh 1 (cosec  ) = log[ cosec  ] 2 60 Example: 31  1 1 3     1 1   =  i log  3  1   i log   1   =  i log        2 2 4  2  4  2   2 D YG *** Basic Level The value of cosh 2 x is ST 1. U Formulae and Transformation of Hyperbolic Functions [Rajasthan PET 1985, 86, 88, 90, 2002] (a) cosh 2 x  sinh 2 x 2. tan( x  y ) equals 92] (d) None of these [Rajasthan PET 1990, 92] (b) 4 sinh 3 z  3 sinh z (c) 3 sinh z  4 sinh 3 z Which of the following statement is true (a) sinh 2 x  cosh 2 x  1 4. (c) 1  2 sinh 2 x sinh 3 z equals (a) 3 sinh z  4 sinh 3 z 3. (b) 1  2 cosh 2 x (b) sinh 2 x  cosh 2 x  1 (d) None of these [Rajasthan PET 1988, 94] (c) sech 2 x  tanh 2 x  1 (d) coth 2 x  cosech 2 x  1 [Rajasthan PET 1990, 91, Hyperbolic Functions 165 (a) (d) sechx (b) coth 2 (c) tanh  (d) tanh 2 e 2  1 is e 2  1 U D YG (b) e 2 x (c) 1 (d) – 1 (c) i (d) – 1 If cosh z  sec  , then sinh z equals (b) cot  (c) tan  (d) tan  2 If cosec   coth x , then the value of tan  is (c) tanh x (d) cosech x x (c) sin 2   2 (d) cos 2 U (b) sinh x ST y If cosh y  sec x , then the value of tanh 2   is 2 x (a) tan 2   2 (b) cot 2 x 2 u  x  u  log tan    ,then the value of tanh is 4 2 2   (a) cot 15. 1 (cosh 2 x  1) 2 (d) e  in 1  tanh x is equal to 1  tanh x (a) cosh x 14. (d) (c) e in (b) e 5 (a) cosec  13. 1 (cosh 2 x  1) 2 (b) e  n   1  tanh     is equal to  1  tanh   (a) e 2 x 12. tanh x  tanh y 1  tanh x tanh y E3 The value of (a) e 10 11. (c) (c) sin x 5 10. (d) 60 (b) cosh 2 x  1 The value of (cosh   sinh  )n is (a) coth  9. tanh x  tanh y 1  tanh x tanh y [Rajasthan PET 1991] (b) cosech x (a) e n  8. (c) Which of the following functions is not defined at x  0 (a) tanh x 7. tanh x  tanh y 1  tanh x tanh y sinh 2 x equals (a) cosh 2 x  1 6. (b) ID 5. tanh x  tanh y 1  tanh x tanh y x 2 (b)  cot x 2 x 2 [Rajasthan PET 1997] (c)  tan x 2 (d) tan x 2 x x If tan   coth    1, then the value of cos x cosh x is 2 2 (a) 1 (b) – 1 (c) cos 2 x (d) sinh 2 x 166 Hyperbolic Functions co sech x cosech 2 x  1 equals (a) tanh x (b) coth x (c) sech x Advance Level If f (x )  cosh x  sinh x and f ( p)  f (x ). f (y ) , then the value of p is (a) f (x1 ). f (x 2 )........ f (x n ) 19. (c) 0 If sin x cosh y  cos  and cos x sinh y  sin  , then sinh 2 y equals (a) sin 2 x If tan   tanh x cot y and tan   tanh x tan y, then (a) cosh 2 x  cos 2y cosh 2 x  cos 2y (c) cos 2 x  cosh 2y cos 2 x  cosh 2y (d) 1 (c) cos 2 x (b) cosh 2 x (d) 1 sin 2 equals sin 2 (b) D YG 20. (b) f (x1 )  f (x 2 ) ......  f (x n ) (d) None of these E3 If f (x )  cosh x  sinh x , then f (x1  x 2 ......  x n ) is equal to ID 18. (c) x  y (b) x  y (a) xy U 17. (d) cosh x 60 16. cosh 2 x  cos 2y cosh 2 x  cos 2y (d) None of these Relation between Hyperbolic and Circular Functions Basic Level  i  The value of cosech   is 6 U 21. ST (a) – 2 22. (d) 2 i (b) icosh x (c) cos x (d)  cos x (c) (d) None of these  i  i tanh    coth   is equal to 4    4  (a) 0 24. (c) – 2 i i   sinh  x   equals 2  (a) icosh x 23. (b) 2 (b) 2 2 cos(i5 x ) equals (a) icosh x [Rajasthan PET 1989] (b) icosh x (c) cosh x (d)  cosh x Hyperbolic Functions 167 If sin( x  iy)  A  iB, then A equals (a) sinh x cos y (b) (c) cos  cosh  e x  e x 2 (c) (c) 2 cosh 2 x 4 i n D YG (b) e x  e x 2i (d) sin  sinh  (d) e x  e x 2i (d) cosh 2 x (c) n i 4 (d)  i (b)  (c) 2i (b)  i (c) (d)  i The period of cosh( 4 x ) is U i 2 (d) 2 Advance Level ST If tan(  i ) = sin( x  iy ) , then the value of coth y sinh 2 is (a) tan x. cot 2 35. (d) 2 cos  cosh  The period of e z is (a) 2 i 34. (c) cosh  cos  (b) cos  cos   nx  The period of coth   is  4  n 1 sin 2 x sinh 2 y 2 [Rajasthan PET 1995] (b) – 1 i (d) E3 e x  e x 2 (a) 2 33. (b) 2 sinh  sin  sin 2 (ix )  cosh 2 x is equal to (a) 32. 1 sinh 2 x sin 2 y 2 [Rajasthan PET 2000] The value of sinh( x  2ni) is (a) 1 31. (c) Real part of cosh(  i ) is (a) 30. 1 cos 2 x cosh 2 y 2 Imaginary part of cosh(  i )  cosh(  i )  (a) cosh  cos  29. (b) 60 1 cosh 2 x cos 2 y 2 (a) 2 sinh  sinh  28. (d) cosh x sin y The imaginary part of sin 2 (x  iy) is (a) 27. (c) cos x sinh y ID 26. (b) sin x cosh y U 25. (b) tan x sin 2 (c) cot x sin 2 (d) None of these (c)  cot  coth  (d)  tan  tanh  If cos(  i )   (cos   i sin ) ,then value of tan  is (a) tanh  tan  (b)  tanh  tan   Inverse Hyperbolic Functions Basic Level 36. sinh 1 x  2000] [Rajasthan PET 1987, 93, 96, 168 Hyperbolic Functions 37. (c) log( x  x 2  1) (b) log( x  x 2  1 ) (a) log( x  1  x 2 ) (d) None of these cosh 1 x  [Rajasthan PET 1988, 90, 92, 2002] (a) log( x  x 2  1) tanh 1 x  60 38. (d) log( x  x 2  1 ) (c) log( x  x 2  1 ) (b) log( x  x 2  1 ) [Rajasthan 92, 99] (b) (b) 1  x 1 log   2  x 1      1  1  x2 (b) log   x  (b) 1  Z 1  log   2  Z 1 (b) log tan x 2 U x 2 (c) 1  Z 1 log   2  Z 1  (c) log cot x If cosech –1 (1)  x  iy, then the value of y is     [Rajasthan PET 1991] 1  1  x2 (d) log   x      Z2 (d)  log   Z 2 (b) 0 (d) None of these [Rajasthan PET 1986] (c) log(1  2 ) (d) – 1 The value of tanh 1 (2 1 ) is (a) log 2 (b) log 2 1 (c) log 3 (d) None of these If log( 2  3 )  cosh 1 K then K equals (a) 1 47. 1  1  x2 (c) log   x  (d) None of these sec h 1 (sin x ) equals ST 46.     1  x 1  log   2  x 1 Z The value of 2 coth 1   is 2 (a) 1 45. (c) cosech -1 x equals (a) log cot 44. (d) None of these ID 1 1  x  log   2 1  x  Z2 (a) log   Z2 43. 1 1  x  log   2 1  x  [Rajasthan PET 1990] D YG 42. (d) (c) log(1  2 ) (b) log( 2  1) 1  1  x2 (a) log   x  1 1  x  log   2 1  x  [Rajasthan PET 1989] coth 1 x equals (a) 41. (c) The value of sinh 1 (1) is (a) 0 40. 1  x 1  log   2  x 1 U 39. 1  x 1 log   2  x 1  E3 (a) PET 1988, 91, (b) 0 (c) 2 (d) None of these (b)  tanh 1 x (c) tanh 1 (ix ) (d) None of these  i tan 1 (ix ) equals (a) tanh 1 x Hyperbolic Functions 169  1   is equal to 2 tan 1    2 If tanh x  (a) 3 , then the value of x is 4 (b)  7 7 1 1 tanh 1    tanh 1   is equal to 2 3 1 sinh 1   is equal to 2  1   (b) tanh 1    5 (a) tanh 1 ( 5 ) 52.    log tan    is equal to 4 2 If sin 1 ( A  iB)  x  iy, then tan x tanh y (c) tanh x tanh y   (b) tanh 1  tanh  2  ST 54. 1 If x  log   y  5 (d) tanh 1   6  2   (d) tanh 1    5   (d) 2 tan 1  tanh  2  [Rajasthan PET 1987] (b) tanh x tan y (d) cos x cosh y Advance Level (b) 2ri  log( x  x 2  1 ) (c) ri  (1)r log( x  x 2  1 ) (d) 2ri  (1)r log( x  x 2  1 )  1  1  , then y is equal to  y2  (a) tanh x 56. (d)  log 7 The general value of cosh 1 x is (a) 2ri  log( x  x 2  1 ) 55.   (c) 2 tanh 1  tan  2  A equals B U (a) (c) tanh 1 ( 3 ) D YG   (a) tanh 1  tan  2  53. 1 (c) tanh 1   6 7  (b) tanh 1   5 5 (a) tanh 1   7  51. (c) log 7  2 1  (d) cosh 1   2 1    E3 50. (c) cosh 1 (3) ID 49. (b) cosh 1 ( 3 ) 60 1 (a) cosh 1   3 U 48. (b) cosh x (c) sinh x (d) cosech x The imaginary part of tan 1 (cos   i sin  ) is (a) tanh 1 (sin  ) (b) tanh 1 () (c) 1 tanh 1 (sin  ) 2 (d) None of these 170 Hyperbolic Functions If cosh 1 (p  iq)  u  iv, then the equation with roots cos 2 u and cosh 2 v (a) x 2  x (p 2  q 2 )  p 2  0 58. (b) x 2  x (p 2  q 2  1)  1  0 (c) x 2  x (p 2  q 2  1)  1  0 (d) x 2  x (p 2  q 2  1)  p 2  0   ix  The value of log tan    is 4 2 (a) i tan 1 (sinh x ) (c) i tan 1 (cosh x ) (b)  i tan 1 (sinh x ) ST U 170 D YG U ID E3 *** (d) None of these 60 57. Hyperbolic Functions Assignment (Basic & Advance Level) Hyperbolic Functions 171 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 c c d b c b a c a a d b a d a c c a c a 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 c a b c b d b a b a b c c c d b d d b b 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 a d a b c c a c c a b c a b d c d a ST U D YG U ID E3 60 1