1. Cosh 2 x  Cosh2 x  1 Sinh 2 x  Cosh2 x  1
2-4- Hyperbolic functions : Hyperbolic functions are used to describe the motions of waves in elastic solids ; the shapes of electric power lines
; temperature distributions in metal fins that cool pipes …etc.
The hyperbolic sine (Sinh) and hyperbolic cosine (Cosh) are defined by the following equations :
1. Sinhu  1 ( eu  eu ) and
Coshu  1 ( eu  eu ) 2 2
Sinhu
eu  e u
2. tanhu  
Coshu eu  e u
and Cothu  
Sinhu eu  eu
Coshu
eu  eu
1 2
1 2
3. Sechu  
and Cschu  
Coshu
eu  e u
Sinhu
eu  e u
4. Cosh2u  Sinh2u  1
5. tanh2 u  Sech2u  1 and
Coth2u  Csch2u  1
Coshu  Sinhu  e u and
Cosh( u )  Coshu and
Cosh0  1 and Sinh0  0
Sinh( x  y )  Sinhx .Coshy  Coshx.Sinhy
Cosh( x  y )  Coshx .Coshy  Sinhx.Sinhy
Sinh2 x  2.Sinhx .Coshx
Cosh2 x  Cosh 2 x  Sinh 2 x
Coshu  Sinhu  e  u
Sinh( u )   Sinhu
Cosh 2 x  Cosh2 x  1 Sinh 2 x  Cosh2 x  1
and
2 2
y=Sinhx
y=Cschx
y=Cschx ١٨

2. y  Sinhx Dx : x and Ry : y y  Coshx Ry : y  1 y  tanh x
y=Sechx
y=Cothx 1
y=tanhx -1
y=Cothx
y  Sinhx
Dx : x
and
Ry : y
y  Coshx
Ry : y  1
y  tanh x
Ry : 1  y  1
y  Cothx
Dx : x  0
Ry : y  1 or y  1
y  Sechx
Ry : 0  y  1
y  Cschx
Ry : y  0
EX-16- Let tanh u = - 7 / 25 , determine the values of the remaining five hyperbolic functions .
Sol.- ١٩

3.  49 eln x  e  ln x e ln x  e ln x e ln x ln x x 2 1  x 2  1
1   25
Cothu 
tanh u 7
 49
tanh 2 u  Sech2 u  1
Sech2 u  1  Sechu  24
625 25
1  25
Sechu 24
Coshu 
tanh u  Sinhu
  7  Sinhu  Sinhu   7
Coshu 25 25 24 24
1   24
Cschu 
Sinhu 7
EX-17- Rewrite the following expressions in terms of exponentials .
Write the final result as simply as you can :
a ) 2Cosh(ln x )
c ) Cosh5 x  Sinh5 x
b ) tanh(ln x )
d ) ( Sinhx  Coshx )4
Sol.-
eln x  e  ln x 1 x
a ) 2Cosh(ln x )  2.
 x  2
e ln x  e ln x
b ) tanh(ln x )   e 
e ln x ln x
x  1
x 2 x
 1
x 
1  x 2  1 x
e5 x  e 5 x e5 x  e 5 x
c ) Cosh5 x  Sinh5 x  
 e5 x
2 2
e  x e x
 e x
 e  x 4
d ) ( Sinhx  Coshx )4      
 e 4 x
2 2
EX-18- Solve the equation for x : Cosh x = Sinh x + 1 / 2 .
Sol. - Coshx  Sinhx  1  e  x  1   x  ln 1  ln 2  x  ln 2
2 2
EX-19 – Verify the following identity :
Sinh(u+v)=Sinh u. Cosh v + Cosh u.Sinh v
then verify Sinh(u-v)=Sinh u. Cosh v - Cosh u.Sinh v
Sol.- ٢٠