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Cosh 2 x  Cosh2 x  1 Sinh 2 x  Cosh2 x  1

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Presentation on theme: "Cosh 2 x  Cosh2 x  1 Sinh 2 x  Cosh2 x  1"— Presentation transcript:

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


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