prove addition form for coshx

Question

I need help with proofs in calc 2

Answer 1

So you want to show that  cosh (x + y) = cosh x  cosh y  +  sinh x  sinh y.    Well be aware that

                            cosh x =  (e^x + e^-x)/2    and    sinh x = (e^x- e^-x)/2   for any real number  x

 

Now,   2 cosh (x + y)  = (e^x+y+ e^-(x+y)) = (e^x e^y+ e^-x e^-y)   ...........................(A)

                          4 cosh x  cosh y = (e^x + e^-x) (e^y + e^-y) = (e^x e^y + e^x e^-y + e^-x e^y + e^-xe^-y)  

                         4 sinh x  sinh y =  (e^x e^y - e^x e^-y- e^-x e^y + e^-x e^-y)

                         4 cosh x  cosh y  +  4 sinh x  sinh y  =  2e^x e^y + 2 e^-x e^-y = 2(e^x+y+ e^-(x+y))

Now it follows that  cosh x cosh y + sinh x  sinh y = (e^x+y + e^-(x+y))/2 ..............................(B)  

Observe that equation (A) and (B) are identical and so

                                               cosh (x + y) = (e^x+y + e^-(x+y))/2 = cosh x  cosh y  + sinh x  sinh y                                      

Answer 2

Hi!

Happy Easter! Here is the proof using extension of well known trigonometrical relations to complex plane - see complex analysis))))))))))))

cosh(x+y) = cos(ix+iy) = cos(ix)cos(iy) - sin(ix)sin(iy) = cosh(x)cosh(y) + sinh(x)sinh(y)

+++ IHM

Answer 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A law of exponents is a^(-b) = 1/(a^b), lets apply this to the equation above

[exp(x+y) + exp(-x-y)]/2 = (exp^2(x+y) + 1)/[exp(x+y))*2]

Simplifying,

(exp(x+y) + 1/exp(x+y))/2 =[(exp^2(x+y) + 1)/[2*exp(x+y)]

[exp(x+ y)]

[exp(2x)exp(2y) + 1)]/[2*(exp(x)exp(y))] =

{[exp(x)exp(x)exp(y)exp(y) + 1]/[2*((exp(x)exp(y)]}

={[exp(x)exp(x)exp(y)exp(y) + 1]/[2*((exp(x)exp(y) + exp(-x)exp(-y)]}

 

 

 

Answer 4

Aubrey, as the starting point for the proof you need to substitue the x  for "s + t," that is, 

cosh x = cosh (s + t).

Note: This equivalence will be important during the last steps of this demonstration.

For the next step, it's preponderant to remember the identity for cosh (s + t):

cosh (s + t) = cosh s cosh t + senh s senh t

(Please, don't confuse the identity of cosh (s + t) with the identity of cos (s + t) in which the latter deals with subtraction instead of addition.)

Using the equivalences involving the expression with the constant e for both sinh and cosh, make the adequate substitutions for the above identity:

cosh (s + t) = cosh s cosh t + senh s senh t

= [½(e^s + e^-s)][½(e^t + e^-t)] + [½(e^s - e^-s)][½(e^t - e^-t)]

Multiply ½ times every term in each parenthesis using the distributive property:

= (½e^s + ½e^-s)(½e^t + ½e^-t) + (½e^s - ½e^-s)(½e^t - ½e^-t)

Use FOIL:

= (¼e^s+t + ¼e^s-t + ¼e^-s+t + ¼e^-s-t) + (¼e^s+t - ¼e^s-t - ¼e^-s+t + ¼e^-s-t)

Since we're dealing with addition the parentheses symbols can be ignored. Remember: when multiplying terms with the same bases, the exponents are added. This is the result after dealing with all the like terms in the above expression:

= ½e^s+t + ½e^-s-t 

Next, factor the exponent in the second term using ^-1 as the common factor:

= ½e^s+t + ½e^-(s+t) 

Earlier, I recalled the importance of the equivalence x = s + t. Now it's time to rewrite the above expression using the original variable:

= ½e^x + ½e^-x 

Voilà! We're arrived at the solution to this proof. By the way, if you want to prove the addition form for sinh, it can be done in a similar way. Hope that this proof can be helpful to you.