From cosh2A = 1+2(sinh^2)A, how do you prove that (sinh^4)A + (cosh^4)A=(cosh4A + 3)/4 and also (cosh^4)A - (sinh^4)A = cosh2A?

1 Answer
Feb 7, 2017

see below

Explanation:

Part I .
sinh^4A+cosh^4A=(cosh4A+3)/4

Use the formulas:
cosh 2A=2cosh^2A-1 -->Solve for cosh^2A

cosh 2A=1+2sinh^2A --->Solve for sinh^2A

Left Hand Side:

sinh^4A+cosh^4A= (sinh^2 A)^2 +(cosh^2A)^2

=((cosh2A-1)/2)^2+ ((cosh2A+1)/2)^2 -->FOIL

=(cosh^2 2A -2cosh2A+1)/4 +(cosh^2 2A+2cosh2A+1)/4

=(cosh^2 2A -2cosh2A+1 +cosh^2 2A+2cosh2A+1)/4

=(cosh^2 2A -cancel(2cosh2A)+1 +cosh^2 2A+cancel(2cosh2A)+1)/4

=(cosh^2 2A +1 +cosh^2 2A+1)/4

=(2cosh^2 2A +2)/4

Note: cosh 4A = 2cosh^2 2A -1 :. cosh 4A+1 = 2cosh^2 2A

=(cosh 4A+1 +2)/4

=(cosh 4A+3)/4

:.= Right Hand Side

Part II
cosh^4A-sinh^4A=cosh 2A

Use the properties :
cosh^2A-sinh^2A =1 and cosh 2A=1+2sinh^2A

Left Hand Side :

cosh^4A-sinh^4A=(cosh^2A+sinh^2A)(cosh^2A-sinh^2A)

=(cosh^2A+sinh^2A)*1

=cosh^2A+sinh^2A

=1+sinh^2A+sinh^2A

=1+2sinh^2A

=cosh 2A

:.= Right Hand Side