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