How do you verify #cos^4x - sin^4x = 1 - 2sin^2x#?
↳Redirected from
"How does implicit differentiation work?"
Note that #cos^4x=(cos^2x)^2#
#L.H.S.#
#=cos^4x-sin^4x#
#=(cos^2x)^2-(sin^2x)^2#
#=(cos^2x+sin^2x)(cos^2x-sin^2x)#
#=(cos^2x+sin^2x)(cos^2x+sin^2x-2sin^2x)#
#=(1)(1-2sin^2x)#
#=1-2sin^2x#
#=R.H.S.#
To prove that #cos^4x-sin^4x=1-2sin^2x#, we'll need the Pythagorean identity and a variation on the Pythagorean identity:
#color(white)=>cos^2x+sin^2x=1#
#=>cos^2x=1-sin^2x#
I'll start with the left-hand side and manipulate it until it looks like the right-hand side using these two identities:
#LHS=cos^4x-sin^4x#
#color(white)(LHS)=(cos^2x)^2-(sin^2x)^2#
#color(white)(LHS)=(color(red)(cos^2x+sin^2x))(cos^2x-sin^2x)#
#color(white)(LHS)=color(red)1*(cos^2x-sin^2x)#
#color(white)(LHS)=color(blue)(cos^2x)-sin^2x#
#color(white)(LHS)=color(blue)(1-sin^2x)-sin^2x#
#color(white)(LHS)=1-2sin^2x#
#color(white)(LHS)=RHS#
That's the proof. Hope this helped!
To verify: #cos^4x - sin^4x = 1-2sin^2x#
Let #theta = sin^2x -> cos^2x = 1-theta#
#:. LHS = (1-theta)^2 - theta^2#
#= 1-2theta+theta^2-theta^2#
#= 1-2theta#
Undo substitution:
#LHS = 1-2sin^2x = RHS#
