How do you verify #cos^4x - sin^4x = 1 - 2sin^2x#?

3 Answers
Mar 19, 2018

As below:

Explanation:

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.#

Mar 19, 2018

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!

Mar 19, 2018

See below

Explanation:

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#