Prove the given Trig identity

sin^4(θ)-cos^4(θ)=1-2cos^2(θ)

Please give the steps on how you did it too... thank you very much!

sin^4(θ)-cos^4(θ)=1-2cos^2(θ)

Taking the left hand side:

= sin^4(θ)-cos^4(θ)
= (sin^2 θ - cos^2 θ) (sin^2 θ + cos^2 θ)

(since (a-b)^4 can be re-written as (a-b)^2 (a+b)^2)

=(sin^2 θ - cos^2 θ) (sin^2 θ + cos^2 θ)
= (sin^2 θ - cos^2 θ) (1) (Since sin^2 θ + cos^2 θ is equal to 1)
= sin^2 θ - cos^2 θ
= 1 - cos^2 θ - cos^2 θ (since sin^2 θ can be written as 1 - cos^2 θ)
= 1 - 2cos^2 θ (Proved)
___________

I hope this helps!

You're very welcome! I'm really glad that my answer helped you out. :)

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Use identity: sin^2θ + cos^2θ = 1
cos^2θ = 1 - sin^2θ *other form of above identity
sin^2θ = 1 - cos^2θ *other form of identity

Problem: sin^4θ-cos^4θ = 1-2cos^2θ
Step 1: (sin^2θ-cos^2θ)(sin^2θ+cos^2θ) = *Working with LEFT side. Factor out left side
Step 2: (1-cos^2θ - cos^2θ)(1) = *Use substitution: (sin^2θ+cos^2θ) becomes 1 & substitute sin^2θ for 1-cos^2θ
Step 3: 1-2cos^2θ = 1-2cos^2θ *Multiply the 1 and Combine like terms

VERIFIED!!

hope that made sense :)

Factor the left side as the difference of two squares

LHS = (sin^2€ + sin^2€)(sin^2€ - cos^2€) = -cos(2€) =
1 - 2cos^2€ = RHS

Sorry, can't make a theta on my phone! :-(

sin^4(θ)-cos^4(θ)=

(sin^2(θ))2-cos^2(θ)

sin^2(θ)+cos^2(θ)=1
sin^2(θ)=1-cos(θ)

(1-cos^2(θ))^2-cos^4(θ)
1-2cos^2(θ)+cos^4(θ)-cos^4(θ)
1-2cos^2(θ)=1-2cos^2(θ)