Question

Prove sin⁴Θ+cos⁴Θ=1-2cos²Θ+2cos⁴Θ

Answer 1

Answer:

Answered by Waymaker

• LHS =1+cos22θ

• =1+(cos2θ−sin2θ)2

• =1+cos4θ+sin4θ−2sin2θcos2θ

• =12+cos4θ+sin4θ−2sin2θcos2θ

• =(sin2θ+cos2θ)2+cos4θ+sin4θ−2sin2θcos2θ

• =sin4θ+cos4θ+2sin2θcos2θ+cos4θ+sin4θ−2sin2θcos2θ

• =2(sin4θ+cos4θ)

• = RHS

Answer 2

let in place of thete we take x

=sin^4x+cos^4x

=We know that sin^2x+cos^2x=1

=sin^2x=1-cos^2x

=(sin^2)^2+cos^4x

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

=1+cos^4x-2cos^2x+cos^4x

=1-2cos^2x+2cos^4x(proof)