Verified answer

sin² x + cos⁴x =

sin² x + cos² xcos² x=

sin² x + (1- sin² x)(1- sin² x)=

sin² x+1-2sin² x+sin⁴x=

1-sin² x+sin⁴x=

cos² x + sin⁴x

Clearly,

[(sin x)^2 - (cos x)^2] = [(sin x)^2 - (cos x)^2]

[(sin x)^2 - (cos x)^2] = [(sin x)^2 - (cos x)^2] * [1]

Now replace 1 with (sin x)^2 + (cos x)^2

[(sin x)^2 - (cos x)^2] = [(sin x)^2 - (cos x)^2] [(sin x)^2 + (cos x)^2]

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

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

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

Done!

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

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

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

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

=> cos^2(x) + sin^2(x)( 1 - cos^2(x))

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

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

sin² x + cos⁴x = cos² x + sin⁴x

LHS

sin² x + cos²x (cos²x) =

sin² x + cos²x (1 - sin²x) =

sin²x + cos²x - cos²x*sin²x =

1 - cos²x sin²x =

1 - (1-sin²x)*(sin²x) =

1- sin²x + sin⁴x =

(1- sin²x) + sin⁴x=

cos²x + sin⁴x => proved . .

hence, LHS(left hand side) = RHS (right hand side)