We have to prove that (sin x)^4 + (cos x)^4 + (sin 2x)^2/2 = 1

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

=> (sin x)^4 + (cos x)^4 + (2*sin x * cos x)^2/2

=> (sin x)^4 + (cos x)^4 + 4*(sin x)^2*(cos x)^2/2

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

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

=> 1^2

=> 1

**This proves that (sin x)^4 + (cos x)^4 + (sin 2x)^2/2 = 1**

We'll use the Pythagorean identity to solve the problem.

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

If we'll raise to square both sides, we'll get:

[(sin x)^2 + (cos x)^2]^2 = 1^2

(sin x)^4 + (cos x)^4 + 2 (sin x)^2 *(cos x)^2 = 1

We'll keep the sum (sin x)^4 + (cos x)^4 to the left:

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

We'll also apply the double angle identity:

sin 2x = 2 sin x*cos x

We'll raise to square both sides:

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

We'll divide by 2:

[(sin 2x)^2]/2 =2 (sin x)^2 *(cos x)^2

We'll re-write the identity to be proved:

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

We'll eliminate like terms:

1 = 1

**We'll get equal values both sides, therefore the identity (sin x)^4 + (cos x)^4 + [(sin 2x)^2]/2 = 1 is verified.**