write sin4xcos2 as the sum or difference of two functions. answers: 1/2(cos6x+cos2x), 1/2(cos2x-cos6x), 1/2(sin6x+sin2x), sin6x-sin2x

To write sin4xcos2 as the sum or difference of two functions, we can use the trigonometric identity:

sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

In this case, we can rewrite sin4xcos2 as:

sin4xcos2 = (sin2x)(cos2x)

Now, let's apply the trigonometric identity to the expression sin2x(1 + cos4x):

sin2x(1 + cos4x) = sin2x⋅1 + sin2x⋅cos4x

= sin2x + sin2x⋅cos4x

Using the trigonometric identity sin2x = 1/2(1 - cos2x), we can substitute it in the equation:

= 1/2(1 - cos2x) + 1/2(1 - cos2x)⋅cos4x

= 1/2 - 1/2cos2x + 1/2⋅cos4x - 1/2cos2x⋅cos4x

Now, let's simplify the expression:

= 1/2 - 1/2cos2x + 1/2⋅cos4x - 1/2cos2xcos4x

We can factor out 1/2 from the first two terms and the last two terms:

= 1/2(1 - cos2x) + 1/2(cos4x - cos2xcos4x)

Finally, we can rewrite the trigonometric identity cos2x - cos4x = 2cos2x - 1:

= 1/2(1 - cos2x) + 1/2(2cos2x - 1)

This simplifies to:

= 1/2(1 - cos2x + 2cos2x - 1)

= 1/2(cos2x - cos2x + 1)

= 1/2(1)

= 1/2

Therefore, the correct answer is 1/2.