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.