Final answer:
Using trigonometric identities, specifically product-to-sum, the given function sin(2x)cos(3x) can be transformed, but none of the options match the transformation. Option (C) corresponds to the derivative of the given function.
Step-by-step explanation:
To find which of the given options is equivalent to f(x) = sin(2x)\u00b7cos(3x), we can use trigonometric identities to transform f(x). Specifically, we'll use a product-to-sum identity:
Using product-to-sum identities:
- sin(a)cos(b) = 1/2 [ sin(a+b) + sin(a-b) ]
Applying the above identity to f(x):
f(x) = sin(2x)cos(3x)
= 1/2 [ sin(2x+3x) + sin(2x-3x) ]
= 1/2 [ sin(5x) + sin(-x) ]
= 1/2 [ sin(5x) - sin(x) ]
Now we expand sin(5x) and sin(x) using sin(A \u2212 B) identity:
sin(A \u2212 B) = sinAcosB \u2212 cosAsinB
sin(5x) \u2212 sin(x) = sin(5x)cos(0) - cos(5x)sin(0) - [sin(x)cos(0) - cos(x)sin(0)]
= sin(5x) \u2212 sin(x)
= sin(4x+1x) - sin(x)
= sin4xcos1x + cos4xsin1x - sin(x)
Which is now in the form of sum of sine and cosine functions. None of the given options seem to match this final form, which means there might be a typo in the question or in the given options. However, upon closer inspection, it can be seen that option (C) with the right sine and cosine coefficients actually represent derivatives of the original function, which are often required in calculus.
Therefore, the answer to this question depends on whether we're looking for an equivalent expression of sin(2x)cos(3x) in a different form, or if we're looking for the derivative of the function. If we were looking for the derivative, option (C) is the correct answer as the derivative of sin(2x)cos(3x) yields:
f'(x) = 2cos(2x)\u00b7sin(3x) - 3sin(2x)\u00b7cos(3x)