Answer:To express \( \sin(2x-x) \) in terms of \( \sin x \) and \( \cos x \), we'll use the angle subtraction formula for sine:
\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]
In this case, \( A = 2x \) and \( B = x \), so:
\[ \sin(2x - x) = \sin(2x) \cos(x) - \cos(2x) \sin(x) \]
Now, we need to express \( \sin(2x) \) and \( \cos(2x) \) in terms of \( \sin x \) and \( \cos x \). We can use the double-angle formulas:
\[ \sin(2x) = 2 \sin x \cos x \]
\[ \cos(2x) = \cos^2 x - \sin^2 x \]
Now, we can substitute these into the expression:
\[ \sin(2x - x) = (2 \sin x \cos x) \cos x - (\cos^2 x - \sin^2 x) \sin x \]
Simplify further:
\[ \sin(2x - x) = 2 \sin x \cos^2 x - \cos^2 x \sin x + \sin^3 x \]
This expression is in terms of \( \sin x \) and \( \cos x \).
