Answer:
Let's express each expression as a product:
1) \( \sin(a - \frac{2\pi}{3}) - \sin(a + \frac{2\pi}{3}) \)
Using the identity \( \sin(A - B) - \sin(A + B) = -2\sin(B)\cos(A) \):
\( -2\sin(\frac{2\pi}{3})\cos(a) \)
\( -2\left(\frac{\sqrt{3}}{2}\right)\cos(a) \)
\( -\sqrt{3}\cos(a) \)
2) \( \sin(a + b) + \sin(a - b) \)
Using the identity \( \sin(A + B) + \sin(A - B) = 2\sin(A)\cos(B) \):
\( 2\sin(a)\cos(b) \)
3) \( \cos(2a - \frac{\pi}{16}) + \cos(\frac{9\pi}{16} - 2a) \)
Using the identity \( \cos(A - B) + \cos(A + B) = 2\cos(A)\cos(B) \):
\( 2\cos\left(\frac{\pi}{32}\right)\cos\left(2a - \frac{7\pi}{32}\right) \)
4) \( 1 - \cos(a) \)
This is already a product: \( (1)(1) - (\cos(a)) \)
5) \( 1 + \sin(a) \)
This is already a product: \( (1)(1) + (\sin(a)) \)
6) \( 1 - 2\cos^2(a) \)
Using the identity \( \cos^2(A) = 1 - \sin^2(A) \):
\( 1 - 2(1 - \sin^2(a)) \)
\( 1 - 2 + 2\sin^2(a) \)
\( 2\sin^2(a) - 1 \)
7) \( \sin(4b) + \sin(10b) + \sin(22b) + \sin(16b) \)
These terms cannot be expressed as a simple product.
