To determine if the expression \( \sin(a+b)\sin(a-b) = \sin^2(b) - \sin^2(a) \) is an identity, we can simplify the left-hand side using the product-to-sum identities.
The product-to-sum formula states:
\[ \sin(x)\sin(y) = \frac{1}{2} [\cos(x-y) - \cos(x+y)] \]
Applying this to \( \sin(a+b)\sin(a-b) \):
\[ \sin(a+b)\sin(a-b) = \frac{1}{2} [\cos((a+b) - (a-b)) - \cos((a+b) + (a-b))] \] \[ = \frac{1}{2} [\cos(2b) - \cos(2a)] \]
Now let's simplify the right-hand side \( \sin^2(b) - \sin^2(a) \). Using the identity \( \sin^2(x) = 1 - \cos^2(x) \):
\[ \sin^2(b) - \sin^2(a) = (1 - \cos^2(b)) - (1 - \cos^2(a)) = \cos^2(a) - \cos^2(b) \]
So, now we compare:
The left side is \( \frac{1}{2} [\cos(2b) - \cos(2a)] \) and the right side is \( \cos^2(a) - \cos^2(b) \).
To check if the original statement is true, we need these two to be equal, which they generally are not.
Specifically, we can evaluate at specific angles (for instance, let \( a = 0 \) and \( b = \frac{\pi}{4} \)):
- Left-hand side \( \sin(0+\frac{\pi}{4})\sin(0-\frac{\pi}{4}) = \sin(\frac{\pi}{4})\sin(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \cdot -\frac{\sqrt{2}}{2} = -\frac{1}{2} \)
- Right-hand side \( \sin^2(\frac{\pi}{4}) - \sin^2(0) = \left(\frac{\sqrt{2}}{2}\right)^2 - 0 = \frac{1}{2} \)
Since \(-\frac{1}{2} \neq \frac{1}{2}\), we see that the equality does not hold.
Therefore, the statement is False.