Question: Verify the identity.
$(\sin x - \cos x)^2 = 1 - \sin 2x$
Part: 0 / 2
Part 1 of 2
Square the binomial.
$(\sin x - \cos x)^2 = \sin^2 x - \square + \cos^2 x$
1. **State the problem:** Verify the trigonometric identity:
$$(\sin x - \cos x)^2 = 1 - \sin 2x$$
2. **Square the binomial on the left side:**
Use the formula for squaring a binomial: $$(a - b)^2 = a^2 - 2ab + b^2$$
So,
$$ (\sin x - \cos x)^2 = \sin^2 x - 2 \sin x \cos x + \cos^2 x $$
3. **Fill in the blank:**
The missing term is $2 \sin x \cos x$.
4. **Simplify using Pythagorean identity:**
Recall that:
$$ \sin^2 x + \cos^2 x = 1 $$
Substitute this into the expression:
$$ 1 - 2 \sin x \cos x $$
5. **Use double-angle identity:**
Recall that:
$$ \sin 2x = 2 \sin x \cos x $$
So,
$$ 1 - 2 \sin x \cos x = 1 - \sin 2x $$
6. **Conclusion:**
We have shown that:
$$ (\sin x - \cos x)^2 = 1 - \sin 2x $$
which verifies the identity.
**Final answer:** The missing term in the expansion is $2 \sin x \cos x$.