To verify that the given equation [tex]\((\sin x - \cos x)^2 = 1 - 2 \sin x \cos x\)[/tex] is an identity, we need to show that both sides of the equation are equal for all values of [tex]\(x\)[/tex]. Let's proceed with a step-by-step approach.
### Step 1: Expand the left side of the equation
Start with the left side of the equation:
[tex]\[
(\sin x - \cos x)^2
\][/tex]
Expand the expression using the algebraic identity [tex]\((a - b)^2 = a^2 - 2ab + b^2\)[/tex]:
[tex]\[
(\sin x - \cos x)^2 = \sin^2 x - 2 \sin x \cos x + \cos^2 x
\][/tex]
### Step 2: Use the Pythagorean identity
Recall the Pythagorean identity in trigonometry:
[tex]\[
\sin^2 x + \cos^2 x = 1
\][/tex]
Using this identity, we can substitute [tex]\(1\)[/tex] for [tex]\(\sin^2 x + \cos^2 x\)[/tex]:
[tex]\[
\sin^2 x - 2 \sin x \cos x + \cos^2 x = (\sin^2 x + \cos^2 x) - 2 \sin x \cos x
\][/tex]
Since [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex], we have:
[tex]\[
1 - 2 \sin x \cos x
\][/tex]
### Step 3: Compare the expanded left side with the right side
We have now simplified the left side of the original equation:
[tex]\[
\sin^2 x - 2 \sin x \cos x + \cos^2 x = 1 - 2 \sin x \cos x
\][/tex]
Notice that the right side of the equation is exactly:
[tex]\[
1 - 2 \sin x \cos x
\][/tex]
Since both the simplified left side and the right side of the equation are identical, we have shown that:
[tex]\[
(\sin x - \cos x)^2 = 1 - 2 \sin x \cos x
\][/tex]
### Conclusion
The equation [tex]\((\sin x - \cos x)^2 = 1 - 2 \sin x \cos x\)[/tex] holds true for all values of [tex]\(x\)[/tex], and thus it is an identity.
