Proving a Trigonometric Identity
To prove the identity
\[
\frac{\sin^3x - \cos^3x}{\sin x - \cos x} = \frac{1 + \sin 2x}{2}
\]
let's start by factoring the numerator using the difference of cubes formula:
\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]
where \( a = \sin x \) and \( b = \cos x \).
Thus,
\[
\sin^3x - \cos^3x = (\sin x - \cos x)(\sin^2x + \sin x\cos x + \cos^2x)
\]
We know that \(\sin^2x + \cos^2x = 1\), so we can simplify as follows:
\[
\sin^3x - \cos^3x = (\sin x - \cos x)(1 + \sin x\cos x)
\]
Now, divide both sides by \(\sin x - \cos x\):
\[
\frac{\sin^3x - \cos^3x}{\sin x - \cos x} = \frac{(\sin x - \cos x)(1 + \sin x\cos x)}{\sin x - \cos x} = 1 + \sin x\cos x
\]
We also know the double-angle identity for sine:
\[
\sin 2x = 2\sin x\cos x
\]
So, rewriting \( \sin x\cos x \) in terms of the double-angle identity:
\[
\frac{1 + \sin 2x}{2}
\]
This means:
\[
1 + \sin x\cos x = \frac{1 + \sin 2x}{2}
\]
Hence, we've shown that:
\[
\frac{\sin^3x - \cos^3x}{\sin x - \cos x} = \frac{1 + \sin 2x}{2}
\]
And the identity is proved.
