Question - Proving a Trigonometric Identity
Solution:
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.
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