Problem 92 - Precalculus Essentials

Question

Verify each identity. $$\frac{\sin ^{3} x-\cos ^{3} x}{\sin x-\cos x}=1+\sin x \cos x$$

Step-by-Step Solution

Answer

The given trigonometric identity is verified to be true. \(\frac{\sin ^{3} x-\cos ^{3} x}{\sin x-\cos x}\) simplifies into \(1 + \sin x \cos x\), the right-hand side of the identity.

Step 1: Start with the more complex side

Begin the verification with the more complex side. In this case it's \( \frac{\sin ^{3} x-\cos ^{3} x}{\sin x-\cos x} \). It is more complicated than right side.

Step 2: Apply Difference of Cubes Formula

Rewrite the numerator as \((\sin x - \cos x)(\sin^2 x + \sin x \cos x + \cos^2 x)\) by applying the formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) . Remember that \(\sin^2 x + \cos^2 x = 1\).

Step 3: Simplify the Expression

Now, cancel out the common factor in the numerator and denominator which is \(\sin x - \cos x\). The left side of the identity becomes \(1 + \sin x \cos x\), and thus equals the right hand side of the original identity.