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Chapter 5: Problem 92

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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 \(\sin^2 x + \cos^2 x = 1\).
03

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.

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Most popular questions from this chapter

Find the exact value of each expression. Do not use a calculator. $$\sin \left[\sin ^{-1} \frac{3}{5}-\cos ^{-1}\left(-\frac{4}{5}\right)\right]$$

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