Precalculus Examples

$\frac{\sin^{3} (b) + \cos^{3} (b)}{\sin (b) + \cos (b)} = 1 - \sin (b) \cos (b)$

Step 1

Start on the left side.

$\frac{\sin^{3} (b) + \cos^{3} (b)}{\sin (b) + \cos (b)}$

Step 2

Factor.

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Step 2.1

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = \sin (b)$ and $b = \cos (b)$ .

$\frac{(\sin (b) + \cos (b)) (\sin^{2} (b) - \sin (b) \cos (b) + \cos^{2} (b))}{\sin (b) + \cos (b)}$

Step 2.2

Simplify.

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Step 2.2.1

Rearrange terms.

$\frac{(\sin (b) + \cos (b)) (- \sin (b) \cos (b) + \sin^{2} (b) + \cos^{2} (b))}{\sin (b) + \cos (b)}$

Step 2.2.2

Apply pythagorean identity.

$\frac{(\sin (b) + \cos (b)) (- \sin (b) \cos (b) + 1)}{\sin (b) + \cos (b)}$

Step 3

Cancel the common factor of $\sin (b) + \cos (b)$ .

$- \sin (b) \cos (b) + 1$

Step 4

Reorder terms.

$1 - \cos (b) \sin (b)$

Step 5

Rewrite $1 - \cos (b) \sin (b)$ as $1 - \sin (b) \cos (b)$ .

$1 - \sin (b) \cos (b)$

Step 6

Because the two sides have been shown to be equivalent, the equation is an identity.

$\frac{\sin^{3} (b) + \cos^{3} (b)}{\sin (b) + \cos (b)} = 1 - \sin (b) \cos (b)$ is an identity