Trigonometry Examples

\frac{\sin^{2} (x) - \cos^{2} (x)}{\sin (x) - \cos (x)} = \sin (x) + \cos (x)

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

Step 1

Start on the left side.

\frac{\sin^{2} (x) - \cos^{2} (x)}{\sin (x) - \cos (x)}

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

Step 2

Simplify.

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Step 2.1

Since both terms are perfect squares, factor using the difference of squares formula,

a^{2} - b^{2} = (a + b) (a - b)

$a^{2} - b^{2} = (a + b) (a - b)$ where

a = \sin (x)

$a = \sin (x)$ and

b = \cos (x)

$b = \cos (x)$ .

\frac{(\sin (x) + \cos (x)) (\sin (x) - \cos (x))}{\sin (x) - \cos (x)}

$\frac{(\sin (x) + \cos (x)) (\sin (x) - \cos (x))}{\sin (x) - \cos (x)}$

Step 2.2

Cancel the common factor of

\sin (x) - \cos (x)

$\sin (x) - \cos (x)$ .

\sin (x) + \cos (x)

$\sin (x) + \cos (x)$

\sin (x) + \cos (x)

$\sin (x) + \cos (x)$

Step 3

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

\frac{\sin^{2} (x) - \cos^{2} (x)}{\sin (x) - \cos (x)} = \sin (x) + \cos (x)

$\frac{\sin^{2} (x) - \cos^{2} (x)}{\sin (x) - \cos (x)} = \sin (x) + \cos (x)$ is an identity