Trigonometry Examples

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

Step 1

Start on the left side.

$\frac{\sin (x)}{1 + \cos (x)}$

Step 2

Multiply $\frac{\sin (x)}{1 + \cos (x)}$ by $\frac{1 - \cos (x)}{1 - \cos (x)}$ .

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

Step 3

Combine.

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

Step 4

Simplify numerator.

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Step 4.1

Apply the distributive property.

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

Step 4.2

Multiply $\sin (x)$ by $1$ .

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

Step 4.3

Reorder factors in $\sin (x) + \sin (x) (- \cos (x))$ .

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

Step 5

Simplify denominator.

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Step 5.1

Expand $(1 + \cos (x)) (1 - \cos (x))$ using the FOIL Method.

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Step 5.1.1

Apply the distributive property.

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

Step 5.1.2

Apply the distributive property.

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

Step 5.1.3

Apply the distributive property.

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

Step 5.2

Simplify and combine like terms.

$\frac{\sin (x) - \sin (x) \cos (x)}{1 - \cos^{2} (x)}$

Step 6

Apply pythagorean identity.

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

Step 7

Simplify.

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Step 7.1

Factor $\sin (x)$ out of $\sin (x) - \sin (x) \cos (x)$ .

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Step 7.1.1

Multiply by $1$ .

$\frac{\sin (x) \cdot 1 - \sin (x) \cos (x)}{{\sin (x)}^{2}}$

Step 7.1.2

Factor $\sin (x)$ out of $- \sin (x) \cos (x)$ .

$\frac{\sin (x) \cdot 1 + \sin (x) (- \cos (x))}{{\sin (x)}^{2}}$

Step 7.1.3

Factor $\sin (x)$ out of $\sin (x) \cdot 1 + \sin (x) (- \cos (x))$ .

$\frac{\sin (x) (1 - \cos (x))}{{\sin (x)}^{2}}$

Step 7.2

Cancel the common factors.

$\frac{1 - \cos (x)}{\sin (x)}$

Step 8

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

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