Trigonometry Examples

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

Step 1

Start on the left side.

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

Step 2

Combine.

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

Step 3

Simplify numerator.

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Step 3.1

Apply the distributive property.

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

Step 3.2

Multiply $\sin (θ)$ by $1$ .

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

Step 3.3

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

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

Step 4

Simplify denominator.

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Step 4.1

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

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Step 4.1.1

Apply the distributive property.

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

Step 4.1.2

Apply the distributive property.

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

Step 4.1.3

Apply the distributive property.

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

Step 4.2

Simplify and combine like terms.

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

Step 5

Apply pythagorean identity.

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

Step 6

Simplify.

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Step 6.1

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

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Step 6.1.1

Multiply by $1$ .

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

Step 6.1.2

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

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

Step 6.1.3

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

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

Step 6.2

Cancel the common factors.

$\frac{1 - \cos (θ)}{\sin (θ)}$

Step 7

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

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