Trigonometry Examples

$\frac{\sin (x - y)}{\sin (x + y)} = \frac{\tan (x) - \tan (y)}{\tan (x) + \tan (y)}$

Step 1

Start on the left side.

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

Step 2

Apply the sum of angles identity.

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

Step 3

Apply the sum of angles identity.

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

Step 4

Simplify the numerator.

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Step 4.1

Since $\cos (- y)$ is an even function, rewrite $\cos (- y)$ as $\cos (y)$ .

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

Step 4.2

Since $\sin (- y)$ is an odd function, rewrite $\sin (- y)$ as $- \sin (y)$ .

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

Step 4.3

Rewrite using the commutative property of multiplication.

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

Step 5

Simplify.

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

Rewrite $- (\cos (x) \sin (y) - \cos (y) \sin (x))$ as $- 1 (\cos (x) \sin (y) - \cos (y) \sin (x))$ .

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

Step 5.5

Move the negative in front of the fraction.

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

Step 6

Rewrite $- \frac{\cos (x) \sin (y) - \cos (y) \sin (x)}{\cos (x) \sin (y) + \cos (y) \sin (x)}$ as $\frac{\tan (x) - \tan (y)}{\tan (x) + \tan (y)}$ .

$\frac{\tan (x) - \tan (y)}{\tan (x) + \tan (y)}$

Step 7

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

$\frac{\sin (x - y)}{\sin (x + y)} = \frac{\tan (x) - \tan (y)}{\tan (x) + \tan (y)}$ is an identity