Precalculus Examples

$\cos (a - b) - \cos (a + b) = 2 \sin (a) \sin (b)$

Step 1

Start on the left side.

$\cos (a - b) - \cos (a + b)$

Step 2

Apply the sum of angles identity $\cos (x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)$ .

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

Step 3

Apply the sum of angles identity $\cos (x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)$ .

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

Step 4

Simplify the expression.

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Step 4.1

Simplify each term.

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Step 4.1.1

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

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

Step 4.1.2

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

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

Step 4.1.3

Multiply $- \sin (a) (- \sin (b))$ .

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Step 4.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 4.1.3.2

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

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

Step 4.1.4

Apply the distributive property.

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

Step 4.1.5

Multiply $- (- \sin (a) \sin (b))$ .

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Step 4.1.5.1

Multiply $- 1$ by $- 1$ .

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

Step 4.1.5.2

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

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

Step 4.2

Combine the opposite terms in $\cos (a) \cos (b) + \sin (a) \sin (b) - \cos (a) \cos (b) + \sin (a) \sin (b)$ .

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Step 4.2.1

Subtract $\cos (a) \cos (b)$ from $\cos (a) \cos (b)$ .

$\sin (a) \sin (b) + 0 + \sin (a) \sin (b)$

Step 4.2.2

Add $\sin (a) \sin (b)$ and $0$ .

$\sin (a) \sin (b) + \sin (a) \sin (b)$

Step 4.3

Add $\sin (a) \sin (b)$ and $\sin (a) \sin (b)$ .

$2 \sin (a) \sin (b)$

Step 5

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

$\cos (a - b) - \cos (a + b) = 2 \sin (a) \sin (b)$ is an identity