Trigonometry Examples

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

Step 1

Start on the left side.

$\sin (a + B) + \sin (a - B)$

Step 2

Apply the sum of angles identity.

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

Step 3

Apply the sum of angles identity.

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

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 $\cos (B)$ .

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

Step 4.1.2

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

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

Step 4.1.3

Rewrite using the commutative property of multiplication.

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

Step 4.2

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

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Step 4.2.1

Subtract $\cos (a) \sin (B)$ from $\cos (a) \sin (B)$ .

$\sin (a) \cos (B) + \sin (a) \cos (B) + 0$

Step 4.2.2

Add $\sin (a) \cos (B) + \sin (a) \cos (B)$ and $0$ .

$\sin (a) \cos (B) + \sin (a) \cos (B)$

Step 4.3

Add $\sin (a) \cos (B)$ and $\sin (a) \cos (B)$ .

$2 \sin (a) \cos (B)$

Step 5

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

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