Trigonometry Examples

\sin (A + B) = \cos (\frac{π}{2} - (A + B))

$\sin (A + B) = \cos (\frac{π}{2} - (A + B))$

Step 1

Start on the right side.

\cos (\frac{π}{2} - (A + B))

$\cos (\frac{π}{2} - (A + B))$

Step 2

Apply the difference of angles identity

\cos (x - y) = \cos (x) \cos (y) + \sin (x) \sin (y)

$\cos (x - y) = \cos (x) \cos (y) + \sin (x) \sin (y)$ .

\cos (\frac{π}{2}) \cos (A + B) + \sin (\frac{π}{2}) \sin (A + B)

$\cos (\frac{π}{2}) \cos (A + B) + \sin (\frac{π}{2}) \sin (A + B)$

Step 3

Simplify the expression.

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Step 3.1

Simplify each term.

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Step 3.1.1

The exact value of

\cos (\frac{π}{2})

$\cos (\frac{π}{2})$ is

0

$0$ .

0 \cos (A + B) + \sin (\frac{π}{2}) \sin (A + B)

$0 \cos (A + B) + \sin (\frac{π}{2}) \sin (A + B)$

Step 3.1.2

Multiply

0

$0$ by

\cos (A + B)

$\cos (A + B)$ .

0 + \sin (\frac{π}{2}) \sin (A + B)

$0 + \sin (\frac{π}{2}) \sin (A + B)$

Step 3.1.3

The exact value of

\sin (\frac{π}{2})

$\sin (\frac{π}{2})$ is

1

$1$ .

0 + 1 \sin (A + B)

$0 + 1 \sin (A + B)$

Step 3.1.4

Multiply

\sin (A + B)

$\sin (A + B)$ by

1

$1$ .

0 + \sin (A + B)

$0 + \sin (A + B)$

0 + \sin (A + B)

$0 + \sin (A + B)$

Step 3.2

Add

0

$0$ and

\sin (A + B)

$\sin (A + B)$ .

\sin (A + B)

$\sin (A + B)$

\sin (A + B)

$\sin (A + B)$

Step 4

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

\sin (A + B) = \cos (\frac{π}{2} - (A + B))

$\sin (A + B) = \cos (\frac{π}{2} - (A + B))$ is an identity