Trigonometry Examples

\cos (x + \frac{π}{2}) = - \sin (x)

$\cos (x + \frac{π}{2}) = - \sin (x)$

Step 1

Start on the left side.

\cos (x + \frac{π}{2})

$\cos (x + \frac{π}{2})$

Step 2

Apply the sum 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 (x) \cos (\frac{π}{2}) - \sin (x) \sin (\frac{π}{2})

$\cos (x) \cos (\frac{π}{2}) - \sin (x) \sin (\frac{π}{2})$

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

$0$ .

\cos (x) \cdot 0 - \sin (x) \sin (\frac{π}{2})

$\cos (x) \cdot 0 - \sin (x) \sin (\frac{π}{2})$

Step 3.1.2

Multiply

\cos (x)

$\cos (x)$ by

0

$0$ .

0 - \sin (x) \sin (\frac{π}{2})

$0 - \sin (x) \sin (\frac{π}{2})$

Step 3.1.3

The exact value of

\sin (\frac{π}{2})

$\sin (\frac{π}{2})$ is

1

$1$ .

0 - \sin (x) \cdot 1

$0 - \sin (x) \cdot 1$

Step 3.1.4

Multiply

- 1

$- 1$ by

1

$1$ .

0 - \sin (x)

$0 - \sin (x)$

0 - \sin (x)

$0 - \sin (x)$

Step 3.2

Subtract

\sin (x)

$\sin (x)$ from

0

$0$ .

- \sin (x)

$- \sin (x)$

- \sin (x)

$- \sin (x)$

Step 4

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

\cos (x + \frac{π}{2}) = - \sin (x)

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