Trigonometry Examples

$\sin (x - \frac{π}{2})$

Step 1

Interchange the variables.

$x = \sin (y - \frac{π}{2})$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\sin (y - \frac{π}{2}) = x$ .

$\sin (y - \frac{π}{2}) = x$

Step 2.2

Take the inverse sine of both sides of the equation to extract $y$ from inside the sine.

$y - \frac{π}{2} = \arcsin (x)$

Step 2.3

Add $\frac{π}{2}$ to both sides of the equation.

$y = \arcsin (x) + \frac{π}{2}$

Step 3

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \arcsin (x) + \frac{π}{2}$

Step 4

Verify if $f^{- 1} (x) = \arcsin (x) + \frac{π}{2}$ is the inverse of $f (x) = \sin (x - \frac{π}{2})$ .

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Step 4.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 4.2

Evaluate $f^{- 1} (f (x))$ .

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Step 4.2.1

Set up the composite result function.

$f^{- 1} (f (x))$

Step 4.2.2

Evaluate $f^{- 1} (\sin (x - \frac{π}{2}))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sin (x - \frac{π}{2})) = \arcsin (\sin (x - \frac{π}{2})) + \frac{π}{2}$

Step 4.3

Evaluate $f (f^{- 1} (x))$ .

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Step 4.3.1

Set up the composite result function.

$f (f^{- 1} (x))$

Step 4.3.2

Evaluate $f (\arcsin (x) + \frac{π}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\arcsin (x) + \frac{π}{2}) = \sin ((\arcsin (x) + \frac{π}{2}) - \frac{π}{2})$

Step 4.3.3

Combine the opposite terms in $(\arcsin (x) + \frac{π}{2}) - \frac{π}{2}$ .

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Step 4.3.3.1

Combine the numerators over the common denominator.

$f (\arcsin (x) + \frac{π}{2}) = \sin (\arcsin (x) + \frac{π - π}{2})$

Step 4.3.3.2

Subtract $π$ from $π$ .

$f (\arcsin (x) + \frac{π}{2}) = \sin (\arcsin (x) + \frac{0}{2})$

Step 4.3.4

Simplify the expression.

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Step 4.3.4.1

Divide $0$ by $2$ .

$f (\arcsin (x) + \frac{π}{2}) = \sin (\arcsin (x) + 0)$

Step 4.3.4.2

Add $\arcsin (x)$ and $0$ .

$f (\arcsin (x) + \frac{π}{2}) = \sin (\arcsin (x))$

Step 4.3.5

The functions sine and arcsine are inverses.

$f (\arcsin (x) + \frac{π}{2}) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \arcsin (x) + \frac{π}{2}$ is the inverse of $f (x) = \sin (x - \frac{π}{2})$ .

$f^{- 1} (x) = \arcsin (x) + \frac{π}{2}$