Trigonometry Examples

$\cos (\arcsin (7 x))$

Step 1

Interchange the variables.

$x = \cos (\arcsin (7 y))$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\cos (\arcsin (7 y)) = x$ .

$\cos (\arcsin (7 y)) = x$

Step 2.2

Take the inverse cosine of both sides of the equation to extract $\arcsin (7 y)$ from inside the cosine.

$\arcsin (7 y) = \arccos (x)$

Step 2.3

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

$7 y = \sin (\arccos (x))$

Step 2.4

Simplify the right side.

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Step 2.4.1

Simplify $\sin (\arccos (x))$ .

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Step 2.4.1.1

Write the expression using exponents.

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Step 2.4.1.1.1

Draw a triangle in the plane with vertices $(x, \sqrt{1^{2} - x^{2}})$ , $(x, 0)$ , and the origin. Then $\arccos (x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(x, \sqrt{1^{2} - x^{2}})$ . Therefore, $\sin (\arccos (x))$ is $\sqrt{1 - x^{2}}$ .

$7 y = \sqrt{1 - x^{2}}$

Step 2.4.1.1.2

Rewrite $1$ as $1^{2}$ .

$7 y = \sqrt{1^{2} - x^{2}}$

Step 2.4.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

$7 y = \sqrt{(1 + x) (1 - x)}$

Step 2.5

Divide each term in $7 y = \sqrt{(1 + x) (1 - x)}$ by $7$ and simplify.

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Step 2.5.1

Divide each term in $7 y = \sqrt{(1 + x) (1 - x)}$ by $7$ .

$\frac{7 y}{7} = \frac{\sqrt{(1 + x) (1 - x)}}{7}$

Step 2.5.2

Simplify the left side.

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Step 2.5.2.1

Cancel the common factor of $7$ .

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Step 2.5.2.1.1

Cancel the common factor.

$\frac{7 y}{7} = \frac{\sqrt{(1 + x) (1 - x)}}{7}$

Step 2.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{\sqrt{(1 + x) (1 - x)}}{7}$

Step 3

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

$f^{- 1} (x) = \frac{\sqrt{(1 + x) (1 - x)}}{7}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt{(1 + x) (1 - x)}}{7}$ is the inverse of $f (x) = \cos (\arcsin (7 x))$ .

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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 $f^{- 1} (\cos (\arcsin (7 x)))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\cos (\arcsin (7 x))) = \frac{\sqrt{(1 + \cos (\arcsin (7 x))) (1 - (\cos (\arcsin (7 x))))}}{7}$

Step 4.2.3

Remove parentheses.

$f^{- 1} (\cos (\arcsin (7 x))) = \frac{\sqrt{(1 + \cos (\arcsin (7 x))) (1 - (\cos (\arcsin (7 x))))}}{7}$

Step 4.2.4

Simplify the numerator.

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Step 4.2.4.1

Draw a triangle in the plane with vertices $(\sqrt{1^{2} - {(7 x)}^{2}}, 7 x)$ , $(\sqrt{1^{2} - {(7 x)}^{2}}, 0)$ , and the origin. Then $\arcsin (7 x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(\sqrt{1^{2} - {(7 x)}^{2}}, 7 x)$ . Therefore, $\cos (\arcsin (7 x))$ is $\sqrt{1 - {(7 x)}^{2}}$ .

$f^{- 1} (\cos (\arcsin (7 x))) = \frac{\sqrt{(1 + \sqrt{1 - {(7 x)}^{2}}) (1 - \cos (\arcsin (7 x)))}}{7}$

Step 4.2.4.2

Rewrite $1$ as $1^{2}$ .

$f^{- 1} (\cos (\arcsin (7 x))) = \frac{\sqrt{(1 + \sqrt{1^{2} - {(7 x)}^{2}}) (1 - \cos (\arcsin (7 x)))}}{7}$

Step 4.2.4.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = 7 x$ .

$f^{- 1} (\cos (\arcsin (7 x))) = \frac{\sqrt{(1 + \sqrt{(1 + 7 x) (1 - (7 x))}) (1 - \cos (\arcsin (7 x)))}}{7}$

Step 4.2.4.4

Multiply $7$ by $- 1$ .

$f^{- 1} (\cos (\arcsin (7 x))) = \frac{\sqrt{(1 + \sqrt{(1 + 7 x) (1 - 7 x)}) (1 - \cos (\arcsin (7 x)))}}{7}$

Step 4.2.4.5

$f^{- 1} (\cos (\arcsin (7 x))) = \frac{\sqrt{(1 + \sqrt{(1 + 7 x) (1 - 7 x)}) (1 - \sqrt{1 - {(7 x)}^{2}})}}{7}$

Step 4.2.4.6

Rewrite $1$ as $1^{2}$ .

$f^{- 1} (\cos (\arcsin (7 x))) = \frac{\sqrt{(1 + \sqrt{(1 + 7 x) (1 - 7 x)}) (1 - \sqrt{1^{2} - {(7 x)}^{2}})}}{7}$

Step 4.2.4.7

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = 7 x$ .

$f^{- 1} (\cos (\arcsin (7 x))) = \frac{\sqrt{(1 + \sqrt{(1 + 7 x) (1 - 7 x)}) (1 - \sqrt{(1 + 7 x) (1 - (7 x))})}}{7}$

Step 4.2.4.8

Multiply $7$ by $- 1$ .

$f^{- 1} (\cos (\arcsin (7 x))) = \frac{\sqrt{(1 + \sqrt{(1 + 7 x) (1 - 7 x)}) (1 - \sqrt{(1 + 7 x) (1 - 7 x)})}}{7}$

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 $f (\frac{\sqrt{(1 + x) (1 - x)}}{7})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \cos (\arcsin (7 (\frac{\sqrt{(1 + x) (1 - x)}}{7})))$

Step 4.3.3

Combine fractions.

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Step 4.3.3.1

Draw a triangle in the plane with vertices $(\sqrt{1^{2} - {(7 (\frac{\sqrt{(1 + x) (1 - x)}}{7}))}^{2}}, 7 (\frac{\sqrt{(1 + x) (1 - x)}}{7}))$ , $(\sqrt{1^{2} - {(7 (\frac{\sqrt{(1 + x) (1 - x)}}{7}))}^{2}}, 0)$ , and the origin. Then $\arcsin (7 (\frac{\sqrt{(1 + x) (1 - x)}}{7}))$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(\sqrt{1^{2} - {(7 (\frac{\sqrt{(1 + x) (1 - x)}}{7}))}^{2}}, 7 (\frac{\sqrt{(1 + x) (1 - x)}}{7}))$ . Therefore, $\cos (\arcsin (7 (\frac{\sqrt{(1 + x) (1 - x)}}{7})))$ is $\sqrt{1 - {(\frac{7 \sqrt{(1 + x) (1 - x)}}{7})}^{2}}$ .

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{1 - {(\frac{7 \sqrt{(1 + x) (1 - x)}}{7})}^{2}}$

Step 4.3.3.2

Rewrite $1$ as $1^{2}$ .

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{1^{2} - {(\frac{7 \sqrt{(1 + x) (1 - x)}}{7})}^{2}}$

Step 4.3.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = \frac{7 \sqrt{(1 + x) (1 - x)}}{7}$ .

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{(1 + \frac{7 \sqrt{(1 + x) (1 - x)}}{7}) (1 - \frac{7 \sqrt{(1 + x) (1 - x)}}{7})}$

Step 4.3.5

Simplify.

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Step 4.3.5.1

Write $1$ as a fraction with a common denominator.

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{(\frac{7}{7} + \frac{7 \sqrt{(1 + x) (1 - x)}}{7}) (1 - \frac{7 \sqrt{(1 + x) (1 - x)}}{7})}$

Step 4.3.5.2

Combine the numerators over the common denominator.

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{\frac{7 + 7 \sqrt{(1 + x) (1 - x)}}{7} \cdot (1 - \frac{7 \sqrt{(1 + x) (1 - x)}}{7})}$

Step 4.3.5.3

Rewrite $\frac{7 + 7 \sqrt{(1 + x) (1 - x)}}{7}$ in a factored form.

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Step 4.3.5.3.1

Factor $7$ out of $7 + 7 \sqrt{(1 + x) (1 - x)}$ .

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Step 4.3.5.3.1.1

Factor $7$ out of $7$ .

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{\frac{7 \cdot 1 + 7 \sqrt{(1 + x) (1 - x)}}{7} \cdot (1 - \frac{7 \sqrt{(1 + x) (1 - x)}}{7})}$

Step 4.3.5.3.1.2

Factor $7$ out of $7 \cdot 1 + 7 \sqrt{(1 + x) (1 - x)}$ .

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{\frac{7 (1 + \sqrt{(1 + x) (1 - x)})}{7} \cdot (1 - \frac{7 \sqrt{(1 + x) (1 - x)}}{7})}$

Step 4.3.5.3.2

Reduce the expression by cancelling the common factors.

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Step 4.3.5.3.2.1

Reduce the expression $\frac{7 (1 + \sqrt{(1 + x) (1 - x)})}{7}$ by cancelling the common factors.

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Step 4.3.5.3.2.1.1

Cancel the common factor.

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{\frac{7 (1 + \sqrt{(1 + x) (1 - x)})}{7} \cdot (1 - \frac{7 \sqrt{(1 + x) (1 - x)}}{7})}$

Step 4.3.5.3.2.1.2

Rewrite the expression.

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{\frac{1 + \sqrt{(1 + x) (1 - x)}}{1} \cdot (1 - \frac{7 \sqrt{(1 + x) (1 - x)}}{7})}$

Step 4.3.5.3.2.2

Divide $1 + \sqrt{(1 + x) (1 - x)}$ by $1$ .

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{(1 + \sqrt{(1 + x) (1 - x)}) (1 - \frac{7 \sqrt{(1 + x) (1 - x)}}{7})}$

Step 4.3.5.4

Cancel the common factor of $7$ .

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Step 4.3.5.4.1

Cancel the common factor.

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{(1 + \sqrt{(1 + x) (1 - x)}) (1 - \frac{7 \sqrt{(1 + x) (1 - x)}}{7})}$

Step 4.3.5.4.2

Divide $\sqrt{(1 + x) (1 - x)}$ by $1$ .

$f (\frac{\sqrt{(1 + x) (1 - x)}}{7}) = \sqrt{(1 + \sqrt{(1 + x) (1 - x)}) (1 - \sqrt{(1 + x) (1 - x)})}$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt{(1 + x) (1 - x)}}{7}$ is the inverse of $f (x) = \cos (\arcsin (7 x))$ .

$f^{- 1} (x) = \frac{\sqrt{(1 + x) (1 - x)}}{7}$