Find the Exact Value sec(arcsin(minus one divide two))

$\sec (\arcsin (- \frac{1}{2}))$

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

$\frac{1}{\sqrt{1^{2} - {(- \frac{1}{2})}^{2}}}$

Clarify the denominator.

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One raised to any power equals one.

$\frac{1}{\sqrt{1 - {(- \frac{1}{2})}^{2}}}$

Apply the power rule ${(a b)}^{n} = a^{n} b^{n}$ for exponent distribution.

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Apply the product rule to $- \frac{1}{2}$ .

$\frac{1}{\sqrt{1 - ({(- 1)}^{2} {(\frac{1}{2})}^{2})}}$

Use the product rule for $\frac{1}{2}$ .

$\frac{1}{\sqrt{1 - ({(- 1)}^{2} \frac{1^{2}}{2^{2}})}}$

First, multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Move ${(- 1)}^{2}$ .

$\frac{1}{\sqrt{1 + {(- 1)}^{2} \cdot - 1 \frac{1^{2}}{2^{2}}}}$

Multiply ${(- 1)}^{2}$ by $- 1$ .

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Raise $- 1$ to the power of $1$ .

$\frac{1}{\sqrt{1 + {(- 1)}^{2} \cdot {(- 1)}^{1} \frac{1^{2}}{2^{2}}}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to Add exponents.

$\frac{1}{\sqrt{1 + {(- 1)}^{2 + 1} \frac{1^{2}}{2^{2}}}}$

Add $2$ and $1$ .

$\frac{1}{\sqrt{1 + {(- 1)}^{3} \frac{1^{2}}{2^{2}}}}$

Raise $- 1$ to the power of $3$ .

$\frac{1}{\sqrt{1 - \frac{1^{2}}{2^{2}}}}$

One to any power is one.

$\frac{1}{\sqrt{1 - \frac{1}{2^{2}}}}$

Raise $2$ to the power of $2$ .

$\frac{1}{\sqrt{1 - \frac{1}{4}}}$

To write $\frac{1}{1}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{1}{\sqrt{\frac{1}{1} \cdot \frac{4}{4} - \frac{1}{4}}}$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Add.

$\frac{1}{\sqrt{\frac{1 \cdot 4}{1 \cdot 4} - \frac{1}{4}}}$

Multiply $4$ by $1$ .

$\frac{1}{\sqrt{\frac{1 \cdot 4}{4} - \frac{1}{4}}}$

Add numerators over the common denominator.

$\frac{1}{\sqrt{\frac{1 \cdot 4 - 1}{4}}}$

Clarify the numerator.

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First, multiply $1$ by $4$ .

$\frac{1}{\sqrt{\frac{4 - 1}{4}}}$

Subtract $1$ from $4$ .

$\frac{1}{\sqrt{\frac{3}{4}}}$

Should be rewritten $\sqrt{\frac{3}{4}}$ as $\frac{\sqrt{3}}{\sqrt{4}}$ .

$\frac{1}{\frac{\sqrt{3}}{\sqrt{4}}}$

Clarify the denominator.

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Rewrite $4$ as $2^{2}$ .

$\frac{1}{\frac{\sqrt{3}}{\sqrt{2^{2}}}}$

Withdraw terms from under the radical, assuming positive real numbers.

$\frac{1}{\frac{\sqrt{3}}{2}}$

Multiply the numerator by the denominator's reciprocal.

$1 \frac{2}{\sqrt{3}}$

Multiply $\frac{2}{\sqrt{3}}$ by $1$ .

$\frac{2}{\sqrt{3}}$

Multiply $\frac{2}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$

Add and then simplify the denominator.

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Multiply $\frac{2}{\sqrt{3}}$ and $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\frac{2 \sqrt{3}}{\sqrt{3} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$\frac{2 \sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$\frac{2 \sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{2 \sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Add $1$ and $1$ .

$\frac{2 \sqrt{3}}{{\sqrt{3}}^{2}}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$\frac{2 \sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

Use the power rule when multiplying the exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{2 \sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\frac{2 \sqrt{3}}{3^{\frac{2}{2}}}$

The common factor should be canceled of $2$ .

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Cancel the common factor.

$\frac{2 \sqrt{3}}{3^{\frac{2}{2}}}$

First, divide $1$ by $1$ .

$\frac{2 \sqrt{3}}{3^{1}}$

Solve the exponent.

$\frac{2 \sqrt{3}}{3}$

The result can be displayed in a variety of ways.

Exact Form:

$\frac{2 \sqrt{3}}{3}$

Decimal Form:

$1.15470053 \dots$

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