Precalculus Examples

$\csc (\frac{5 π}{12})$

Step 1

Replace $\csc (\frac{5 π}{12})$ with an equivalent expression $\frac{1}{\sin (\frac{5 π}{12})}$ using the fundamental identities.

$\frac{1}{\sin (\frac{5 π}{12})}$

Step 2

Use a sum or difference formula on the denominator.

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Step 2.1

First, split the angle into two angles where the values of the six trigonometric functions are known. In this case, $\frac{5 π}{12}$ can be split into $\frac{π}{6} + \frac{π}{4}$ .

$\frac{1}{\sin (\frac{π}{6} + \frac{π}{4})}$

Step 2.2

Use the sum formula for sine to simplify the expression. The formula states that $\sin (A + B) = \sin (A) \cos (B) + \cos (A) \sin (B)$ .

$\frac{1}{\sin (\frac{π}{6}) \cos (\frac{π}{4}) + \cos (\frac{π}{6}) \sin (\frac{π}{4})}$

Step 2.3

Remove parentheses.

$\frac{1}{\sin (\frac{π}{6}) \cos (\frac{π}{4}) + \cos (\frac{π}{6}) \sin (\frac{π}{4})}$

Step 2.4

Simplify each term.

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Step 2.4.1

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$\frac{1}{\frac{1}{2} \cdot \cos (\frac{π}{4}) + \cos (\frac{π}{6}) \sin (\frac{π}{4})}$

Step 2.4.2

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$\frac{1}{\frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \cos (\frac{π}{6}) \sin (\frac{π}{4})}$

Step 2.4.3

Multiply $\frac{1}{2} \cdot \frac{\sqrt{2}}{2}$ .

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Step 2.4.3.1

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

$\frac{1}{\frac{\sqrt{2}}{2 \cdot 2} + \cos (\frac{π}{6}) \sin (\frac{π}{4})}$

Step 2.4.3.2

Multiply $2$ by $2$ .

$\frac{1}{\frac{\sqrt{2}}{4} + \cos (\frac{π}{6}) \sin (\frac{π}{4})}$

Step 2.4.4

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$\frac{1}{\frac{\sqrt{2}}{4} + \frac{\sqrt{3}}{2} \cdot \sin (\frac{π}{4})}$

Step 2.4.5

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$\frac{1}{\frac{\sqrt{2}}{4} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}}$

Step 2.4.6

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

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Step 2.4.6.1

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

$\frac{1}{\frac{\sqrt{2}}{4} + \frac{\sqrt{3} \sqrt{2}}{2 \cdot 2}}$

Step 2.4.6.2

Combine using the product rule for radicals.

$\frac{1}{\frac{\sqrt{2}}{4} + \frac{\sqrt{3 \cdot 2}}{2 \cdot 2}}$

Step 2.4.6.3

Multiply $3$ by $2$ .

$\frac{1}{\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{2 \cdot 2}}$

Step 2.4.6.4

Multiply $2$ by $2$ .

$\frac{1}{\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}}$

Step 3

Simplify.

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Step 3.1

Combine the numerators over the common denominator.

$\frac{1}{\frac{\sqrt{2} + \sqrt{6}}{4}}$

Step 3.2

Multiply the numerator by the reciprocal of the denominator.

$1 \frac{4}{\sqrt{2} + \sqrt{6}}$

Step 3.3

Multiply $\frac{4}{\sqrt{2} + \sqrt{6}}$ by $1$ .

$\frac{4}{\sqrt{2} + \sqrt{6}}$

Step 3.4

Multiply $\frac{4}{\sqrt{2} + \sqrt{6}}$ by $\frac{\sqrt{2} - \sqrt{6}}{\sqrt{2} - \sqrt{6}}$ .

$\frac{4}{\sqrt{2} + \sqrt{6}} \cdot \frac{\sqrt{2} - \sqrt{6}}{\sqrt{2} - \sqrt{6}}$

Step 3.5

Multiply $\frac{4}{\sqrt{2} + \sqrt{6}}$ by $\frac{\sqrt{2} - \sqrt{6}}{\sqrt{2} - \sqrt{6}}$ .

$\frac{4 (\sqrt{2} - \sqrt{6})}{(\sqrt{2} + \sqrt{6}) (\sqrt{2} - \sqrt{6})}$

Step 3.6

Expand the denominator using the FOIL method.

$\frac{4 (\sqrt{2} - \sqrt{6})}{{\sqrt{2}}^{2} - \sqrt{12} + \sqrt{12} - {\sqrt{6}}^{2}}$

Step 3.7

Simplify.

$\frac{4 (\sqrt{2} - \sqrt{6})}{- 4}$

Step 3.8

Simplify the expression.

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Step 3.8.1

Move the negative one from the denominator of $\frac{\sqrt{2} - \sqrt{6}}{- 1}$ .

$- 1 \cdot (\sqrt{2} - \sqrt{6})$

Step 3.8.2

Rewrite $- 1 \cdot (\sqrt{2} - \sqrt{6})$ as $- (\sqrt{2} - \sqrt{6})$ .

$- (\sqrt{2} - \sqrt{6})$

Step 3.9

Apply the distributive property.

$- \sqrt{2} - - \sqrt{6}$

Step 3.10

Multiply $- - \sqrt{6}$ .

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Step 3.10.1

Multiply $- 1$ by $- 1$ .

$- \sqrt{2} + 1 \sqrt{6}$

Step 3.10.2

Multiply $\sqrt{6}$ by $1$ .

$- \sqrt{2} + \sqrt{6}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$- \sqrt{2} + \sqrt{6}$

Decimal Form:

$1.03527618 \dots$