Precalculus Examples

$\sin (15)$

Step 1

Convert the radian measure to degrees.

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Step 1.1

To convert radians to degrees, multiply by $\frac{180}{π}$ , since a full circle is $360 °$ or $2 π$ radians.

$(\sin (15)) \cdot \frac{180 °}{π}$

Step 1.2

The exact value of $\sin (15)$ is $\frac{\sqrt{6} - \sqrt{2}}{4}$ .

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Step 1.2.1

Split $15$ into two angles where the values of the six trigonometric functions are known.

$\sin (45 - 30) \cdot \frac{180}{π}$

Step 1.2.2

Separate negation.

$\sin (45 - (30)) \cdot \frac{180}{π}$

Step 1.2.3

Apply the difference of angles identity.

$(\sin (45) \cos (30) - \cos (45) \sin (30)) \cdot \frac{180}{π}$

Step 1.2.4

The exact value of $\sin (45)$ is $\frac{\sqrt{2}}{2}$ .

$(\frac{\sqrt{2}}{2} \cos (30) - \cos (45) \sin (30)) \cdot \frac{180}{π}$

Step 1.2.5

The exact value of $\cos (30)$ is $\frac{\sqrt{3}}{2}$ .

$(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \cos (45) \sin (30)) \cdot \frac{180}{π}$

Step 1.2.6

The exact value of $\cos (45)$ is $\frac{\sqrt{2}}{2}$ .

$(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \sin (30)) \cdot \frac{180}{π}$

Step 1.2.7

The exact value of $\sin (30)$ is $\frac{1}{2}$ .

$(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \cdot \frac{180}{π}$

Step 1.2.8

Simplify $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$ .

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Step 1.2.8.1

Simplify each term.

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Step 1.2.8.1.1

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

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Step 1.2.8.1.1.1

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

$(\frac{\sqrt{2} \sqrt{3}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \cdot \frac{180}{π}$

Step 1.2.8.1.1.2

Combine using the product rule for radicals.

$(\frac{\sqrt{2 \cdot 3}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \cdot \frac{180}{π}$

Step 1.2.8.1.1.3

Multiply $2$ by $3$ .

$(\frac{\sqrt{6}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \cdot \frac{180}{π}$

Step 1.2.8.1.1.4

Multiply $2$ by $2$ .

$(\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \cdot \frac{180}{π}$

Step 1.2.8.1.2

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

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Step 1.2.8.1.2.1

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

$(\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{2 \cdot 2}) \cdot \frac{180}{π}$

Step 1.2.8.1.2.2

Multiply $2$ by $2$ .

$(\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}) \cdot \frac{180}{π}$

Step 1.2.8.2

Combine the numerators over the common denominator.

$\frac{\sqrt{6} - \sqrt{2}}{4} \cdot \frac{180}{π}$

Step 1.3

Cancel the common factor of $4$ .

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Step 1.3.1

Factor $4$ out of $180$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} \cdot \frac{4 (45)}{π}$

Step 1.3.2

Cancel the common factor.

$\frac{\sqrt{6} - \sqrt{2}}{4} \cdot \frac{4 \cdot 45}{π}$

Step 1.3.3

Rewrite the expression.

$(\sqrt{6} - \sqrt{2}) \cdot \frac{45}{π}$

Step 1.4

Apply the distributive property.

$\sqrt{6} \frac{45}{π} - \sqrt{2} \frac{45}{π}$

Step 1.5

Combine $\sqrt{6}$ and $\frac{45}{π}$ .

$\frac{\sqrt{6} \cdot 45}{π} - \sqrt{2} \frac{45}{π}$

Step 1.6

Combine $\frac{45}{π}$ and $\sqrt{2}$ .

$\frac{\sqrt{6} \cdot 45}{π} - \frac{45 \sqrt{2}}{π}$

Step 1.7

Move $45$ to the left of $\sqrt{6}$ .

$\frac{45 \sqrt{6}}{π} - \frac{45 \sqrt{2}}{π}$

Step 1.8

$π$ is approximately equal to $3.14159265$ .

$\frac{45 \sqrt{6}}{3.14159265} - \frac{45 \sqrt{2}}{3.14159265}$

Step 1.9

Convert to a decimal.

$14.82923894 °$

Step 2

The angle is in the first quadrant.

Quadrant $1$

Step 3