Trigonometry Examples

$\sin (15) + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1

Simplify $\sin (15) + \tan (30) \cdot \cos (15)$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

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

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Step 1.1.1.1

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

$\sin (45 - 30) + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.2

Separate negation.

$\sin (45 - (30)) + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.3

Apply the difference of angles identity.

$\sin (45) \cos (30) - \cos (45) \sin (30) + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.4

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

$\frac{\sqrt{2}}{2} \cos (30) - \cos (45) \sin (30) + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.5

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

$\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \cos (45) \sin (30) + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.6

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

$\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \sin (30) + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.7

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

$\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.8

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

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Step 1.1.1.8.1

Simplify each term.

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Step 1.1.1.8.1.1

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

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Step 1.1.1.8.1.1.1

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

$\frac{\sqrt{2} \sqrt{3}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.8.1.1.2

Combine using the product rule for radicals.

$\frac{\sqrt{2 \cdot 3}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.8.1.1.3

Multiply $2$ by $3$ .

$\frac{\sqrt{6}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.8.1.1.4

Multiply $2$ by $2$ .

$\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.8.1.2

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

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Step 1.1.1.8.1.2.1

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

$\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{2 \cdot 2} + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.8.1.2.2

Multiply $2$ by $2$ .

$\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.1.8.2

Combine the numerators over the common denominator.

$\frac{\sqrt{6} - \sqrt{2}}{4} + \tan (30) \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.2

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

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot \cos (15) = \sqrt{\frac{6}{3}}$

Step 1.1.3

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

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Step 1.1.3.1

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

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot \cos (45 - 30) = \sqrt{\frac{6}{3}}$

Step 1.1.3.2

Separate negation.

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot \cos (45 - (30)) = \sqrt{\frac{6}{3}}$

Step 1.1.3.3

Apply the difference of angles identity $\cos (x - y) = \cos (x) \cos (y) + \sin (x) \sin (y)$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\cos (45) \cos (30) + \sin (45) \sin (30)) = \sqrt{\frac{6}{3}}$

Step 1.1.3.4

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

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\frac{\sqrt{2}}{2} \cos (30) + \sin (45) \sin (30)) = \sqrt{\frac{6}{3}}$

Step 1.1.3.5

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

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \sin (45) \sin (30)) = \sqrt{\frac{6}{3}}$

Step 1.1.3.6

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

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \sin (30)) = \sqrt{\frac{6}{3}}$

Step 1.1.3.7

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

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) = \sqrt{\frac{6}{3}}$

Step 1.1.3.8

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

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Step 1.1.3.8.1

Simplify each term.

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Step 1.1.3.8.1.1

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

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Step 1.1.3.8.1.1.1

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

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\frac{\sqrt{2} \sqrt{3}}{2 \cdot 2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) = \sqrt{\frac{6}{3}}$

Step 1.1.3.8.1.1.2

Combine using the product rule for radicals.

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\frac{\sqrt{2 \cdot 3}}{2 \cdot 2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) = \sqrt{\frac{6}{3}}$

Step 1.1.3.8.1.1.3

Multiply $2$ by $3$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\frac{\sqrt{6}}{2 \cdot 2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) = \sqrt{\frac{6}{3}}$

Step 1.1.3.8.1.1.4

Multiply $2$ by $2$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) = \sqrt{\frac{6}{3}}$

Step 1.1.3.8.1.2

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

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Step 1.1.3.8.1.2.1

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

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2 \cdot 2}) = \sqrt{\frac{6}{3}}$

Step 1.1.3.8.1.2.2

Multiply $2$ by $2$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}) = \sqrt{\frac{6}{3}}$

Step 1.1.3.8.2

Combine the numerators over the common denominator.

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3}}{3} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} = \sqrt{\frac{6}{3}}$

Step 1.1.4

Multiply $\frac{\sqrt{3}}{3} \cdot \frac{\sqrt{6} + \sqrt{2}}{4}$ .

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Step 1.1.4.1

Multiply $\frac{\sqrt{3}}{3}$ by $\frac{\sqrt{6} + \sqrt{2}}{4}$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3} (\sqrt{6} + \sqrt{2})}{3 \cdot 4} = \sqrt{\frac{6}{3}}$

Step 1.1.4.2

Multiply $3$ by $4$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3} (\sqrt{6} + \sqrt{2})}{12} = \sqrt{\frac{6}{3}}$

Step 1.1.5

Apply the distributive property.

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3} \sqrt{6} + \sqrt{3} \sqrt{2}}{12} = \sqrt{\frac{6}{3}}$

Step 1.1.6

Combine using the product rule for radicals.

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3 \cdot 6} + \sqrt{3} \sqrt{2}}{12} = \sqrt{\frac{6}{3}}$

Step 1.1.7

Combine using the product rule for radicals.

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3 \cdot 6} + \sqrt{3 \cdot 2}}{12} = \sqrt{\frac{6}{3}}$

Step 1.1.8

Simplify each term.

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Step 1.1.8.1

Multiply $3$ by $6$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{18} + \sqrt{3 \cdot 2}}{12} = \sqrt{\frac{6}{3}}$

Step 1.1.8.2

Rewrite $18$ as $3^{2} \cdot 2$ .

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Step 1.1.8.2.1

Factor $9$ out of $18$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{9 (2)} + \sqrt{3 \cdot 2}}{12} = \sqrt{\frac{6}{3}}$

Step 1.1.8.2.2

Rewrite $9$ as $3^{2}$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{3^{2} \cdot 2} + \sqrt{3 \cdot 2}}{12} = \sqrt{\frac{6}{3}}$

Step 1.1.8.3

Pull terms out from under the radical.

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{3 \sqrt{2} + \sqrt{3 \cdot 2}}{12} = \sqrt{\frac{6}{3}}$

Step 1.1.8.4

Multiply $3$ by $2$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{3 \sqrt{2} + \sqrt{6}}{12} = \sqrt{\frac{6}{3}}$

Step 1.2

To write $\frac{\sqrt{6} - \sqrt{2}}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} \cdot \frac{3}{3} + \frac{3 \sqrt{2} + \sqrt{6}}{12} = \sqrt{\frac{6}{3}}$

Step 1.3

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.1

Multiply $\frac{\sqrt{6} - \sqrt{2}}{4}$ by $\frac{3}{3}$ .

$\frac{(\sqrt{6} - \sqrt{2}) \cdot 3}{4 \cdot 3} + \frac{3 \sqrt{2} + \sqrt{6}}{12} = \sqrt{\frac{6}{3}}$

Step 1.3.2

Multiply $4$ by $3$ .

$\frac{(\sqrt{6} - \sqrt{2}) \cdot 3}{12} + \frac{3 \sqrt{2} + \sqrt{6}}{12} = \sqrt{\frac{6}{3}}$

Step 1.4

Combine the numerators over the common denominator.

$\frac{(\sqrt{6} - \sqrt{2}) \cdot 3 + 3 \sqrt{2} + \sqrt{6}}{12} = \sqrt{\frac{6}{3}}$

Step 1.5

Simplify the numerator.

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Step 1.5.1

Apply the distributive property.

$\frac{\sqrt{6} \cdot 3 - \sqrt{2} \cdot 3 + 3 \sqrt{2} + \sqrt{6}}{12} = \sqrt{\frac{6}{3}}$

Step 1.5.2

Move $3$ to the left of $\sqrt{6}$ .

$\frac{3 \cdot \sqrt{6} - \sqrt{2} \cdot 3 + 3 \sqrt{2} + \sqrt{6}}{12} = \sqrt{\frac{6}{3}}$

Step 1.5.3

Multiply $3$ by $- 1$ .

$\frac{3 \cdot \sqrt{6} - 3 \sqrt{2} + 3 \sqrt{2} + \sqrt{6}}{12} = \sqrt{\frac{6}{3}}$

Step 1.5.4

Add $3 \sqrt{6}$ and $\sqrt{6}$ .

$\frac{4 \sqrt{6} - 3 \sqrt{2} + 3 \sqrt{2}}{12} = \sqrt{\frac{6}{3}}$

Step 1.5.5

Add $- 3 \sqrt{2}$ and $3 \sqrt{2}$ .

$\frac{4 \sqrt{6} + 0}{12} = \sqrt{\frac{6}{3}}$

Step 1.5.6

Add $4 \sqrt{6}$ and $0$ .

$\frac{4 \sqrt{6}}{12} = \sqrt{\frac{6}{3}}$

Step 1.6

Cancel the common factor of $4$ and $12$ .

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Step 1.6.1

Factor $4$ out of $4 \sqrt{6}$ .

$\frac{4 (\sqrt{6})}{12} = \sqrt{\frac{6}{3}}$

Step 1.6.2

Cancel the common factors.

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Step 1.6.2.1

Factor $4$ out of $12$ .

$\frac{4 \sqrt{6}}{4 \cdot 3} = \sqrt{\frac{6}{3}}$

Step 1.6.2.2

Cancel the common factor.

$\frac{4 \sqrt{6}}{4 \cdot 3} = \sqrt{\frac{6}{3}}$

Step 1.6.2.3

Rewrite the expression.

$\frac{\sqrt{6}}{3} = \sqrt{\frac{6}{3}}$

Step 2

Divide $6$ by $3$ .

$\frac{\sqrt{6}}{3} = \sqrt{2}$

Step 3