Trigonometry Examples

$\tan (- \frac{7 π}{12})$

Step 1

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\tan (\frac{17 π}{12})$

Step 2

The exact value of $\tan (\frac{17 π}{12})$ is $\sqrt{7 + 4 \sqrt{3}}$ .

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Step 2.1

Rewrite $\frac{17 π}{12}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$\tan (\frac{\frac{17 π}{6}}{2})$

Step 2.2

Apply the tangent half-angle identity.

$\pm \sqrt{\frac{1 - \cos (\frac{17 π}{6})}{1 + \cos (\frac{17 π}{6})}}$

Step 2.3

Change the $\pm$ to $+$ because tangent is positive in the third quadrant.

$\sqrt{\frac{1 - \cos (\frac{17 π}{6})}{1 + \cos (\frac{17 π}{6})}}$

Step 2.4

Simplify $\sqrt{\frac{1 - \cos (\frac{17 π}{6})}{1 + \cos (\frac{17 π}{6})}}$ .

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Step 2.4.1

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\sqrt{\frac{1 - \cos (\frac{5 π}{6})}{1 + \cos (\frac{17 π}{6})}}$

Step 2.4.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$\sqrt{\frac{1 - - \cos (\frac{π}{6})}{1 + \cos (\frac{17 π}{6})}}$

Step 2.4.3

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

$\sqrt{\frac{1 - - \frac{\sqrt{3}}{2}}{1 + \cos (\frac{17 π}{6})}}$

Step 2.4.4

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

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Step 2.4.4.1

Multiply $- 1$ by $- 1$ .

$\sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{1 + \cos (\frac{17 π}{6})}}$

Step 2.4.4.2

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

$\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 + \cos (\frac{17 π}{6})}}$

Step 2.4.5

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

$\sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{1 + \cos (\frac{17 π}{6})}}$

Step 2.4.6

Combine the numerators over the common denominator.

$\sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 + \cos (\frac{17 π}{6})}}$

Step 2.4.7

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 + \cos (\frac{5 π}{6})}}$

Step 2.4.8

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$\sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 - \cos (\frac{π}{6})}}$

Step 2.4.9

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

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

Step 2.4.10

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

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

Step 2.4.11

Combine the numerators over the common denominator.

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

Step 2.4.12

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.4.13

Cancel the common factor of $2$ .

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Step 2.4.13.1

Cancel the common factor.

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

Step 2.4.13.2

Rewrite the expression.

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

Step 2.4.14

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

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

Step 2.4.15

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

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

Step 2.4.16

Expand the denominator using the FOIL method.

$\sqrt{(2 + \sqrt{3}) \frac{2 + \sqrt{3}}{4 + 2 \sqrt{3} - 2 \sqrt{3} - {\sqrt{3}}^{2}}}$

Step 2.4.17

Simplify.

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

Step 2.4.18

Divide $2 + \sqrt{3}$ by $1$ .

$\sqrt{(2 + \sqrt{3}) (2 + \sqrt{3})}$

Step 2.4.19

Expand $(2 + \sqrt{3}) (2 + \sqrt{3})$ using the FOIL Method.

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Step 2.4.19.1

Apply the distributive property.

$\sqrt{2 (2 + \sqrt{3}) + \sqrt{3} (2 + \sqrt{3})}$

Step 2.4.19.2

Apply the distributive property.

$\sqrt{2 \cdot 2 + 2 \sqrt{3} + \sqrt{3} (2 + \sqrt{3})}$

Step 2.4.19.3

Apply the distributive property.

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

Step 2.4.20

Simplify and combine like terms.

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Step 2.4.20.1

Simplify each term.

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Step 2.4.20.1.1

Multiply $2$ by $2$ .

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

Step 2.4.20.1.2

Move $2$ to the left of $\sqrt{3}$ .

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

Step 2.4.20.1.3

Combine using the product rule for radicals.

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

Step 2.4.20.1.4

Multiply $3$ by $3$ .

$\sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{9}}$

Step 2.4.20.1.5

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

$\sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{3^{2}}}$

Step 2.4.20.1.6

Pull terms out from under the radical, assuming positive real numbers.

$\sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + 3}$

Step 2.4.20.2

Add $4$ and $3$ .

$\sqrt{7 + 2 \sqrt{3} + 2 \sqrt{3}}$

Step 2.4.20.3

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

$\sqrt{7 + 4 \sqrt{3}}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$\sqrt{7 + 4 \sqrt{3}}$

Decimal Form:

$3.73205080 \dots$