Precalculus Examples

$\cos (\frac{13 π}{6}) - \tan (\frac{8 π}{3})$

Step 1

Simplify each term.

Tap for more steps...

Step 1.1

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

$\cos (\frac{π}{6}) - \tan (\frac{8 π}{3})$

Step 1.2

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

$\frac{\sqrt{3}}{2} - \tan (\frac{8 π}{3})$

Step 1.3

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

$\frac{\sqrt{3}}{2} - \tan (\frac{2 π}{3})$

Step 1.4

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

$\frac{\sqrt{3}}{2} - - \tan (\frac{π}{3})$

Step 1.5

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

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

Step 1.6

Multiply $- - \sqrt{3}$ .

Tap for more steps...

Step 1.6.1

Multiply $- 1$ by $- 1$ .

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

Step 1.6.2

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

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

Step 2

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

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

Step 3

Combine fractions.

Tap for more steps...

Step 3.1

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

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

Step 3.2

Combine the numerators over the common denominator.

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

Step 4

Simplify the numerator.

Tap for more steps...

Step 4.1

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

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

Step 4.2

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

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

Step 5

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$2.59807621 \dots$