Algebra Examples

$\tan (\frac{π}{6}) + \sin (\frac{5 π}{3}) \cdot \cos (- \frac{3 π}{4})$

Step 1

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

$\frac{\sqrt{3}}{3} + \sin (\frac{5 π}{3}) \cdot \cos (- \frac{3 π}{4})$

Step 2

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

$\frac{\sqrt{3}}{3} - \sin (\frac{π}{3}) \cdot \cos (- \frac{3 π}{4})$

Step 3

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

$\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{2} \cdot \cos (- \frac{3 π}{4})$

Step 4

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

$\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{2} \cdot \cos (\frac{5 π}{4})$

Step 5

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 third quadrant.

$\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{2} \cdot (- \cos (\frac{π}{4}))$

Step 6

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

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

Step 7

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

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Step 7.1

Multiply $- 1$ by $- 1$ .

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

Step 7.2

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

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

Step 7.3

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

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

Step 7.4

Combine using the product rule for radicals.

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

Step 7.5

Multiply $3$ by $2$ .

$\frac{\sqrt{3}}{3} + \frac{\sqrt{6}}{2 \cdot 2}$

Step 7.6

Multiply $2$ by $2$ .

$\frac{\sqrt{3}}{3} + \frac{\sqrt{6}}{4}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{\sqrt{3}}{3} + \frac{\sqrt{6}}{4}$

Decimal Form:

$1.18972270 \dots$