Trigonometry Examples

$\sin (\frac{4 π}{3})$

Step 1

The angle $\frac{4 π}{3}$ is an angle where the values of the six trigonometric functions are known. Because this is the case, add $0$ to keep the value the same.

$\sin (\frac{4 π}{3} + 0)$

Step 2

Use the sum formula for sine to simplify the expression. The formula states that $\sin (A + B) = \sin (A) \cos (B) + \cos (A) \sin (B)$ .

$\sin (\frac{4 π}{3}) \cos (0) + \cos (\frac{4 π}{3}) \sin (0)$

Step 3

Remove parentheses.

$\sin (\frac{4 π}{3}) \cos (0) + \cos (\frac{4 π}{3}) \sin (0)$

Step 4

Simplify each term.

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Step 4.1

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 third quadrant.

$- \sin (\frac{π}{3}) \cos (0) + \cos (\frac{4 π}{3}) \sin (0)$

Step 4.2

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

$- \frac{\sqrt{3}}{2} \cos (0) + \cos (\frac{4 π}{3}) \sin (0)$

Step 4.3

The exact value of $\cos (0)$ is $1$ .

$- \frac{\sqrt{3}}{2} \cdot 1 + \cos (\frac{4 π}{3}) \sin (0)$

Step 4.4

Multiply $- 1$ by $1$ .

$- \frac{\sqrt{3}}{2} + \cos (\frac{4 π}{3}) \sin (0)$

Step 4.5

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.

$- \frac{\sqrt{3}}{2} - \cos (\frac{π}{3}) \sin (0)$

Step 4.6

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$- \frac{\sqrt{3}}{2} - \frac{1}{2} \sin (0)$

Step 4.7

The exact value of $\sin (0)$ is $0$ .

$- \frac{\sqrt{3}}{2} - \frac{1}{2} \cdot 0$

Step 4.8

Multiply $- \frac{1}{2} \cdot 0$ .

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Step 4.8.1

Multiply $0$ by $- 1$ .

$- \frac{\sqrt{3}}{2} + 0 (\frac{1}{2})$

Step 4.8.2

Multiply $0$ by $\frac{1}{2}$ .

$- \frac{\sqrt{3}}{2} + 0$

Step 5

Add $- \frac{\sqrt{3}}{2}$ and $0$ .

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$- 0.86602540 \dots$