Trigonometry Examples

$\cos (120)$

Step 1

The angle $120$ 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.

$\cos (120 + 0)$

Step 2

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

$\cos (0) \cdot \cos (120) - \sin (0) \cdot \sin (120)$

Step 3

Remove parentheses.

$\cos (0) \cdot \cos (120) - \sin (0) \cdot \sin (120)$

Step 4

Simplify each term.

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Step 4.1

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

$1 \cdot \cos (120) - \sin (0) \cdot \sin (120)$

Step 4.2

Multiply $\cos (120)$ by $1$ .

$\cos (120) - \sin (0) \cdot \sin (120)$

Step 4.3

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.

$- \cos (60) - \sin (0) \cdot \sin (120)$

Step 4.4

The exact value of $\cos (60)$ is $\frac{1}{2}$ .

$- \frac{1}{2} - \sin (0) \cdot \sin (120)$

Step 4.5

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

$- \frac{1}{2} - 0 \cdot \sin (120)$

Step 4.6

Multiply $- 1$ by $0$ .

$- \frac{1}{2} + 0 \cdot \sin (120)$

Step 4.7

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$- \frac{1}{2} + 0 \cdot \sin (60)$

Step 4.8

The exact value of $\sin (60)$ is $\frac{\sqrt{3}}{2}$ .

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

Step 4.9

Multiply $0$ by $\frac{\sqrt{3}}{2}$ .

$- \frac{1}{2} + 0$

Step 5

Add $- \frac{1}{2}$ and $0$ .

$- \frac{1}{2}$