Precalculus Examples

$\sin (150 °)$

Step 1

The angle $150 °$ 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.

$\sin (150 ° + 0)$

Step 2

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

$\sin (150 °) \cos (0) + \cos (150 °) \sin (0)$

Step 3

Remove parentheses.

$\sin (150 °) \cos (0) + \cos (150 °) \sin (0)$

Step 4

Simplify each term.

Tap for more steps...

Step 4.1

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

$\sin (30) \cos (0) + \cos (150 °) \sin (0)$

Step 4.2

The exact value of $\sin (30)$ is $\frac{1}{2}$ .

$\frac{1}{2} \cos (0) + \cos (150 °) \sin (0)$

Step 4.3

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

$\frac{1}{2} \cdot 1 + \cos (150 °) \sin (0)$

Step 4.4

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

$\frac{1}{2} + \cos (150 °) \sin (0)$

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 second quadrant.

$\frac{1}{2} - \cos (30) \sin (0)$

Step 4.6

The exact value of $\cos (30)$ is $\frac{\sqrt{3}}{2}$ .

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

Step 4.7

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

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

Step 4.8

Multiply $- \frac{\sqrt{3}}{2} \cdot 0$ .

Tap for more steps...

Step 4.8.1

Multiply $0$ by $- 1$ .

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

Step 4.8.2

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

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

Step 5

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

$\frac{1}{2}$