The value of
`sin 600^(@) cos 330^(@)+ cos 120^(@) sin 150^(@)` is

Options

A. `-1`
B. 1
C. `(1)/(sqrt2)`
D. `(sqrt3)/(2)`

Answer

To solve the expression \( \sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ \), we can break it down step by step.

### Step 1: Calculate \( \sin 600^\circ \)

To find \( \sin 600^\circ \), we can use the periodic property of sine:
\[ 600^\circ = 360^\circ + 240^\circ \]
Thus,
\[ \sin 600^\circ = \sin(360^\circ + 240^\circ) = \sin 240^\circ \]
From the unit circle, we know:
\[ \sin 240^\circ = -\sin 60^\circ = -\frac{\sqrt{3}}{2} \]

### Step 2: Calculate \( \cos 330^\circ \)

Next, we find \( \cos 330^\circ \):
\[ \cos 330^\circ = \cos(360^\circ - 30^\circ) = \cos 30^\circ \]
From the unit circle, we know:
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]

### Step 3: Calculate \( \cos 120^\circ \)

Now, we calculate \( \cos 120^\circ \):
\[ \cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ \]
From the unit circle, we know:
\[ \cos 60^\circ = \frac{1}{2} \implies \cos 120^\circ = -\frac{1}{2} \]

### Step 4: Calculate \( \sin 150^\circ \)

Next, we find \( \sin 150^\circ \):
\[ \sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ \]
From the unit circle, we know:
\[ \sin 30^\circ = \frac{1}{2} \]

### Step 5: Substitute the values into the expression

Now we can substitute the values we found into the original expression:
\[ \sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ = \left(-\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2}\right) \left(\frac{1}{2}\right) \]

### Step 6: Simplify the expression

Calculating each term:
1. \( -\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{3}{4} \)
2. \( -\frac{1}{2} \cdot \frac{1}{2} = -\frac{1}{4} \)

Combining these:
\[ -\frac{3}{4} - \frac{1}{4} = -\frac{4}{4} = -1 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{-1} \]