Find the value of `sin 750^(@)cos300^(@)+cos1470^(@)sin(-1020^(@))`

Answer

To solve the expression \( \sin 750^\circ \cos 300^\circ + \cos 1470^\circ \sin (-1020^\circ) \), we will evaluate each trigonometric function step by step.

### Step 1: Evaluate \( \sin 750^\circ \)

We can simplify \( 750^\circ \) as follows:
\[ 750^\circ = 720^\circ + 30^\circ = 2 \times 360^\circ + 30^\circ \]
Since \( \sin \) has a periodicity of \( 360^\circ \), we have:
\[ \sin 750^\circ = \sin 30^\circ \]
From trigonometric values, we know:
\[ \sin 30^\circ = \frac{1}{2} \]

### Step 2: Evaluate \( \cos 300^\circ \)

We can express \( 300^\circ \) as:
\[ 300^\circ = 360^\circ - 60^\circ \]
Since cosine is positive in the fourth quadrant:
\[ \cos 300^\circ = \cos 60^\circ \]
And from trigonometric values:
\[ \cos 60^\circ = \frac{1}{2} \]

### Step 3: Evaluate \( \cos 1470^\circ \)

We can simplify \( 1470^\circ \) as follows:
\[ 1470^\circ = 1440^\circ + 30^\circ = 4 \times 360^\circ + 30^\circ \]
Thus:
\[ \cos 1470^\circ = \cos 30^\circ \]
From trigonometric values:
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]

### Step 4: Evaluate \( \sin (-1020^\circ) \)

We can simplify \( -1020^\circ \) as follows:
\[ -1020^\circ = -1080^\circ + 60^\circ = -3 \times 360^\circ + 60^\circ \]
Thus:
\[ \sin (-1020^\circ) = \sin 60^\circ \]
From trigonometric values:
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

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

Now substituting all the evaluated values back into the original expression:
\[ \sin 750^\circ \cos 300^\circ + \cos 1470^\circ \sin (-1020^\circ) = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) \]

Calculating each term:
\[ = \frac{1}{4} + \frac{3}{4} = \frac{1 + 3}{4} = \frac{4}{4} = 1 \]

### Final Answer
The value of the expression \( \sin 750^\circ \cos 300^\circ + \cos 1470^\circ \sin (-1020^\circ) \) is \( 1 \).

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