(a) [tex] \sin \left(45^{\circ} - 30^{\circ}\right) = \sin 45^{\circ} \cdot \cos 30^{\circ} - \cos 45^{\circ} \cdot \sin 30^{\circ} [/tex]



Answer :

Sure! Let's solve the expression \(\sin \left(45^{\circ} - 30^{\circ}\right) = \sin 45^{\circ} \cdot \cos 30^{\circ} - \cos 45^{\circ} \cdot \sin 30^{\circ}\) step by step.

### Step 1: Understand the Expression

We need to verify the formula:
[tex]\[ \sin \left(45^{\circ} - 30^{\circ}\right) = \sin 45^{\circ} \cdot \cos 30^{\circ} - \cos 45^{\circ} \cdot \sin 30^{\circ} \][/tex]

### Step 2: Calculate the Individual Trigonometric Values

#### For \( \sin 45^\circ \):
[tex]\[ \sin 45^\circ \approx 0.7071 \][/tex]

#### For \( \cos 30^\circ \):
[tex]\[ \cos 30^\circ \approx 0.8660 \][/tex]

#### For \( \cos 45^\circ \):
[tex]\[ \cos 45^\circ \approx 0.7071 \][/tex]

#### For \( \sin 30^\circ \):
[tex]\[ \sin 30^\circ \approx 0.5000 \][/tex]

### Step 3: Substitute the Values into the Expression

Substitute these values into the given equation:

[tex]\[ \sin 45^\circ \cdot \cos 30^\circ - \cos 45^\circ \cdot \sin 30^\circ \][/tex]

Substitute the known values:

[tex]\[ 0.7071 \cdot 0.8660 - 0.7071 \cdot 0.5000 \][/tex]

### Step 4: Perform the Multiplication

Calculate each product:

[tex]\[ 0.7071 \cdot 0.8660 \approx 0.6124 \][/tex]

[tex]\[ 0.7071 \cdot 0.5000 \approx 0.3536 \][/tex]

### Step 5: Subtract the Results

Subtract the second product from the first product:

[tex]\[ 0.6124 - 0.3536 \approx 0.2588 \][/tex]

### Step 6: Verify the Result

Thus, we have:

[tex]\[ \sin \left(45^{\circ} - 30^{\circ} \right) = 0.2588 \][/tex]

After evaluating all the steps, the values calculated are:
[tex]\[ \sin 45^\circ \approx 0.7071, \quad \cos 30^\circ \approx 0.8660, \quad \cos 45^\circ \approx 0.7071, \quad \sin 30^\circ \approx 0.5000 \][/tex]

Finally, we conclude:

[tex]\[ \sin \left(45^{\circ} - 30^{\circ} \right) = \sin 45^{\circ} \cdot \cos 30^{\circ} - \cos 45^{\circ} \cdot \sin 30^{\circ} \approx 0.2588 \][/tex]

Therefore, the given trigonometric identity holds true.