Value of `sin 47^(@) + sin 61^(@)- sin 11 ^(@)-sin 25 ^(@)` is

Options

A. `cos 7 ^(@)`
B. `sin 7 ^(@)`
C. `sin 61^(@)`
D. `-sin 25 ^(@)`

Answer

To solve the expression \( \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ \), we can use the sine subtraction formula and some trigonometric identities. Here’s the step-by-step solution:

### Step 1: Group the Sine Terms
We can group the terms in pairs:
\[ (\sin 47^\circ - \sin 11^\circ) + (\sin 61^\circ - \sin 25^\circ) \]

### Step 2: Apply the Sine Subtraction Formula
We will use the formula for the difference of sines:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]

#### For \( \sin 47^\circ - \sin 11^\circ \):
Let \( A = 47^\circ \) and \( B = 11^\circ \):
\[ \sin 47^\circ - \sin 11^\circ = 2 \cos\left(\frac{47^\circ + 11^\circ}{2}\right) \sin\left(\frac{47^\circ - 11^\circ}{2}\right) \]
Calculating the averages:
\[ \frac{47^\circ + 11^\circ}{2} = \frac{58^\circ}{2} = 29^\circ \]
\[ \frac{47^\circ - 11^\circ}{2} = \frac{36^\circ}{2} = 18^\circ \]
Thus,
\[ \sin 47^\circ - \sin 11^\circ = 2 \cos(29^\circ) \sin(18^\circ) \]

#### For \( \sin 61^\circ - \sin 25^\circ \):
Let \( A = 61^\circ \) and \( B = 25^\circ \):
\[ \sin 61^\circ - \sin 25^\circ = 2 \cos\left(\frac{61^\circ + 25^\circ}{2}\right) \sin\left(\frac{61^\circ - 25^\circ}{2}\right) \]
Calculating the averages:
\[ \frac{61^\circ + 25^\circ}{2} = \frac{86^\circ}{2} = 43^\circ \]
\[ \frac{61^\circ - 25^\circ}{2} = \frac{36^\circ}{2} = 18^\circ \]
Thus,
\[ \sin 61^\circ - \sin 25^\circ = 2 \cos(43^\circ) \sin(18^\circ) \]

### Step 3: Combine the Results
Now we can combine the results:
\[ \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2 \cos(29^\circ) \sin(18^\circ) + 2 \cos(43^\circ) \sin(18^\circ) \]
Factoring out \( 2 \sin(18^\circ) \):
\[ = 2 \sin(18^\circ) \left( \cos(29^\circ) + \cos(43^\circ) \right) \]

### Step 4: Use the Cosine Addition Formula
Using the cosine addition formula:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Let \( A = 29^\circ \) and \( B = 43^\circ \):
\[ \cos(29^\circ) + \cos(43^\circ) = 2 \cos\left(\frac{29^\circ + 43^\circ}{2}\right) \cos\left(\frac{29^\circ - 43^\circ}{2}\right) \]
Calculating the averages:
\[ \frac{29^\circ + 43^\circ}{2} = \frac{72^\circ}{2} = 36^\circ \]
\[ \frac{29^\circ - 43^\circ}{2} = \frac{-14^\circ}{2} = -7^\circ \]
Thus,
\[ \cos(29^\circ) + \cos(43^\circ) = 2 \cos(36^\circ) \cos(-7^\circ) = 2 \cos(36^\circ) \cos(7^\circ) \]

### Step 5: Substitute Back
Now substituting back:
\[ \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2 \sin(18^\circ) \cdot 2 \cos(36^\circ) \cos(7^\circ) \]
\[ = 4 \sin(18^\circ) \cos(36^\circ) \cos(7^\circ) \]

### Step 6: Final Simplification
Using the identity \( \sin(18^\circ) = \frac{\sqrt{5}-1}{4} \) and \( \cos(36^\circ) = \frac{\sqrt{5}+1}{4} \):
\[ = 4 \cdot \frac{\sqrt{5}-1}{4} \cdot \frac{\sqrt{5}+1}{4} \cdot \cos(7^\circ) \]
\[ = (\sqrt{5}-1)(\sqrt{5}+1) \cdot \cos(7^\circ) \]
\[ = (5 - 1) \cdot \cos(7^\circ) = 4 \cos(7^\circ) \]

Thus, the final answer is:
\[ \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 4 \cos(7^\circ) \]