If `cos(theta+31^(@))=sin47^(@)` , what is the value of `sin5theta` ?

Options

A. `(1)/(2)`
B. `(1)/(sqrt(2))`
C. `(sqrt(3))/(2)`
D. 0

Answer

To solve the equation \( \cos(\theta + 31^\circ) = \sin(47^\circ) \) and find the value of \( \sin(5\theta) \), we can follow these steps:

### Step 1: Use the co-function identity
We know that \( \sin(90^\circ - x) = \cos(x) \). Therefore, we can rewrite \( \sin(47^\circ) \) as:
\[ \sin(47^\circ) = \cos(90^\circ - 47^\circ) = \cos(43^\circ) \]
So, we can rewrite the original equation as:
\[ \cos(\theta + 31^\circ) = \cos(43^\circ) \]

### Step 2: Set up the equation
Since the cosine function is equal, we can set up the following equations:
\[ \theta + 31^\circ = 43^\circ + 360^\circ n \quad \text{or} \quad \theta + 31^\circ = -43^\circ + 360^\circ n \]
where \( n \) is any integer. For simplicity, we will consider \( n = 0 \).

### Step 3: Solve for \( \theta \)
From the first equation:
\[ \theta + 31^\circ = 43^\circ \]
Subtract \( 31^\circ \) from both sides:
\[ \theta = 43^\circ - 31^\circ = 12^\circ \]

From the second equation:
\[ \theta + 31^\circ = -43^\circ \]
Subtract \( 31^\circ \) from both sides:
\[ \theta = -43^\circ - 31^\circ = -74^\circ \]
Since we are looking for a principal angle, we will use \( \theta = 12^\circ \).

### Step 4: Find \( \sin(5\theta) \)
Now, we need to calculate \( \sin(5\theta) \):
\[ 5\theta = 5 \times 12^\circ = 60^\circ \]
Thus, we find:
\[ \sin(5\theta) = \sin(60^\circ) \]

### Step 5: Calculate \( \sin(60^\circ) \)
We know that:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Final Answer
So, the value of \( \sin(5\theta) \) is:
\[ \sin(5\theta) = \frac{\sqrt{3}}{2} \]

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