If `sin(2a+45^(@))=cos(30^(@)-a)`, where `0^(@)lt90^(@)` , then the value of a is :

Options

A. `0^(@)`
B. `15^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

To solve the equation \( \sin(2a + 45^\circ) = \cos(30^\circ - a) \), we can follow these steps:

### Step 1: Use the co-function identity
We know that \( \cos(90^\circ - \theta) = \sin(\theta) \). Therefore, we can rewrite \( \cos(30^\circ - a) \) as:
\[ \cos(30^\circ - a) = \sin(90^\circ - (30^\circ - a)) = \sin(60^\circ + a) \]

### Step 2: Set the sine functions equal
Now we can rewrite the original equation:
\[ \sin(2a + 45^\circ) = \sin(60^\circ + a) \]

### Step 3: Equate the angles
Since the sine function is equal, we can set the angles equal to each other:
\[ 2a + 45^\circ = 60^\circ + a \]

### Step 4: Solve for \( a \)
Now, we can solve for \( a \):
1. Subtract \( a \) from both sides:
\[ 2a - a + 45^\circ = 60^\circ \]
This simplifies to:
\[ a + 45^\circ = 60^\circ \]

2. Subtract \( 45^\circ \) from both sides:
\[ a = 60^\circ - 45^\circ \]
Thus:
\[ a = 15^\circ \]

### Final Answer
The value of \( a \) is \( 15^\circ \).

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