Evaluate :
`(cos3A-3cos4A)/(sin3A+2sin4A)," when "A=15^(@)`.

Answer

To evaluate the expression \((\cos 3A - 3\cos 4A) / (\sin 3A + 2\sin 4A)\) when \(A = 15^\circ\), we will follow these steps:

### Step 1: Substitute \(A\) in the expression
We start by substituting \(A = 15^\circ\) into the expression:
\[ \cos(3A) = \cos(3 \times 15^\circ) = \cos(45^\circ) \]
\[ \cos(4A) = \cos(4 \times 15^\circ) = \cos(60^\circ) \]
\[ \sin(3A) = \sin(3 \times 15^\circ) = \sin(45^\circ) \]
\[ \sin(4A) = \sin(4 \times 15^\circ) = \sin(60^\circ) \]

### Step 2: Write the expression with substituted values
Now we can rewrite the expression:
\[ \frac{\cos(45^\circ) - 3\cos(60^\circ)}{\sin(45^\circ) + 2\sin(60^\circ)} \]

### Step 3: Use known values of trigonometric functions
We know the following values:
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}}, \quad \cos(60^\circ) = \frac{1}{2}, \quad \sin(45^\circ) = \frac{1}{\sqrt{2}}, \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Step 4: Substitute the known values into the expression
Substituting these values into the expression gives:
\[ \frac{\frac{1}{\sqrt{2}} - 3 \times \frac{1}{2}}{\frac{1}{\sqrt{2}} + 2 \times \frac{\sqrt{3}}{2}} \]

### Step 5: Simplify the numerator and denominator
Calculating the numerator:
\[ \frac{1}{\sqrt{2}} - \frac{3}{2} = \frac{1 - \frac{3\sqrt{2}}{2}}{\sqrt{2}} = \frac{2 - 3\sqrt{2}}{2\sqrt{2}} \]

Calculating the denominator:
\[ \frac{1}{\sqrt{2}} + \sqrt{3} = \frac{1 + \sqrt{6}}{\sqrt{2}} \]

### Step 6: Combine the results
Now we can combine the results:
\[ \frac{\frac{2 - 3\sqrt{2}}{2\sqrt{2}}}{\frac{1 + \sqrt{6}}{\sqrt{2}}} = \frac{2 - 3\sqrt{2}}{2(1 + \sqrt{6})} \]

### Step 7: Rationalize if necessary
To rationalize, we can multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{(2 - 3\sqrt{2})(1 - \sqrt{6})}{2((1 + \sqrt{6})(1 - \sqrt{6}))} \]
The denominator simplifies to:
\[ 2(1 - 6) = -10 \]

### Final Expression
Thus, the final expression becomes:
\[ \frac{(2 - 3\sqrt{2})(1 - \sqrt{6})}{-10} \]