If `5cosx+12cosy=13`, then the maximum value of `5sinx+12siny` is (A) `12` (B) `sqrt(120)` (C) `sqrt(20)` (D) 13

Options

A. 12
B. `sqrt(120)`
C. `sqrt20`
D. 13

Answer

To solve the problem, we need to find the maximum value of \( 5\sin x + 12\sin y \) given that \( 5\cos x + 12\cos y = 13 \).

### Step 1: Square the given equation
We start with the equation:
\[ 5\cos x + 12\cos y = 13 \]
We square both sides:
\[ (5\cos x + 12\cos y)^2 = 13^2 \]
This expands to:
\[ 25\cos^2 x + 144\cos^2 y + 120\cos x \cos y = 169 \]

### Step 2: Assume a new variable
Let:
\[ a = 5\sin x + 12\sin y \]
We will also square this expression:
\[ (5\sin x + 12\sin y)^2 = a^2 \]
Expanding this gives:
\[ 25\sin^2 x + 144\sin^2 y + 120\sin x \sin y = a^2 \]

### Step 3: Combine the equations
Now, we will add the squared equations from Step 1 and Step 2:
\[ (25\cos^2 x + 144\cos^2 y + 120\cos x \cos y) + (25\sin^2 x + 144\sin^2 y + 120\sin x \sin y) = 169 + a^2 \]

### Step 4: Use the Pythagorean identity
Using the identity \( \cos^2 x + \sin^2 x = 1 \) and \( \cos^2 y + \sin^2 y = 1 \), we can simplify:
\[ 25(\cos^2 x + \sin^2 x) + 144(\cos^2 y + \sin^2 y) + 120(\cos x \cos y + \sin x \sin y) = 169 + a^2 \]
This simplifies to:
\[ 25 + 144 + 120(\cos x \cos y + \sin x \sin y) = 169 + a^2 \]
\[ 169 + 120\cos(x - y) = 169 + a^2 \]

### Step 5: Isolate \( a^2 \)
Subtracting 169 from both sides gives:
\[ 120\cos(x - y) = a^2 \]

### Step 6: Find the maximum value of \( a \)
The maximum value of \( \cos(x - y) \) is 1. Therefore:
\[ a^2 = 120 \cdot 1 = 120 \]
Taking the square root gives:
\[ a = \sqrt{120} \]

### Conclusion
Thus, the maximum value of \( 5\sin x + 12\sin y \) is:
\[ \sqrt{120} \]