Question

If \( 7\sin x + 15\sin y = 17 \), then the maximum value of \( 7\cos x + 15\cos y \) is:

• A. \( \sqrt{190} \)
• B. \( \sqrt{195} \)
• C. \( \sqrt{200} \)
• D. \( \sqrt{205} \)

Hint:

For expressions involving \[ a\sin x+b\sin y \] and \[ a\cos x+b\cos y, \] always square and add them. The identity \[ (\text{sine part})^2+(\text{cosine part})^2 = a^2+b^2+2ab\cos(x-y) \] usually leads directly to the answer.

Answer 1

The Correct Option is B

Concept: For any real numbers \(a,b\), \[ (a\sin x+b\sin y)^2+(a\cos x+b\cos y)^2 = a^2+b^2+2ab\cos(x-y). \] Since \[ -1\le \cos(x-y)\le 1, \] this identity can be used to determine the maximum possible value of the required expression.

Step 1: Introduce the required quantity. Let \[ M=7\cos x+15\cos y. \] Given, \[ 7\sin x+15\sin y=17. \] Therefore, \[ (7\sin x+15\sin y)^2=17^2=289. \]

Step 2: Square and add the sine and cosine expressions. \[ 289+M^2 = (7\sin x+15\sin y)^2 + (7\cos x+15\cos y)^2. \] Expanding, \[ 289+M^2 = 49(\sin^2x+\cos^2x) + 225(\sin^2y+\cos^2y) + 210(\sin x\sin y+\cos x\cos y). \] Using \[ \sin^2\theta+\cos^2\theta=1 \] and \[ \sin x\sin y+\cos x\cos y=\cos(x-y), \] we get \[ 289+M^2 = 49+225+210\cos(x-y). \] Thus, \[ 289+M^2 = 274+210\cos(x-y). \] \[ M^2 = 210\cos(x-y)-15. \]

Step 3: Use the existence condition. Since \(M^2\ge0\), \[ 210\cos(x-y)-15\ge0 \] which gives \[ \cos(x-y)\ge \frac1{14}. \] To maximize \(M\), we must maximize \(M^2\). Since \[ \cos(x-y)\le1, \] maximum occurs when \[ \cos(x-y)=1. \] Therefore, \[ M^2 = 210(1)-15 = 195. \] Hence, \[ M=\sqrt{195}. \] However, for \(\cos(x-y)=1\), we have \(x=y\), and then \[ 22\sin x=17. \] This is possible since \(17<22\). Therefore the maximum value is \[ \boxed{\sqrt{195}}. \] Hence the mathematically correct answer is \(\boxed{(B)}\)