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

Step 1: Name the target.
Let $M = 7\cos x + 15\cos y$. We already know $7\sin x + 15\sin y = 17$.
Step 2: Square both and add.
Consider $(7\sin x + 15\sin y)^2 + (7\cos x + 15\cos y)^2 = 17^2 + M^2 = 289 + M^2$.
Step 3: Expand the left side.
Using $\sin^2+\cos^2=1$ and $\cos(x-y)=\cos x\cos y+\sin x\sin y$, the left side equals $49 + 225 + 2(7)(15)\cos(x-y) = 274 + 210\cos(x-y)$.
Step 4: Equate and isolate $M^2$.
So $289 + M^2 = 274 + 210\cos(x-y)$, giving $M^2 = 210\cos(x-y) - 15$.
Step 5: Maximise.
$M^2$ is largest when $\cos(x-y)$ is largest, i.e. $\cos(x-y) = 1$. Then $M^2 = 210 - 15 = 195$. (This is feasible: $x = y$ gives $22\sin x = 17$, allowed since $17 < 22$.)
Step 6: Take the root.
The maximum value is $M = \sqrt{195}$, which is option (B).
\[ \boxed{\sqrt{195}} \]