Question

If $\sin x - \sin y = \frac{27}{65}$ and $\cos x - \cos y = -\frac{21}{65}$, then $\sin(x+y)= $

• A. \( \frac{16}{65} \)
• B. \( -\frac{16}{65} \)
• C. \( \frac{63}{65} \)
• D. \( -\frac{63}{65} \)

Hint:

When solving trigonometric problems involving sums and differences of trigonometric functions, remember to utilize sum-to-product or product-to-sum identities. These identities can simplify complex expressions into simpler forms that are easier to manipulate. Also, pay attention to the signs of trigonometric ratios to determine the correct quadrant, which affects the sign of the square roots.

Answer 1

The Correct Option is C

We are given two equations: 1) $\sin x - \sin y = \frac{27}{65}$ 2) $\cos x - \cos y = -\frac{21}{65}$ We use the sum-to-product trigonometric identities: $\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ Applying these identities to the given equations: 1) $2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) = \frac{27}{65}$ (Equation P) 2) $-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) = -\frac{21}{65}$ (Equation Q) To eliminate $\sin\left(\frac{x-y}{2}\right)$, we divide Equation P by Equation Q. Assuming $\sin\left(\frac{x-y}{2}\right) \neq 0$ (if it were zero, then $\frac{27}{65}=0$ and $-\frac{21}{65}=0$, which is a contradiction): $\frac{2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)}{-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)} = \frac{27/65}{-21/65}$ $-\frac{\cos\left(\frac{x+y}{2}\right)}{\sin\left(\frac{x+y}{2}\right)} = -\frac{27}{21}$ $-\cot\left(\frac{x+y}{2}\right) = -\frac{9}{7}$ $\cot\left(\frac{x+y}{2}\right) = \frac{9}{7}$ From $\cot\left(\frac{x+y}{2}\right) = \frac{9}{7}$, we can find $\sin\left(\frac{x+y}{2}\right)$ and $\cos\left(\frac{x+y}{2}\right)$. Consider a right triangle where the adjacent side is 9 and the opposite side is 7. The hypotenuse is $\sqrt{9^2 + 7^2} = \sqrt{81+49} = \sqrt{130}$. Since $\cot\left(\frac{x+y}{2}\right)>0$, $\frac{x+y}{2}$ must be in Quadrant I or Quadrant III. If $\frac{x+y}{2}$ is in Quadrant I: $\sin\left(\frac{x+y}{2}\right) = \frac{7}{\sqrt{130}}$ $\cos\left(\frac{x+y}{2}\right) = \frac{9}{\sqrt{130}}$ If $\frac{x+y}{2}$ is in Quadrant III: $\sin\left(\frac{x+y}{2}\right) = -\frac{7}{\sqrt{130}}$ $\cos\left(\frac{x+y}{2}\right) = -\frac{9}{\sqrt{130}}$ We need to find $\sin(x+y)$. We use the double angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$. Let $\theta = \frac{x+y}{2}$. $\sin(x+y) = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)$ Case 1: $\frac{x+y}{2}$ in Quadrant I $\sin(x+y) = 2 \left(\frac{7}{\sqrt{130}}\right) \left(\frac{9}{\sqrt{130}}\right) = 2 \times \frac{63}{130} = \frac{126}{130} = \frac{63}{65}$. Case 2: $\frac{x+y}{2}$ in Quadrant III $\sin(x+y) = 2 \left(-\frac{7}{\sqrt{130}}\right) \left(-\frac{9}{\sqrt{130}}\right) = 2 \times \frac{63}{130} = \frac{126}{130} = \frac{63}{65}$. In both cases, the value of $\sin(x+y)$ is $\frac{63}{65}$. The final answer is $\boxed{\frac{63}{65}}$.