Question 38

If $\sin x - \sin y = -\frac{1}{3}$ and $\cos x - \cos y = \frac{1}{2}$ , then the value of $\sin(x + y)$ = ....

Recall the concept of identities

\sin x - \sin y = 2 \cos \frac{x + y}{2} \sin \frac{x - y}{2}

\cos x - \cos y = -2 \sin \frac{x + y}{2} \sin \frac{x - y}{2}

Then we will get two equations from what is known, namely

2 \cos \frac{x + y}{2} \sin \frac{x - y}{2} = -\frac{1}{3} \quad \ldots \text{(1)}

-2 \sin \frac{x + y}{2} \sin \frac{x - y}{2} = \frac{1}{2} \quad \ldots \text{(2)}

Equation $\text{(2)}$ divided by equation $\text{(1)}$

\frac{\text{(2)}}{\text{(1)}}: \quad \frac{-2 \sin \frac{x + y}{2} \sin \frac{x - y}{2}}{2 \cos \frac{x + y}{2} \sin \frac{x - y}{2}} = \frac{\frac{1}{2}}{-\frac{1}{3}}

-\tan \frac{x + y}{2} = -\frac{3}{2}

\tan \frac{x + y}{2} = \frac{3}{2}

The opposite side is $3$ , and the adjacent side is $2$ . Then, using Pythagoras, the hypotenuse is obtained

\text{Hyp} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}

Thus

\sin \frac{x + y}{2} = \frac{\text{opp}}{\text{hyp}} = \frac{3}{\sqrt{13}}

\cos \frac{x + y}{2} = \frac{\text{adj}}{\text{hyp}} = \frac{2}{\sqrt{13}}

Then the value of $\sin(x + y)$

\sin(x + y) = 2 \sin \frac{x + y}{2} \cos \frac{x + y}{2}

\sin(x + y) = 2 \cdot \frac{3}{\sqrt{13}} \cdot \frac{2}{\sqrt{13}}

\sin(x + y) = \frac{12}{13}

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