Question

Ex. 8) If $\sin A + \sin B = x$ and $\cos A + \cos B = y$ then show that $\sin (A+B) = \frac{2xy}{x^2+y^2}$

Answer 1

$\sin (A+B) = \frac{2xy}{x^2+y^2}$

Answer 2

Solution:

Express the sums using the sum‐to‐product formulas:

$$\sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} = x, \quad \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} = y.$$

Then,

$$x^2+y^2 = 4\cos^2\frac{A-B}{2}\Bigl(\sin^2\frac{A+B}{2}+\cos^2\frac{A+B}{2}\Bigr)=4\cos^2\frac{A-B}{2}.$$

Next, multiply $x$ and $y$:

$$xy = \Bigl(2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\Bigr)\Bigl(2\cos\frac{A+B}{2}\cos\frac{A-B}{2}\Bigr)=4\sin\frac{A+B}{2}\cos\frac{A+B}{2}\cos^2\frac{A-B}{2}.$$

Recall the double-angle formula:

$$\sin(A+B)=2\sin\frac{A+B}{2}\cos\frac{A+B}{2}.$$

Thus,

$$xy = 2\sin(A+B)\cos^2\frac{A-B}{2}.$$

Now substitute for $\cos^2\frac{A-B}{2}$ from the expression for $x^2+y^2$:

$$\cos^2\frac{A-B}{2} = \frac{x^2+y^2}{4}.$$

So,

$$xy = 2\sin(A+B) \cdot \frac{x^2+y^2}{4} = \frac{1}{2}\sin(A+B)(x^2+y^2).$$

Finally, solving for $\sin(A+B)$:

$$\sin(A+B) = \frac{2xy}{x^2+y^2}.$$

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Core Explanation:

1. Write sums using sum‐to‐product formulas.
2. Compute $x^2+y^2=4\cos^2\frac{A-B}{2}$.
3. Multiply $x$ and $y$ and express in terms of $\sin(A+B)$.
4. Substitute and simplify to obtain the result.