8. $$ \sin (x-y)+\sin (x+y)=2 \sin x \cos y $$

Question

UpStudy AI · Accepted Answer

We start with the left-hand side:

$$
\sin (x-y)+\sin (x+y)
$$

Using the sum-to-product formula for sine functions, we have

$$
\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)
$$

Let

$$
A = x-y \quad \text{and} \quad B = x+y.
$$

Then, calculate

$$
\frac{A+B}{2} = \frac{(x-y)+(x+y)}{2} = \frac{2x}{2} = x
$$

and

$$
\frac{A-B}{2} = \frac{(x-y)-(x+y)}{2} = \frac{-2y}{2} = -y.
$$

Substituting these back into the sum-to-product formula, we obtain

$$
\sin (x-y)+\sin (x+y) = 2 \sin x \cos(-y).
$$

Since cosine is an even function, we know that

$$
\cos(-y)=\cos y.
$$

Thus, the right-hand side simplifies to

$$
2 \sin x \cos y.
$$

So, we have shown that

$$
\sin (x-y)+\sin (x+y) = 2 \sin x \cos y,
$$

which completes the proof.
$$ \sin (x-y) + \sin (x+y) = 2 \sin x \cos y $$

8. \( \sin (x-y)+\sin (x+y)=2 \sin x \cos y \)