$$ \sin ( heta + \omega ) \cdot \sin ( heta - \omega ) = ( \sin heta + \sin \omega ) ( \sin heta - \sin \omega ) $$

Question

$$ \sin ( 	heta + \omega ) \cdot \sin ( 	heta - \omega ) = ( \sin 	heta + \sin \omega ) ( \sin 	heta - \sin \omega ) $$

UpStudy AI · Accepted Answer

Yes, the identity you've provided is correct. Let's verify and understand why:

$$
\sin(\theta + \omega) \cdot \sin(\theta - \omega) = (\sin \theta + \sin \omega)(\sin \theta - \sin \omega)
$$

### **Step 1: Expand the Right-Hand Side (RHS)**

The RHS is a product of two binomials, which resembles the difference of squares formula:

$$
(\sin \theta + \sin \omega)(\sin \theta - \sin \omega) = \sin^2 \theta - \sin^2 \omega
$$

### **Step 2: Expand the Left-Hand Side (LHS)**

The LHS involves the product of sine functions with sum and difference arguments. We can use the **product-to-sum identities** for sine functions to simplify it:

$$
\sin A \cdot \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]
$$

Let $$ A = \theta + \omega $$ and $$ B = \theta - \omega $$. Applying the identity:

$$
\sin(\theta + \omega) \cdot \sin(\theta - \omega) = \frac{1}{2} [\cos((\theta + \omega) - (\theta - \omega)) - \cos((\theta + \omega) + (\theta - \omega))]
$$

Simplify the arguments inside the cosine functions:

$$
= \frac{1}{2} [\cos(2\omega) - \cos(2\theta)]
$$

### **Step 3: Express $$ \sin^2 \theta $$ and $$ \sin^2 \omega $$ in Terms of Cosine**

Using the **double-angle identity** for sine:

$$
\sin^2 x = \frac{1 - \cos(2x)}{2}
$$

Apply this to both $$ \sin^2 \theta $$ and $$ \sin^2 \omega $$:

$$
\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}
$$
$$
\sin^2 \omega = \frac{1 - \cos(2\omega)}{2}
$$

Subtracting these:

$$
\sin^2 \theta - \sin^2 \omega = \frac{1 - \cos(2\theta)}{2} - \frac{1 - \cos(2\omega)}{2} = \frac{\cos(2\omega) - \cos(2\theta)}{2}
$$

### **Step 4: Compare Both Sides**

From the LHS:

$$
\sin(\theta + \omega) \cdot \sin(\theta - \omega) = \frac{\cos(2\omega) - \cos(2\theta)}{2}
$$

From the RHS:

$$
(\sin \theta + \sin \omega)(\sin \theta - \sin \omega) = \sin^2 \theta - \sin^2 \omega = \frac{\cos(2\omega) - \cos(2\theta)}{2}
$$

Both sides are equal:

$$
\sin(\theta + \omega) \cdot \sin(\theta - \omega) = (\sin \theta + \sin \omega)(\sin \theta - \sin \omega)
$$

### **Conclusion**

The identity holds true because both sides simplify to the same expression involving double-angle cosine terms. This demonstrates the powerful relationships between trigonometric functions and how different forms can represent the same mathematical reality.
The identity $$ \sin(\theta + \omega) \cdot \sin(\theta - \omega) = (\sin \theta + \sin \omega)(\sin \theta - \sin \omega) $$ is correct because both sides simplify to $$ \frac{\cos(2\omega) - \cos(2\theta)}{2} $$.

\( \sin ( \theta + \omega ) \cdot \sin ( \theta - \omega ) = ( \sin \theta + \sin \omega ) ( \sin \theta - \sin \omega ) \)

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