change option B's equation to sin3x-sinx/(cos3x+cosx=tanx

Let's analyze the modified equation in Option B:

### Modified Option B: 
\[
\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \tan x
\]

#### Step 1: Simplify the numerator using the sum-to-product identity:
\[
\sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right)
\]
For \( A = 3x \) and \( B = x \), this becomes:
\[
\sin 3x - \sin x = 2 \cos\left(\frac{3x + x}{2}\right)\sin\left(\frac{3x - x}{2}\right) = 2 \cos(2x) \sin(x)
\]

#### Step 2: Simplify the denominator using the sum-to-product identity:
\[
\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
\]
For \( A = 3x \) and \( B = x \), this becomes:
\[
\cos 3x + \cos x = 2 \cos\left(\frac{3x + x}{2}\right)\cos\left(\frac{3x - x}{2}\right) = 2 \cos(2x) \cos(x)
\]

#### Step 3: Substitute the simplified expressions back into the original equation:
\[
\frac{2 \cos(2x) \sin(x)}{2 \cos(2x) \cos(x)} = \frac{\sin(x)}{\cos(x)} = \tan x
\]

Since the simplification works out perfectly, the modified equation **is an identity**. 

So, for the modified version, **Option B is an identity** as well.

Would you like further details or have any other questions?

Here are 5 related questions:
1. How does the sum-to-product identity differ from the product-to-sum identity?
2. Can you derive the sum-to-product identity from basic trigonometric identities?
3. What are the implications of proving that a trigonometric equation is an identity?
4. How can you use trigonometric identities to solve equations involving multiple angles?
5. What is the significance of the tangent function in trigonometric identities?

**Tip:** Always simplify expressions using trigonometric identities to see if they match a standard form. This approach helps verify identities efficiently.

Learn how to solve the trigonometric equation sin(3x) - sin(x) / (cos(3x) + cos(x)) = tan(x) using sum-to-product identities. This problem involves advanced trigonometric concepts and identities. Suitable for advanced high school students and beyond.

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