Simplifying Trigonometric Expressions: \( \sin(A+B) \sin(A-B) \)

This problem asks us to simplify the product of two sine functions, \( \sin(A+B) \) and \( \sin(A-B) \). To do this, we can use the sum and difference identities for the sine function.

Key Trigonometric Identities for Simplification

We will use the following fundamental identities:

• Sum Identity: \( \sin(X+Y) = \sin X \cos Y + \cos X \sin Y \)
• Difference Identity: \( \sin(X-Y) = \sin X \cos Y - \cos X \sin Y \)

Let \( X = A \) and \( Y = B \). Applying these identities to the given expression:

\( \sin(A+B) = \sin A \cos B + \cos A \sin B \)

\( \sin(A-B) = \sin A \cos B - \cos A \sin B \)

Step-by-Step Simplification

Now, we multiply these two expanded forms:

\( \sin(A+B) \sin(A-B) = (\sin A \cos B + \cos A \sin B)(\sin A \cos B - \cos A \sin B) \)

This expression is in the form \( (x + y)(x - y) \), where \( x = \sin A \cos B \) and \( y = \cos A \sin B \). We know that \( (x+y)(x-y) = x^2 - y^2 \).

Applying this difference of squares formula:

\( (\sin A \cos B)^2 - (\cos A \sin B)^2 \)

Expanding the squares:

\( \sin^2 A \cos^2 B - \cos^2 A \sin^2 B \)

We want to express this result in terms of either \( \sin^2 A \) and \( \sin^2 B \) or \( \cos^2 A \) and \( \cos^2 B \). Let's use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \) to replace \( \cos^2 B \) and \( \cos^2 A \).

Substitute \( \cos^2 B = 1 - \sin^2 B \):

\( \sin^2 A (1 - \sin^2 B) - \cos^2 A \sin^2 B \)

Distribute \( \sin^2 A \):

\( \sin^2 A - \sin^2 A \sin^2 B - \cos^2 A \sin^2 B \)

Now substitute \( \cos^2 A = 1 - \sin^2 A \) in the second term:

\( \sin^2 A - \sin^2 A \sin^2 B - (1 - \sin^2 A) \sin^2 B \)

Distribute \( \sin^2 B \):

\( \sin^2 A - \sin^2 A \sin^2 B - (\sin^2 B - \sin^2 A \sin^2 B) \)

Remove the parentheses, remembering to change the signs:

\( \sin^2 A - \sin^2 A \sin^2 B - \sin^2 B + \sin^2 A \sin^2 B \)

The terms \( - \sin^2 A \sin^2 B \) and \( + \sin^2 A \sin^2 B \) cancel each other out.

The simplified expression is:

\( \sin^2 A - \sin^2 B \)

Alternatively, we could have substituted \( \sin^2 \theta = 1 - \cos^2 \theta \) into the expression \( \sin^2 A \cos^2 B - \cos^2 A \sin^2 B \).

Substitute \( \sin^2 A = 1 - \cos^2 A \):

\( (1 - \cos^2 A) \cos^2 B - \cos^2 A \sin^2 B \)

Distribute \( \cos^2 B \):

\( \cos^2 B - \cos^2 A \cos^2 B - \cos^2 A \sin^2 B \)

Notice that the last two terms have a common factor of \( -\cos^2 A \):

\( \cos^2 B - \cos^2 A (\cos^2 B + \sin^2 B) \)

Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we have \( \cos^2 B + \sin^2 B = 1 \).

\( \cos^2 B - \cos^2 A (1) \)

\( \cos^2 B - \cos^2 A \)

This gives another form of the result. Both \( \sin^2 A - \sin^2 B \) and \( \cos^2 B - \cos^2 A \) are valid simplifications. We check the options provided.

Comparing with Options

Let's compare our result \( \sin^2 A - \sin^2 B \) with the given options:

1. \( \sin^2 A - \sin^2 B \)
2. \( \cos^2 A - \cos^2 B \)
3. \( \sin^2 A + \sin^2 B \)
4. \( \cos^2 A \)

Our simplified expression matches the first option.

Conclusion

By expanding \( \sin(A+B) \) and \( \sin(A-B) \) using trigonometric identities and simplifying the product, we arrive at the expression \( \sin^2 A - \sin^2 B \).

Revision Table: Key Trigonometry Identities

Additional Information: Product-to-Sum Identities

The expression \( \sin(A+B) \sin(A-B) \) can also be simplified directly using a product-to-sum identity. While not strictly necessary for this problem if you know the sum/difference identities, it's a useful related concept.

One of the product-to-sum identities is derived from the cosine sum/difference formulas:

\( \cos(X-Y) = \cos X \cos Y + \sin X \sin Y \)

\( \cos(X+Y) = \cos X \cos Y - \sin X \sin Y \)

Subtracting the second from the first:

\( \cos(X-Y) - \cos(X+Y) = (\cos X \cos Y + \sin X \sin Y) - (\cos X \cos Y - \sin X \sin Y) \)

\( \cos(X-Y) - \cos(X+Y) = 2 \sin X \sin Y \)

So, \( \sin X \sin Y = \frac{1}{2}[\cos(X-Y) - \cos(X+Y)] \)

Let \( X = A+B \) and \( Y = A-B \). Then \( X+Y = (A+B) + (A-B) = 2A \) and \( X-Y = (A+B) - (A-B) = 2B \).

Applying the identity:

\( \sin(A+B) \sin(A-B) = \sin X \sin Y \)

\( = \frac{1}{2}[\cos(X-Y) - \cos(X+Y)] \)

\( = \frac{1}{2}[\cos(2B) - \cos(2A)] \)

Now, use the double angle identity \( \cos(2\theta) = 1 - 2 \sin^2 \theta \) (or \( 2 \cos^2 \theta - 1 \)).

\( = \frac{1}{2}[(1 - 2\sin^2 B) - (1 - 2\sin^2 A)] \)

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

\( = \frac{1}{2}[2\sin^2 A - 2\sin^2 B] \)

\( = \frac{1}{2} \times 2 (\sin^2 A - \sin^2 B) \)

\( = \sin^2 A - \sin^2 B \)

This confirms the result obtained using the sum and difference identities.