Question

Prove that :
sin(A+B). sin (A-B) = sin² A-sin ² B .​

Answer 1

Answer:

Certainly! Let's break it down step by step:

1. Start with the left side of the equation: \( \sin(A+B) \cdot \sin(A-B) \).

2. Use the product-to-sum identity \( \sin(A+B) \cdot \sin(A-B) = \frac{1}{2}[\cos((A+B)-(A-B)) - \cos((A+B)+(A-B))] \).

3. Simplify inside the brackets: \( \cos(A+B-A+B) - \cos(A+B+A-B) \).

4. Simplify further: \( \cos(2B) - \cos(2A) \).

5. Apply the identity \( \cos(2\theta) = 1 - 2\sin^2 \theta \): \( 1 - 2\sin^2 B - (1 - 2\sin^2 A) \).

6. Simplify: \( 1 - 2\sin^2 B - 1 + 2\sin^2 A \).

7. Combine like terms: \( 2\sin^2 A - 2\sin^2 B \).

8. Factor out \(2\): \( 2(\sin^2 A - \sin^2 B) \).

9. Which gives us the right side of the equation: \( \sin^2 A - \sin^2 B \).

So, starting from the left side and applying trigonometric identities step by step, we've shown that \( \sin(A+B) \cdot \sin(A-B) = \sin^2 A - \sin^2 B \).

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Answer 2

Answer:

$\text{L.H.S. = sin(A+B). sin(A-B)}\\\text{}\ \ \ \ \ \ \ \ = \text{(sinA cosB + cosA sinB)(sinA cosB - cosA sinB)}\\\text{}\ \ \ \ \ \ \ \ =\text{sin}^2\text{A cos}^2\text{B}-\text{cos}^2\text{A }\text{sin}^2\text{B}\\\text{}\ \ \ \ \ \ \ \ =\text{sin}^2\text{A}(1-\text{sin}^2\text{B})-(1-\text{sin}^2\text{A})\text{sin}^2\text{B}\\\text{ }\ \ \ \ \ \ \ =\text{sin}^2\text{A}-\text{sin}^2\text{A sin}^2\text{B}-\text{sin}^2\text{B + sin}^2\text{A sin}^2\text{B}\\=\text{sin}^2\text{A}-\text{sin}^2\text{B}$