In any Δ ABC, prove the following: a sin A - b sin B = c sin (A - B) - Mathematics and Statistics

Question

In any Δ ABC, prove the following:

a sin A - b sin B = c sin (A - B)

Sum

Solution

By sine rule,

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = k$

∴ a = k sin A, b = k sin B, c = k sin C

LHS = a sin A - b sin B

= k sin A. sin A - k sin B. sin B

= k (sin² A - sin² B)

= k (sin A + sin B)(sin A - sin B)

$= k \times 2 \sin (\frac{A + B}{2}) . \cos (\frac{A - B}{2}) \times 2 \cos (\frac{A + B}{2}) . \sin (\frac{A - B}{2})$

$= k \times 2 \sin (\frac{A + B}{2}) . \cos (\frac{A + B}{2}) \times 2 \sin (\frac{A - B}{2}) . \cos (\frac{A - B}{2})$

= k × sin (A + B) × sin (A - B)

= k sin (π - C). sin (A - B) … [∴ A + B + C = π]

= k sin C. sin (A - B)

= c sin (A - B)

= RHS.

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Q 11.1 Q 10 Q 11.2

Chapter 3: Trigonometric Functions - Miscellaneous exercise 3 [Page 109]

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Balbharati Mathematics and Statistics 1 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board

Chapter 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 11.1 | Page 109

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