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If sinAsinC=sin(A - B)sin(B - C), then show that a2, b2, c2 are in A.P.

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Question

If `(sin "A")/(sin "C") = (sin ("A - B"))/(sin ("B - C"))`, then show that a2, b2, c2 are in A.P.

Sum
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Solution

By sine rule,

`"sin A"/"a" = "sin B"/"b" = "sin C"/"c" = "k"`

∴ sin A = ka, sin B = kb, sin C = kc

Now, `(sin "A")/(sin "C") = (sin ("A - B"))/(sin ("B - C"))` 

∴ sin A . sin (B - C) = sin C. sin (A - B)

∴ sin [π - (B + C)]. sin (B - C)

= sin [π - (A + B)]. sin(A - B)    .....[∵ A + B + C = π]

∴ sin (B + C). sin (B - C) = sin (A + B). sin (A - B)

∴ sin2B - sin2C = sin2A - sin2B

∴ 2 sin2B = sin2A + sin2C

∴ 2k2b2 = k2a2 + k2c2

∴ 2b2 = a2 + c2

Hence, a2, b2, c2 are in A.P.

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

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Balbharati Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 9 | Page 109

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