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In ΔABC, prove that a - bCa + bCc(a - b)2cos2 C2+(a + b)2sin2 C2=c2

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Question

In ΔABC, prove that `("a - b")^2 cos^2  "C"/2 + ("a + b")^2 sin^2  "C"/2 = "c"^2`

Sum
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Solution

LHS = `("a - b")^2 cos^2  "C"/2 + ("a + b")^2 sin^2  "C"/2 = "c"^2`

`= ("a"^2 + "b"^2 - 2"ab")cos^2  "C"/2 + ("a"^2 + "b"^2 + 2"ab")sin^2  "C"/2`

`= ("a"^2 + "b"^2)cos^2  "C"/2 - 2"ab" cos^2  "C"/2 + ("a"^2 + "b"^2) "sin"^2 "C"/2 + 2"ab"  "sin"^2  "C"/2`

`= ("a"^2 + "b"^2)(cos^2  "C"/2 + "sin"^2 "C"/2) - 2"ab" (cos^2  "C"/2 - "sin"^2 "C"/2)`

= a2 + b2 - 2ab cos C

= c2 = RHS

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

APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 7 | Page 109

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