With usual notations prove that 2 a sin 2 C 2 + c sin 2 A 2 = a - b + c .

Given: A △ A B C with opposite sides length a , b , c , respectively To prove : 2 a sin 2 C 2 + c sin 2 A 2 = a - b + c Proof: 2 a sin 2 C 2 + c sin 2 A 2 {LHS} We know that: sin 2 θ = 1 - cos 2 θ 2 ∴ 2 a 1 - cos C 2 + c 1 - cos A 2 {replace θ by C 2 and A 2 } = 2 a 1 - cos C + c 1 - cos A 2 = a 1 - cos C + c 1 - cos A = a + c - a cos C + c cos A = a + c - a a 2 + b 2 - c 2 2 a b + c b 2 + c 2 - a 2 2 b c {cosine rule} = a + c - a 2 + b 2 - c 2 + b 2 + c 2 - a 2 2 b = a + c - 2 b 2 2 b = a - b + c {RHS} Hence proved

In ∆ A B C prove that a - b 2 cos 2 C 2 + a + b 2 sin 2 C 2 = c 2

Consider a - b 2 cos 2 C 2 + a + b 2 sin 2 C 2 = c 2 Taking L . H . S . = a - b 2 cos 2 C 2 + a + b 2 sin 2 C 2 = a 2 + b 2 - 2 a b cos 2 C 2 + a 2 + b 2 + 2 a b sin 2 C 2 = a 2 + b 2 cos 2 C 2 + sin 2 C 2 + 2 a b sin 2 C 2 - cos 2 C 2 = a 2 + b 2 - 2 a b cos 2 C 2 - sin 2 C 2 = a 2 + b 2 - 2 a b cos C We know that, cos C = a 2 + b 2 - c 2 2 a b L . H . S . = a 2 + b 2 - 2 a b a 2 + b 2 - c 2 2 a b = a 2 + b 2 - a 2 + b 2 - c 2 = a 2 + b 2 - a 2 - b 2 + c 2 = c 2 L . H . S . = R . H . S . Hence, proved.

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus - 2023)>Chapter 3 - Trigonometric Functions>Exercise

Exercise

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