Question

using cosine rule prove the sine rule​

Answer 1

Refer to Attachment!!!!

Answer 2

• Using cosine rule, sine rule is proved below .

To Prove :- Using cosine rule prove the sine rule ?

• sin A/a = sin B/b = sin C/c

Formula used :-

cosine rule :-

• cos A = (b² + c² - a²) / 2bc
• cos B = (a² + c² - b²) / 2ac
• cos C = (a² + b² - c²) / 2ab

Algebra formula :-

• (a² + b² - 2ab) = (a - b)²
• (a² + b² + 2ab) = (a + b)²
• a² - b² = (a + b)(a - b)

Trigonometric formula :-

• sin²θ = (1 - cos²θ)

Proof :-

sin A / a :-

→ (sin A/a)²

→ sin² A / a²

using sin²θ = (1 - cos²θ) in the numerator,

→ (1 - cos²A) / a²

putting cosine rule for cos A in numerator now,

→ [1 - {(b² + c² - a²) / 2bc}²] / a²

→ [(2bc)² - (b² + c² - a²)] / (2bc)²•a²

using a² - b² = (a + b)(a - b) in numerator,

→ [(2bc + b² + c² - a²)(2bc - b² - c² + a²) / (2abc)²

→ [{(b² + c² + 2bc) - (a)²}{(a)² - (b² + c² - 2bc)}] / (2abc)²

using (a² + b² + 2ab) = (a + b)² and (a² + b² - 2ab) = (a - b)² in numerator,

→ [{(b + c)² - (a)²}{(a)² - (b - c)²}] / (2abc)²

using a² - b² = (a + b)(a - b) in numerator,

→ [(b + c + a)(b + c - a)(a + b - c)(a - b + c)] / (2abc)² ---------- Equation (1)

sin B / b :-

→ (sin B/b)²

→ sin² B / b²

using sin²θ = (1 - cos²θ) in the numerator,

→ (1 - cos²B) / b²

putting cosine rule for cos B in numerator now,

→ [1 - {(a² + c² - b²) / 2ac}²] / b²

→ [(2ac)² - (a² + c² - b²)] / (2ac)²•b²

using a² - b² = (a + b)(a - b) in numerator,

→ [(2ac + a² + c² - b²)(2ac - a² - c² + b²) / (2abc)²

→ [{(a² + c² + 2ac) - (b)²}{(b)² - (a² + c² - 2ac)}] / (2abc)²

using (a² + b² + 2ab) = (a + b)² and (a² + b² - 2ab) = (a - b)² in numerator,

→ [{(a + c)² - (b)²}{(b)² - (a - c)²}] / (2abc)²

using a² - b² = (a + b)(a - b) in numerator,

→ [(a + c + b)(a + c - b)(b + a - c)(b - a + c)] / (2abc)²

re - arranging the numerator now,

→ [(b + c + a)(b + c - a)(a + b - c)(a - b + c)] / (2abc)² ---------- Equation (2)

sin C / c :-

(sin C/c)²

→ sin² C / c²

using sin²θ = (1 - cos²θ) in the numerator,

→ (1 - cos²C) / c²

putting cosine rule for cos C in numerator now,

→ [1 - {(a² + b² - c²) / 2ab}²] / c²

→ [(2ab)² - (a² + b² - c²)] / (2ab)²•c²

using a² - b² = (a + b)(a - b) in numerator,

→ [(2ab + a² + b² - c²)(2ab - a² - b² + c²) / (2abc)²

→ [{(a² + b² + 2ab) - (c)²}{(c)² - (a² + b² - 2ab)}] / (2abc)²

using (a² + b² + 2ab) = (a + b)² and (a² + b² - 2ab) = (a - b)² in numerator,

→ [{(a + b)² - (c)²}{(c)² - (a - b)²}] / (2abc)²

using a² - b² = (a + b)(a - b) in numerator,

→ [(a + b + c)(a + b - c)(c + a - b)(c - a + b)] / (2abc)²

re - arranging the numerator now,

→ [(b + c + a)(b + c - a)(a + b - c)(a - b + c)] / (2abc)² ---------- Equation (3)

as we can see that,

→ Equation (1) = Equation (2) = Equation (3)

therefore,

→ (sin²A/a²) = (sin²B/b²) = (sin²C/c²)

since sine angles and length of sides of triangles are always positive, taking square root we get,

→ (sin A/a) = (sin B/b) = (sin C/c) (Proved)

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