Question

If 3sinθ+5cosθ=5, prove that 5sinθ-3cosθ=+-3.

Answer 1

Given, 3sinθ + 5 cosθ = 5

Let, 5sinθ - 3cosθ = m

Squaring both equations and adding them gives,

⇒(3sinθ + 5 cosθ)² + (5sinθ - 3cosθ)² = 5² + m²

⇒(3sinθ)² + (5 cosθ)² + 2(3sinθ) (5cosθ) + (5sinθ)² + (3cosθ)² - 2(5sinθ) (3cosθ) = 25 + m²

⇒9sin²θ + 25cos²θ + 30sinθcosθ + 25sin²θ + 9cos²θ - 30sinθcosθ = 25 + m²

⇒9(sin²θ+ cos²θ) + 25 (sin²θ+ cos²θ) = 25 + m²

Trigonometry identity : sin²x + cos²x = 1

⇒9 + 25 = 25 + m²

⇒9 = m²

⇒ m = ±√9

⇒ m = ± 3

Therefore, 5sinθ - 3cosθ = ±3

Answer 2

• Squaring both equations and adding

• Squaring both equations and addingthem gives,

(3sine+5 cose) + (5sine - 3cos0)* = 5?

-(3sin@+ (5 cose + 2(3sine) (5cose) +

(5sine+ (3cose) - 2(5sin) (3cose) = 25+m+m

9sin +25cos e +30sin@cose +

25sine+ 9cose - 30sin@cose = 25 + m2

9(sin+ cos@) + 25 (sin?e+ cos^e) = 25

+m

Trigonometry identity: sin?x + cos?x = 1

»9+25 = 25 + m2

9 m

m=tv9

m =t 3

Therefore, 5sine 3cose = t3