Question

Can you prove this:
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Answer 1

asinθ + bcosθ = c 
taking square both sides, 
(asinθ + bcosθ)² = c² 
⇒a²sin²θ + b²cos²θ + 2absinθ.cosθ = c² --------(1)

Let acosθ - bsinθ = x 
Squaring both sides
(acosθ - bsinθ)² = x² 
⇒a²cos²θ + b²sin²θ -2absinθ.cosθ = x² ------(2)

Add equation (1) and (2),
a²sin²θ + b²cos²θ +2abinθ.cosθ
+ a²cos²θ + b²sin²θ -2absinθ.cosθ = c² + x² 
⇒(a² + b²)cos²θ + (a² +b²)sin²θ = c² + x² 
⇒(a² + b²)[sin²θ + cos²θ ] = c² + x² 
⇒(a² + b²) = c² + x² [∵ sin²x + cos²x = 1 ] 
⇒(a² + b² - c²) = x²
=≥ x=√a^2+b^2-c^2
HENCE PROVED !