If a sin theta + b …
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Gaurav Seth 3 years, 4 months ago
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
Take square root both sides,
Hence, acosθ - bsinθ =
1Thank You