Question

If acosθ – bsinθ = c, prove that asinθ + bcosθ = `\pm \sqrt{a^{2}+b^{2}-c^{2}`

Answer 1

We have,

`(acosθ – bsinθ)^2 + (asinθ + bcosθ)^2`

`= (a^2 cos^2 θ + b^2 sin^2 θ – 2ab sinθcosθ) + (a^2 sin^2 θ + b^2 cos^2θ + 2ab sinθcosθ)`

`= a^2 (cos^2 θ + sin^2 θ) + b^2 (sin^2 θ + cos^2 θ)`

`= a^2 + b^2`

`⇒ c^2 + (asinθ + bcosθ)^2 = a^2 + b^2 [∵ acosθ – bsinθ = c]`

`⇒ (asinθ + bcosθ)^2 = a^2 + b^2 – c^2`

`⇒ asinθ + bcosθ = \pm \sqrt{a^{2}+b^{2}-c^{2}}`