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Question-46472




Question Number 46472 by peter frank last updated on 27/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Oct/18
(asinθ+bcosθ)^2 =c^2   a^2 sin^2 θ+b^2 cos^2 θ+2absinθcosθ=c^2   a^2 (1−cos^2 θ)+b^2 (1−sin^2 θ)+2absinθcosθ=c^2   a^2 +b^2 −c^2 =a^2 cos^2 θ+b^2 sin^2 θ−2absinθcosθ  (a^2 +b^2 −c^2 )=(acosθ−bsinθ)^2   so (acosθ−bsinθ)=±(√(a^2 +b^2 −c^2 ))
$$\left({asin}\theta+{bcos}\theta\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta+{b}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta+\mathrm{2}{absin}\theta{cos}\theta={c}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} \left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)+{b}^{\mathrm{2}} \left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right)+\mathrm{2}{absin}\theta{cos}\theta={c}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} ={a}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta−\mathrm{2}{absin}\theta{cos}\theta \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)=\left({acos}\theta−{bsin}\theta\right)^{\mathrm{2}} \\ $$$${so}\:\left({acos}\theta−{bsin}\theta\right)=\pm\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }\: \\ $$

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