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