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Question Number 106950 by bemath last updated on 08/Aug/20
         ^(@bemath@)   Given  { ((2cos x+7cos y =5)),((2sin x+7sin y = 6)) :}  cos (x−y) =?
$$\:\:\:\:\:\:\:\:\:\:^{@{bemath}@} \\ $$$${Given}\:\begin{cases}{\mathrm{2cos}\:{x}+\mathrm{7cos}\:{y}\:=\mathrm{5}}\\{\mathrm{2sin}\:{x}+\mathrm{7sin}\:{y}\:=\:\mathrm{6}}\end{cases} \\ $$$$\mathrm{cos}\:\left({x}−{y}\right)\:=?\: \\ $$
Answered by john santu last updated on 08/Aug/20
         ⧫JS◊  → { ((4cos^2 x+14cos xcos y+49cos^2 y=25)),((4sin^2 x+14sin xsin y+49sin^2 y=36)) :}  (1)+(2)⇒53+14cos (x−y)=61      cos (x−y) =((61−53)/(14))=(8/(14))=(4/7)
$$\:\:\:\:\:\:\:\:\:\blacklozenge\mathrm{JS}\lozenge \\ $$$$\rightarrow\begin{cases}{\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{14cos}\:\mathrm{xcos}\:\mathrm{y}+\mathrm{49cos}\:^{\mathrm{2}} \mathrm{y}=\mathrm{25}}\\{\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{14sin}\:\mathrm{xsin}\:\mathrm{y}+\mathrm{49sin}\:^{\mathrm{2}} \mathrm{y}=\mathrm{36}}\end{cases} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow\mathrm{53}+\mathrm{14cos}\:\left(\mathrm{x}−\mathrm{y}\right)=\mathrm{61} \\ $$$$\:\:\:\:\mathrm{cos}\:\left(\mathrm{x}−\mathrm{y}\right)\:=\frac{\mathrm{61}−\mathrm{53}}{\mathrm{14}}=\frac{\mathrm{8}}{\mathrm{14}}=\frac{\mathrm{4}}{\mathrm{7}} \\ $$

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