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Question Number 106579 by john santu last updated on 06/Aug/20
Given cos x+cos y=(3/2)+cos (x+y) where x,y ∈ [0,2π ]. find x & y
$$\mathcal{G}\mathrm{iven}\:\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{where}\:\mathrm{x},\mathrm{y}\:\in\:\left[\mathrm{0},\mathrm{2}\pi\:\right].\:\mathrm{find}\:\mathrm{x}\:\&\:\mathrm{y}\: \\ $$
Answered by bobhans last updated on 06/Aug/20
cos x+cos y = (3/2)+cos xcos y−sin xsin y cos x+cos y−cos xcos y+sin xsin y=(3/2) recall → { ((cos x=((1−tan^2 ((x/2)))/(1+tan^2 ((x/2)))))),((sin x=((2tan ((x/2)))/(1+tan^2 ((x/2)))))) :} set tan ((x/2))= p & tan ((y/2))= q ⇔ ((1−p^2 )/(1+p^2 ))+((1−q^2 )/(1+q^2 ))−(((1−p^2 )/(1+p^2 )))(((1−q^2 )/(1+q^2 )))+((4pq)/((1+p^2 )(1+q^2 )))=(3/2) 9p^2 q^2 −8pq +p^2 +q^2 +1 = 0 (3pq−1)^2 + (p−q)^2 = 0 → { ((p= q)),((p=± (1/( (√3))))) :} case(1) { ((p=(1/( (√3)))⇒tan ((x/2))=tan ((π/6)))),((q=(1/( (√3)))⇒tan ((y/2))=tan ((π/6)))) :} → { ((x=(π/3))),((y=(π/3))) :} ⇒((π/3),(π/3)) case(2) → { ((p=−(1/( (√3)))⇒tan ((x/2))=tan (−(π/6)))),((q=−(1/( (√3)))⇒tan ((y/2))=tan (−(π/6)))) :} → { ((x=((5π)/3))),((y=((5π)/3))) :} ⇒ (((5π)/3), ((5π)/3))
$$\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{xcos}\:\mathrm{y}−\mathrm{sin}\:\mathrm{xsin}\:\mathrm{y} \\ $$$$\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}−\mathrm{cos}\:\mathrm{xcos}\:\mathrm{y}+\mathrm{sin}\:\mathrm{xsin}\:\mathrm{y}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{recall}\:\rightarrow\begin{cases}{\mathrm{cos}\:\mathrm{x}=\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}}\\{\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{2tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}}\end{cases} \\ $$$$\mathrm{set}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\mathrm{p}\:\&\:\mathrm{tan}\:\left(\frac{\mathrm{y}}{\mathrm{2}}\right)=\:\mathrm{q} \\ $$$$\Leftrightarrow\:\frac{\mathrm{1}−\mathrm{p}^{\mathrm{2}} }{\mathrm{1}+\mathrm{p}^{\mathrm{2}} }+\frac{\mathrm{1}−\mathrm{q}^{\mathrm{2}} }{\mathrm{1}+\mathrm{q}^{\mathrm{2}} }−\left(\frac{\mathrm{1}−\mathrm{p}^{\mathrm{2}} }{\mathrm{1}+\mathrm{p}^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}−\mathrm{q}^{\mathrm{2}} }{\mathrm{1}+\mathrm{q}^{\mathrm{2}} }\right)+\frac{\mathrm{4pq}}{\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{q}^{\mathrm{2}} \right)}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{9p}^{\mathrm{2}} \mathrm{q}^{\mathrm{2}} −\mathrm{8pq}\:+\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{3pq}−\mathrm{1}\right)^{\mathrm{2}} +\:\left(\mathrm{p}−\mathrm{q}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\rightarrow\begin{cases}{\mathrm{p}=\:\mathrm{q}}\\{\mathrm{p}=\pm\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\end{cases} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\begin{cases}{\mathrm{p}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{6}}\right)}\\{\mathrm{q}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow\mathrm{tan}\:\left(\frac{\mathrm{y}}{\mathrm{2}}\right)=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{6}}\right)}\end{cases} \\ $$$$\rightarrow\begin{cases}{\mathrm{x}=\frac{\pi}{\mathrm{3}}}\\{\mathrm{y}=\frac{\pi}{\mathrm{3}}}\end{cases}\:\Rightarrow\left(\frac{\pi}{\mathrm{3}},\frac{\pi}{\mathrm{3}}\right) \\ $$$$\mathrm{case}\left(\mathrm{2}\right) \\ $$$$\rightarrow\begin{cases}{\mathrm{p}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{tan}\:\left(−\frac{\pi}{\mathrm{6}}\right)}\\{\mathrm{q}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow\mathrm{tan}\:\left(\frac{\mathrm{y}}{\mathrm{2}}\right)=\mathrm{tan}\:\left(−\frac{\pi}{\mathrm{6}}\right)}\end{cases} \\ $$$$\rightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{5}\pi}{\mathrm{3}}}\\{\mathrm{y}=\frac{\mathrm{5}\pi}{\mathrm{3}}}\end{cases}\:\Rightarrow\:\left(\frac{\mathrm{5}\pi}{\mathrm{3}},\:\frac{\mathrm{5}\pi}{\mathrm{3}}\right) \\ $$
Commented by john santu last updated on 06/Aug/20
cool...& jooss ♠
$$\mathrm{cool}…\&\:\mathrm{jooss}\:\spadesuit \\ $$

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