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Question Number 144305 by gsk2684 last updated on 24/Jun/21
if (a−b)sin(θ+φ)=(a+b)sin(θ−φ) and a tan(θ/2) − b tan(φ/2) = c then prove that the following i) sinφ = ((2bc)/(a^2 −b^2 −c^2 )) ii) sinθ = ((2ac)/(a^2 −b^2 +c^2 ))
$$\mathrm{if}\:\left(\mathrm{a}−\mathrm{b}\right)\mathrm{sin}\left(\theta+\phi\right)=\left(\mathrm{a}+\mathrm{b}\right)\mathrm{sin}\left(\theta−\phi\right)\: \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathrm{tan}\frac{\theta}{\mathrm{2}}\:−\:\mathrm{b}\:\mathrm{tan}\frac{\phi}{\mathrm{2}}\:=\:\mathrm{c}\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following} \\ $$$$\left.\mathrm{i}\right)\:\mathrm{sin}\phi\:=\:\frac{\mathrm{2bc}}{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} }\: \\ $$$$\left.\mathrm{ii}\right)\:\mathrm{sin}\theta\:=\:\frac{\mathrm{2ac}}{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }\: \\ $$
Answered by liberty last updated on 24/Jun/21
⇔ (a−b)(sin θ cos φ+cos θ sin φ)= (a+b)(sin θ cos φ−cos θ sin φ) ⇔ a cos θ sin φ −bsin θ cos φ= −acos θ sin φ+bsin θ cos φ ⇔ 2a cos θ sin φ = 2b sin θ cos φ ⇔ (a/b) = ((sin θ cos φ)/(cos θ sin φ)) ⇔ (a/b) = tan θ cot φ ⇔ a tan φ = b tan θ ⇔ a(((2 tan (φ/2))/(1−tan^2 (φ/2))))=b(((2tan (θ/2))/(1−tan^2 (θ/2))))...(i) tan ((θ/2))=(c/a)+(b/a)tan ((φ/2))...(ii) let tan ((φ/2))= x ⇔ a((x/(1−x^2 )))=b((((c/a)+(b/a)x)/(1−(((c+bx)/a))^2 ))) ⇔a((x/(1−x^2 )))=b(((a(c+bx))/(a^2 −c^2 −2bcx−b^2 x^2 ))) ⇔(x/(1−x^2 )) = ((bc+b^2 x)/(a^2 −c^2 −2bcx−b^2 x^2 )) ⇔a^2 x−c^2 x−2bcx^2 −b^2 x^3 =bc+b^2 x−bcx^2 −b^2 x^3 ⇒a^2 x−c^2 x−b^2 x−bcx^2 −bc=0 ⇒bcx^2 +(b^2 +c^2 −a^2 )x+bc =0 ⇒x = ((a^2 −b^2 −c^2 +(√((b^2 +c^2 −a^2 )^2 −4(bc)^2 )))/(2bc)) ⇒x=((a^2 −b^2 −c^2 +(√((b^2 +c^2 −a^2 +2bc)(b^2 +c^2 −a^2 −2bc))))/(2bc)) ⇒x=((a^2 −b^2 −c^2 +(√(((b+c)^2 −a^2 )((b−c)^2 −a^2 ))))/(2bc))
$$\Leftrightarrow\:\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{sin}\:\theta\:\mathrm{cos}\:\phi+\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi\right)= \\ $$$$\:\:\:\:\:\:\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{sin}\:\theta\:\mathrm{cos}\:\phi−\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi\right) \\ $$$$\Leftrightarrow\:\mathrm{a}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi\:−\mathrm{bsin}\:\theta\:\mathrm{cos}\:\phi= \\ $$$$\:\:\:\:−\mathrm{acos}\:\theta\:\mathrm{sin}\:\phi+\mathrm{bsin}\:\:\theta\:\mathrm{cos}\:\:\phi \\ $$$$\Leftrightarrow\:\mathrm{2a}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi\:=\:\:\mathrm{2b}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\phi \\ $$$$\Leftrightarrow\:\frac{\mathrm{a}}{\mathrm{b}}\:=\:\frac{\mathrm{sin}\:\theta\:\mathrm{cos}\:\phi}{\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi} \\ $$$$\Leftrightarrow\:\frac{\mathrm{a}}{\mathrm{b}}\:=\:\mathrm{tan}\:\theta\:\mathrm{cot}\:\phi \\ $$$$\Leftrightarrow\:\mathrm{a}\:\mathrm{tan}\:\phi\:=\:\mathrm{b}\:\mathrm{tan}\:\theta \\ $$$$\Leftrightarrow\:\mathrm{a}\left(\frac{\mathrm{2}\:\mathrm{tan}\:\:\frac{\phi}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\phi}{\mathrm{2}}}\right)=\mathrm{b}\left(\frac{\mathrm{2tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}\right)…\left(\mathrm{i}\right) \\ $$$$\mathrm{tan}\:\left(\frac{\theta}{\mathrm{2}}\right)=\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{a}}\mathrm{tan}\:\left(\frac{\phi}{\mathrm{2}}\right)…\left(\mathrm{ii}\right) \\ $$$$\mathrm{let}\:\mathrm{tan}\:\left(\frac{\phi}{\mathrm{2}}\right)=\:\mathrm{x} \\ $$$$\Leftrightarrow\:\mathrm{a}\left(\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\right)=\mathrm{b}\left(\frac{\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{a}}\mathrm{x}}{\mathrm{1}−\left(\frac{\mathrm{c}+\mathrm{bx}}{\mathrm{a}}\right)^{\mathrm{2}} }\right) \\ $$$$\Leftrightarrow\mathrm{a}\left(\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\right)=\mathrm{b}\left(\frac{\mathrm{a}\left(\mathrm{c}+\mathrm{bx}\right)}{\mathrm{a}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} −\mathrm{2bcx}−\mathrm{b}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }\right) \\ $$$$\Leftrightarrow\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{bc}+\mathrm{b}^{\mathrm{2}} \mathrm{x}}{\mathrm{a}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} −\mathrm{2bcx}−\mathrm{b}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }\: \\ $$$$\Leftrightarrow\mathrm{a}^{\mathrm{2}} \mathrm{x}−\mathrm{c}^{\mathrm{2}} \mathrm{x}−\mathrm{2bcx}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \mathrm{x}^{\mathrm{3}} =\mathrm{bc}+\mathrm{b}^{\mathrm{2}} \mathrm{x}−\mathrm{bcx}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \mathrm{x}^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} \mathrm{x}−\mathrm{c}^{\mathrm{2}} \mathrm{x}−\mathrm{b}^{\mathrm{2}} \mathrm{x}−\mathrm{bcx}^{\mathrm{2}} −\mathrm{bc}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{bcx}^{\mathrm{2}} +\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)\mathrm{x}+\mathrm{bc}\:=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}\:=\:\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} +\sqrt{\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{bc}\right)^{\mathrm{2}} }}{\mathrm{2bc}} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} +\sqrt{\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} +\mathrm{2bc}\right)\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} −\mathrm{2bc}\right)}}{\mathrm{2bc}} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} +\sqrt{\left(\left(\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)\left(\left(\mathrm{b}−\mathrm{c}\right)^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)}}{\mathrm{2bc}} \\ $$
Commented by gsk2684 last updated on 24/Jun/21
thank you very much
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$
Answered by ajfour last updated on 24/Jun/21
((sin (θ+φ))/(sin (θ−φ)))=((a+b)/(a−b)) ⇒ ((sin (θ+φ)+sin (θ−φ))/(sin (θ+φ)−sin (θ−φ))) =((2a)/(2b))=(a/b) ⇒ ((sin θcos φ)/(cos θsin φ))=(a/b) ⇒ (((2tan (θ/2))/(1−tan^2 (θ/2))))(((1−tan^2 (φ/2))/(2tan (φ/2))))=(a/b) ((p(1−q^2 ))/(q(1−p^2 )))=(a/b) ....(I) & atan (θ/2)−btan (φ/2)=c ⇒ ap−bq=c .....(II) ((p{b^2 −(ap−c)^2 })/((ap−c)(1−p^2 )))=a b^2 p−a^2 p^3 +2acp^2 −c^2 p =−a^2 p^3 +acp^2 +a^2 p−ac ⇒ acp^2 +(b^2 −c^2 −a^2 )p+ac=0 ⇒ p+(1/p)=((a^2 +c^2 −b^2 )/(ac)) sin θ=((2p)/(1+p^2 ))=(2/((1/p)+p)) ⇒ sin𝛉=((2ac)/(a^2 −b^2 +c^2 )) similarly from (I), (II) (((c+bq)(1−q^2 ))/(q{a^2 −(c+bq)^2 }))=(1/b) ⇒ −b^2 q^3 −bcq^2 +b^2 q+bc =−b^2 q^3 −2bcq^2 +a^2 q−c^2 q ⇒ bcq^2 +(b^2 −a^2 +c^2 )q+bc=0 ⇒ q+(1/q)=((a^2 −b^2 −c^2 )/(bc)) sin φ=((2q)/(1+q^2 ))=(2/((1/q)+q)) ⇒ sin𝛗=((2bc)/(a^2 −b^2 −c^2 )) ★
$$\frac{\mathrm{sin}\:\left(\theta+\phi\right)}{\mathrm{sin}\:\left(\theta−\phi\right)}=\frac{{a}+{b}}{{a}−{b}} \\ $$$$\:\Rightarrow\:\frac{\mathrm{sin}\:\left(\theta+\phi\right)+\mathrm{sin}\:\left(\theta−\phi\right)}{\mathrm{sin}\:\left(\theta+\phi\right)−\mathrm{sin}\:\left(\theta−\phi\right)} \\ $$$$\:\:\:=\frac{\mathrm{2}{a}}{\mathrm{2}{b}}=\frac{{a}}{{b}} \\ $$$$\Rightarrow\:\frac{\mathrm{sin}\:\theta\mathrm{cos}\:\phi}{\mathrm{cos}\:\theta\mathrm{sin}\:\phi}=\frac{{a}}{{b}} \\ $$$$\Rightarrow\:\left(\frac{\mathrm{2tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}\right)\left(\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\phi}{\mathrm{2}}}{\mathrm{2tan}\:\frac{\phi}{\mathrm{2}}}\right)=\frac{{a}}{{b}} \\ $$$$\frac{{p}\left(\mathrm{1}−{q}^{\mathrm{2}} \right)}{{q}\left(\mathrm{1}−{p}^{\mathrm{2}} \right)}=\frac{{a}}{{b}}\:\:\:\:\:….\left({I}\right) \\ $$$$\& \\ $$$${a}\mathrm{tan}\:\frac{\theta}{\mathrm{2}}−{b}\mathrm{tan}\:\frac{\phi}{\mathrm{2}}={c} \\ $$$$\Rightarrow\:\:{ap}−{bq}={c}\:\:\:\:\:\:\:…..\left({II}\right) \\ $$$$\frac{{p}\left\{{b}^{\mathrm{2}} −\left({ap}−{c}\right)^{\mathrm{2}} \right\}}{\left({ap}−{c}\right)\left(\mathrm{1}−{p}^{\mathrm{2}} \right)}={a} \\ $$$${b}^{\mathrm{2}} {p}−{a}^{\mathrm{2}} {p}^{\mathrm{3}} +\mathrm{2}{acp}^{\mathrm{2}} −{c}^{\mathrm{2}} {p} \\ $$$$\:\:=−{a}^{\mathrm{2}} {p}^{\mathrm{3}} +{acp}^{\mathrm{2}} +{a}^{\mathrm{2}} {p}−{ac} \\ $$$$\Rightarrow\:{acp}^{\mathrm{2}} +\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){p}+{ac}=\mathrm{0} \\ $$$$\Rightarrow\:\:{p}+\frac{\mathrm{1}}{{p}}=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{ac}} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{2}{p}}{\mathrm{1}+{p}^{\mathrm{2}} }=\frac{\mathrm{2}}{\frac{\mathrm{1}}{{p}}+{p}} \\ $$$$\Rightarrow\:\:\boldsymbol{{sin}\theta}=\frac{\mathrm{2}\boldsymbol{{ac}}}{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} } \\ $$$${similarly}\:{from}\:\left({I}\right),\:\left({II}\right) \\ $$$$\frac{\left({c}+{bq}\right)\left(\mathrm{1}−{q}^{\mathrm{2}} \right)}{{q}\left\{{a}^{\mathrm{2}} −\left({c}+{bq}\right)^{\mathrm{2}} \right\}}=\frac{\mathrm{1}}{{b}} \\ $$$$\Rightarrow\:\:−{b}^{\mathrm{2}} {q}^{\mathrm{3}} −{bcq}^{\mathrm{2}} +{b}^{\mathrm{2}} {q}+{bc} \\ $$$$\:\:\:\:=−{b}^{\mathrm{2}} {q}^{\mathrm{3}} −\mathrm{2}{bcq}^{\mathrm{2}} +{a}^{\mathrm{2}} {q}−{c}^{\mathrm{2}} {q} \\ $$$$\Rightarrow\:{bcq}^{\mathrm{2}} +\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){q}+{bc}=\mathrm{0} \\ $$$$\Rightarrow\:{q}+\frac{\mathrm{1}}{{q}}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{bc}} \\ $$$$\mathrm{sin}\:\phi=\frac{\mathrm{2}{q}}{\mathrm{1}+{q}^{\mathrm{2}} }=\frac{\mathrm{2}}{\frac{\mathrm{1}}{{q}}+{q}} \\ $$$$\Rightarrow\:\boldsymbol{{sin}\phi}=\frac{\mathrm{2}\boldsymbol{{bc}}}{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} } \\ $$$$\bigstar \\ $$
Commented by gsk2684 last updated on 24/Jun/21
thank you very much
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

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