Menu Close

if-a-b-sin-a-b-sin-and-a-tan-2-b-tan-2-c-then-prove-that-the-following-i-sin-2bc-a-2-b-2-c-2-ii-sin-2ac-a-2-b-2-c-2-

Tinku Tara 0 Comments
Pin




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 ))
if(a−b)sin(θ+ϕ)=(a+b)sin(θ−ϕ)andatanθ2−btanϕ2=cthenprovethatthefollowingi)sinϕ=2bca2−b2−c2ii)sinθ=2aca2−b2+c2
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))
⇔(a−b)(sinθcosϕ+cosθsinϕ)=(a+b)(sinθcosϕ−cosθsinϕ)⇔acosθsinϕ−bsinθcosϕ=−acosθsinϕ+bsinθcosϕ⇔2acosθsinϕ=2bsinθcosϕ⇔ab=sinθcosϕcosθsinϕ⇔ab=tanθcotϕ⇔atanϕ=btanθ⇔a(2tanϕ21−tan2ϕ2)=b(2tanθ21−tan2θ2)…(i)tan(θ2)=ca+batan(ϕ2)…(ii)lettan(ϕ2)=x⇔a(x1−x2)=b(ca+bax1−(c+bxa)2)⇔a(x1−x2)=b(a(c+bx)a2−c2−2bcx−b2x2)⇔x1−x2=bc+b2xa2−c2−2bcx−b2x2⇔a2x−c2x−2bcx2−b2x3=bc+b2x−bcx2−b2x3⇒a2x−c2x−b2x−bcx2−bc=0⇒bcx2+(b2+c2−a2)x+bc=0⇒x=a2−b2−c2+(b2+c2−a2)2−4(bc)22bc⇒x=a2−b2−c2+(b2+c2−a2+2bc)(b2+c2−a2−2bc)2bc⇒x=a2−b2−c2+((b+c)2−a2)((b−c)2−a2)2bc
Commented by gsk2684 last updated on 24/Jun/21
thank you very much
thankyouverymuch
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 )) ★
sin(θ+ϕ)sin(θ−ϕ)=a+ba−b⇒sin(θ+ϕ)+sin(θ−ϕ)sin(θ+ϕ)−sin(θ−ϕ)=2a2b=ab⇒sinθcosϕcosθsinϕ=ab⇒(2tanθ21−tan2θ2)(1−tan2ϕ22tanϕ2)=abp(1−q2)q(1−p2)=ab….(I)&atanθ2−btanϕ2=c⇒ap−bq=c…..(II)p{b2−(ap−c)2}(ap−c)(1−p2)=ab2p−a2p3+2acp2−c2p=−a2p3+acp2+a2p−ac⇒acp2+(b2−c2−a2)p+ac=0⇒p+1p=a2+c2−b2acsinθ=2p1+p2=21p+p⇒sinθ=2aca2−b2+c2similarlyfrom(I),(II)(c+bq)(1−q2)q{a2−(c+bq)2}=1b⇒−b2q3−bcq2+b2q+bc=−b2q3−2bcq2+a2q−c2q⇒bcq2+(b2−a2+c2)q+bc=0⇒q+1q=a2−b2−c2bcsinϕ=2q1+q2=21q+q⇒sinϕ=2bca2−b2−c2★
Commented by gsk2684 last updated on 24/Jun/21
thank you very much
thankyouverymuch

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *