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Question Number 102390 by Dwaipayan Shikari last updated on 08/Jul/20

asinθ+bcosθ=c  acosecθ+bsecθ=c    prove that    sin2θ=((2ab)/(c^2 −a^2 −b^2 ))

asinθ+bcosθ=cacosecθ+bsecθ=cprovethatsin2θ=2abc2a2b2

Commented by PRITHWISH SEN 2 last updated on 08/Jul/20

Just multiply the two equations

Justmultiplythetwoequations

Commented by bobhans last updated on 09/Jul/20

if a^2 +b^2  ≥ c^2  or a^2 +b^2  ≤ c^2  ?

ifa2+b2c2ora2+b2c2?

Commented by bemath last updated on 09/Jul/20

(2)(a/(sin θ))+(b/(cos θ)) = c   ((acos θ+bsin θ)/((1/2)sin 2θ)) = c ⇒acos θ+bsin θ=(1/2)c sin 2θ  (1)×(2)   (asin θ+bcos θ)(acos θ+bsin θ)=(1/2)c^2  sin 2θ  (1/2)a^2 sin 2θ+absin^2 θ+abcos^2 θ+(1/2)b^2 sin 2θ=(1/2)c^2  sin 2θ  {(1/2)a^2 +(1/2)b^2 }sin 2θ+ab=(1/2)c^2  sin 2θ  ab = {(1/2)c^2 −(1/2)a^2 −(1/2)b^2 } sin 2θ  sin 2θ = ((2ab)/(c^2 −a^2 −b^2 ))

(2)asinθ+bcosθ=cacosθ+bsinθ12sin2θ=cacosθ+bsinθ=12csin2θ(1)×(2)(asinθ+bcosθ)(acosθ+bsinθ)=12c2sin2θ12a2sin2θ+absin2θ+abcos2θ+12b2sin2θ=12c2sin2θ{12a2+12b2}sin2θ+ab=12c2sin2θab={12c212a212b2}sin2θsin2θ=2abc2a2b2

Commented by 1549442205 last updated on 09/Jul/20

Answered by Dwaipayan Shikari last updated on 08/Jul/20

(asinθ+bcosθ)(acosθ+bsinθ).(2/(sin2θ))=c^2   (a^2 sinθcosθ+ab+b^2 sinθcosθ)(2/(sin2θ))=c^2   ((sin2θ)/(sin2θ))(a^2 +b^2 )+((2ab)/(sin2θ))=c^2   sin2θ=((2ab)/(c^2 −a^2 −b^2 ))  (proved)

(asinθ+bcosθ)(acosθ+bsinθ).2sin2θ=c2(a2sinθcosθ+ab+b2sinθcosθ)2sin2θ=c2sin2θsin2θ(a2+b2)+2absin2θ=c2sin2θ=2abc2a2b2(proved)

Commented by bobhans last updated on 09/Jul/20

If  { ((2sin θ+cos θ=2)),((2csc θ+sec θ=2)) :}  ⇔sin 2θ = ((2.2.1)/(4−4−1)) = −4??   it correct ?

If{2sinθ+cosθ=22cscθ+secθ=2sin2θ=2.2.1441=4??itcorrect?

Commented by 1549442205 last updated on 09/Jul/20

but the your system has no roots !  the first eqs implies that sin2θ=((24)/(25))!

buttheyoursystemhasnoroots!thefirsteqsimpliesthatsin2θ=2425!

Commented by PRITHWISH SEN 2 last updated on 09/Jul/20

a,b,and c cannot be an arbitary choice for these  identities. It should follow the certain condition  that      −1≤((2ab)/(c^2 −a^2 −b^2 )) ≤1  such pair can be     a=1   b=2   and  c=5 and so on.

a,b,andccannotbeanarbitarychoicefortheseidentities.Itshouldfollowthecertainconditionthat12abc2a2b21suchpaircanbea=1b=2andc=5andsoon.

Commented by bemath last updated on 09/Jul/20

it does meant c^2  ≥ a^2 +b^2

itdoesmeantc2a2+b2

Answered by 1549442205 last updated on 09/Jul/20

(1)⇔(a/(√(a^2 +b^2 )))sinθ+(b/(√(a^2 +b^2 )))cosθ=(c/(√(a^2 +b^2 )))  ⇔sin(ϕ+θ)=sin[sin^(−1) ((c/(√(a^2 +b^2 ))))](where ϕ=arccos(a/(√(a^2 +b^2 ))))  (with the condition a^2 +b^2 ≥c^2 )  ⇒ϕ+θ=sin^(−1) ((c/(√(a^2 +b^2 ))))+2kπ  ⇒θ=sin^(−1) ((c/(√(a^2 +b^2 ))))+2kπ−ϕ  ⇒sinθ=sin(sin^(−1) ((c/(√(a^2 +b^2 ))))−ϕ)  =(c/(√(a^2 +b^2 )))cosϕ−((√(a^2 +b^2 −c^2 ))/(√(a^2 +b^2 )))sinϕ  =(c/(√(a^2 +b^2 ))).(a/(√(a^2 +b^2 )))−((√(a^2 +b^2 −c^2 ))/(√(a^2 +b^2 ))).(b/(√(a^2 +b^2 )))  =((ac−b(√(a^2 +b^2 −c^2 )))/(a^2 +b^2 ))  ⇒cosθ=(1/(a^2 +b^2 ))(√((a^2 +b^2 )^2 −(ac−b(√(a^2 +b^2 −c^2 )^2 ))))  ⇒sin2θ=2sinθcosθ=2(((ac−b(√(a^2 +b^2 −c^2 )))/(a^2 +b^2 )))((1/(a^2 +b^2 ))(√((a^2 +b^2 )^2 −(ac−b(√(a^2 +b^2 −c^2 )^2 )))) )=f(a,b,c)

(1)aa2+b2sinθ+ba2+b2cosθ=ca2+b2sin(φ+θ)=sin[sin1(ca2+b2)](whereφ=arccosaa2+b2)(withtheconditiona2+b2c2)φ+θ=sin1(ca2+b2)+2kπθ=sin1(ca2+b2)+2kπφsinθ=sin(sin1(ca2+b2)φ)=ca2+b2cosφa2+b2c2a2+b2sinφ=ca2+b2.aa2+b2a2+b2c2a2+b2.ba2+b2=acba2+b2c2a2+b2cosθ=1a2+b2(a2+b2)2(acba2+b2c2)2sin2θ=2sinθcosθ=2(acba2+b2c2a2+b2)(1a2+b2(a2+b2)2(acba2+b2c2)2)=f(a,b,c)

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