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 ))

Commented byPRITHWISH SEN 2 last updated on 08/Jul/20

Just multiply the two equations

Commented bybobhans last updated on 09/Jul/20

if a^2 +b^2 ≥ c^2 or a^2 +b^2 ≤ c^2 ?

Commented bybemath 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 ))

Commented by1549442205 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)

Commented bybobhans 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 ?

Commented by1549442205 last updated on 09/Jul/20

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

Commented byPRITHWISH 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.

Commented bybemath last updated on 09/Jul/20

it does meant c^2 ≥ a^2 +b^2

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)