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Question Number 210375 by Spillover last updated on 08/Aug/24

Commented by som(math1967) last updated on 08/Aug/24

 if sin^(−1) x+sin^(−1) y+sin^(−1) z=𝛑  then x^4 +y^4 +z^4 +4x^2 y^2 z^2    =2(x^2 y^2 +y^2 z^2 +z^2 x^2 )

ifsin1x+sin1y+sin1z=πthenx4+y4+z4+4x2y2z2=2(x2y2+y2z2+z2x2)

Commented by som(math1967) last updated on 08/Aug/24

 ((sinxsiny)/(cosx+cosy))=((2tan(x/2)tan(y/2))/(1−tan^2 (x/2)tan^2 (y/2)))

sinxsinycosx+cosy=2tanx2tany21tan2x2tan2y2

Commented by Spillover last updated on 08/Aug/24

correct.thank you

correct.thankyou

Commented by Spillover last updated on 08/Aug/24

correct.thank you

correct.thankyou

Answered by som(math1967) last updated on 08/Aug/24

(c) tan^2 𝛉=1−m^2   ⇒1+tan^2 θ=2−m^2   ⇒sec^2 θ=2−m^2   ⇒secθ=±(2−m^2 )^(1/2)    secθ+tan^3 θcosecθ  =secθ+((sin^3 θcosecθ)/(cos^3 θ))  =secθ+tan^2 θsecθ  =secθ(1+tan^2 θ)  =±(2−m^2 )^(1/2) (2−m^2 )  =±(2−m^2 )^(3/2)

(c)tan2θ=1m21+tan2θ=2m2sec2θ=2m2secθ=±(2m2)12secθ+tan3θcosecθ=secθ+sin3θcosecθcos3θ=secθ+tan2θsecθ=secθ(1+tan2θ)=±(2m2)12(2m2)=±(2m2)32

Commented by Spillover last updated on 08/Aug/24

great.thanks

great.thanks

Answered by som(math1967) last updated on 08/Aug/24

(e) cosA=((b^2 +c^2 −a^2 )/(2bc))  ⇒2bccosA=b^2 +c^2 −a^2   ⇒2bc(1−2sin^2 (A/2))=b^2 +c^2 −a^2   ⇒a^2 =b^2 +c^2 −2bc+4bcsin^2 (A/2)  ∴ a^2 =(b−c)^2 +4bcsin^2 (A/2)

(e)cosA=b2+c2a22bc2bccosA=b2+c2a22bc(12sin2A2)=b2+c2a2a2=b2+c22bc+4bcsin2A2a2=(bc)2+4bcsin2A2

Commented by Spillover last updated on 08/Aug/24

great.thanks

great.thanks

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