Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 99433 by Dwaipayan Shikari last updated on 20/Jun/20

sin7φ+cos2φ=−2 Find,φ

sin7ϕ+cos2ϕ=−2Find,ϕ

Commented by bobhans last updated on 21/Jun/20

sin 7∅+sin (90−2∅)=−2 2sin (((90+5∅)/2)) cos (((9∅−90)/2)) = −2 sin (((90+5∅)/2)) cos (((9∅−90)/2)) = −1 case (1) { ((sin (((90+5∅)/2)) = −1)),((cos (((9∅−90)/2)) = 1)) :} case(2) { ((sin (((90+5∅)/2)) = 1)),((cos (((9∅−90)/2)) = −1)) :}

sin7∅+sin(90−2∅)=−22sin(90+5∅2)cos(9∅−902)=−2sin(90+5∅2)cos(9∅−902)=−1case(1){sin(90+5∅2)=−1cos(9∅−902)=1case(2){sin(90+5∅2)=1cos(9∅−902)=−1

Answered by mr W last updated on 21/Jun/20

sin 7φ=−1 & cos 2φ=−1 7φ=2mπ−(π/2) ⇒φ=((2mπ)/7)−(π/(14)) 2φ=2nπ+π ⇒φ=nπ+(π/2) nπ+(π/2)=((2mπ)/7)−(π/(14)) 2m−7n=4 ⇒n=2k ⇒φ=2kπ+(π/2)

sin7ϕ=−1&cos2ϕ=−17ϕ=2mπ−π2⇒ϕ=2mπ7−π142ϕ=2nπ+π⇒ϕ=nπ+π2nπ+π2=2mπ7−π142m−7n=4⇒n=2k⇒ϕ=2kπ+π2

Commented by Rio Michael last updated on 21/Jun/20

have a little problem, how did you arrive at the conclusion sin 7∅ = −1 and cos 2∅ = −1

havealittleproblem,howdidyouarriveattheconclusionsin7∅=−1andcos2∅=−1

Commented by ChristopherPeter last updated on 21/Jun/20

jjn

Commented by floor(10²Eta[1]) last updated on 21/Jun/20

it′s the only possible cases, since sina and cosb are between −1 and 1

it′stheonlypossiblecases,sincesinaandcosbarebetween−1and1

Commented by mr W last updated on 21/Jun/20

exactly! if A≥−1, B≥−1 and if A+B=−2 then A=−1 and B=−1.

exactly!ifA⩾−1,B⩾−1andifA+B=−2thenA=−1andB=−1.

Commented by 1549442205 last updated on 21/Jun/20

because { ((sin7φ≥−1)),((cos2φ≥−1)) :} which implies that sin7φ+cos2φ≥−2. Hence,sin7φ+cos2φ=−2 if and only if { ((sin7φ=−1)),((cos2φ=−1)) :}

because{sin7ϕ⩾−1cos2ϕ⩾−1whichimpliesthatsin7ϕ+cos2ϕ⩾−2.Hence,sin7ϕ+cos2ϕ=−2ifandonlyif{sin7ϕ=−1cos2ϕ=−1

Answered by 1549442205 last updated on 21/Jun/20

one other way: sin7φ+cos2φ=−2⇔(1+cos((π/2)−7φ))+(1+cos2φ)=0 ⇔2cos^2 (((π/2)−7φ)/2)+2cos^2 φ=0 ⇔ { ((cos(((π/2)−7φ)/2)=0 (1) )),((cosφ=0 (2))) :}⇔ { (((((π/2)−7φ)/2)=(π/2)+m.π)),((φ=(π/2)+n.π)) :} ⇔ { ((φ=((((−π)/2)+2m.π)/7))),((φ=(π/2)+n.π)) :}⇔((((−π)/2)+2m.π)/7)=(π/2)+nπ 2m=4+7n⇔m=3n+(n/2)+2⇒ { ((n=2k)),((m=7k+2)) :} Thus,𝛗=(𝛑/2)+2k𝛑 is roots of the problem

oneotherway:sin7ϕ+cos2ϕ=−2⇔(1+cos(π2−7ϕ))+(1+cos2ϕ)=0⇔2cos2π2−7ϕ2+2cos2ϕ=0⇔{cosπ2−7ϕ2=0(1)cosϕ=0(2)⇔{π2−7ϕ2=π2+m.πϕ=π2+n.π⇔{ϕ=−π2+2m.π7ϕ=π2+n.π⇔−π2+2m.π7=π2+nπ2m=4+7n⇔m=3n+n2+2⇒{n=2km=7k+2Thus,ϕ=π2+2kπisrootsoftheproblem

Terms of Service

Privacy Policy

Contact: info@tinkutara.com