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For-R-cos-cosx-siny-sinx-x-R-then-find-the-sum-of-the-possible-values-of-sin-siny-




Question Number 27958 by bmind4860 last updated on 17/Jan/18
For α∈R, cosαcosx+siny≥sinx, ∀x∈R,  then find the sum of the possible values  of sinα+siny.
ForαR,cosαcosx+sinysinx,xR,thenfindthesumofthepossiblevaluesofsinα+siny.
Answered by ajfour last updated on 17/Jan/18
    sin y ≥ sin x−cos αcos x  ⇒    ≥ (√(1+cos^2 α)) [(1/( (√(1+cos^2 α))))sin x−((cos α)/( (√(1+cos^2 α))))cos x]  sin y≥ (√(1+cos^2 α)) sin [x−tan^(−1) (cos α)]  since x can be any value,  and sin y ≤ 1 ⇒  cos α=0  ⇒ sin α =±1  then this implies       sin y ≥ sin x  and again for this to be always  true sin y =1  hence sin y+sin α=0 or 2 .
sinysinxcosαcosx1+cos2α[11+cos2αsinxcosα1+cos2αcosx]siny1+cos2αsin[xtan1(cosα)]sincexcanbeanyvalue,andsiny1cosα=0sinα=±1thenthisimpliessinysinxandagainforthistobealwaystruesiny=1hencesiny+sinα=0or2.
Commented by bmind4860 last updated on 17/Jan/18
thank you! would you suggest what do  you think about the following question,  whether it is right or wrong  find general solution, sinx+sin3x+sin(√x)=0.
thankyou!wouldyousuggestwhatdoyouthinkaboutthefollowingquestion,whetheritisrightorwrongfindgeneralsolution,sinx+sin3x+sinx=0.
Commented by ajfour last updated on 17/Jan/18
cant say, i couldn′t find a way .
cantsay,icouldntfindaway.
Commented by bmind4860 last updated on 17/Jan/18
me too that′s why i think it is wrong  and thank you for suggestion
metoothatswhyithinkitiswrongandthankyouforsuggestion

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