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if max {5sinx+3sin(x-q)}=7 then the set of possible values of q is: |
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Answer» 5sinx+3sin(x-q) = 5sinx+3sinxcosq-3cosxsinq =(5+3cosq) sinx-3sinqcosx =asinx-bcosx, [where (5+3cosq)=a and 3sinq =b] = (a2+b2)^1/2[{a/(a2+b2)^1/2}sinx-{b/(a2+b2)^1/2}cosx] = (a2+b2)^1/2[cosØsinx-sinØcosx], where tanØ=b/a =(a2+b2)^1/2[sin(x-Ø)] this will have maximum value for sin(x-Ø)=1 Hence max[5sinx+3sin(x-q)] =7 => (a2+b2)^1/2=7 =>(a2+b2)=49 => ((5+3cosq)2+(3sinq)2)=49 => (25+9cos2q+30cosq+9sin2q)=49 => 34+30cosq=49 {since 9cos2q+9sin2q=9] => cosq=1/2=cos(π/3) So q = 2nπ±π/3, where n belongs to Z |
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