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The maximum value of the expression sinx +ncosx,when x can take any value is

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one year ago

Vikas TU
13783 Points
```							Dear student Sinx + nCosx Multiply and divide by [(coefficient of sinx)^2+(coefficient of cosx)^2]^1/2 i.e. in this condition(n^2+1^2)^1/2Now the equation transforms to[sinx/(n^2+1^2)^1/2+ncosx/(n^2+12)^1/2]×(n^2+1^2)^1/2Now consider a triangle with angle ø.Let the perpendicular length be n and base be 1.So hypotenuse is (n^2+1^2)^1/2Therefore our equation transforms to[cosøsinx+sinøcosx]×(n^2+1^2)^1/2sin(ø+x)×(n^2+1^2)^1/2As max value of sin(x+ø) is 1So maximum value of  sin(ø+x)×(n^2+1^2) ^1/2 is (n^2+1^2)^1/2 which is the maximum value of our original equation.So max value is [(coefficient of sinx)2+(coefficient of cosx)2]^1/2
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one year ago
Rajat
213 Points
```							f=sinx+ncosxf'= cosx -nsinx=0 (for maximum )So tanx = 1/nSo, sinx= 1/√(1+n^2)sinx+ncosx=1/√(1+n^2)+n^2/√(1+n^2)= √(1+n^2)
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one year ago
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