General Physics
The maximum value of the expression sinx +ncosx,when x can take any value is
The maximum value of the expression sinx +ncosx,when x can take any value is
2 Answers
Vikas TU
Dear student
Sinx + nCosx
Multiply and divide by [(coefficient of sinx)^2+(coefficient of cosx)^2]^1/2 i.e. in this condition
Multiply and divide by [(coefficient of sinx)^2+(coefficient of cosx)^2]^1/2 i.e. in this condition
(n^2+1^2)^1/2
Now the equation transforms to
[sinx/(n^2+1^2)^1/2+ncosx/(n^2+12)^1/2]×(n^2+1^2)^1/2
Now consider a triangle with angle ø.
Let the perpendicular length be n and base be 1.
So hypotenuse is (n^2+1^2)^1/2
Therefore our equation transforms to
[cosøsinx+sinøcosx]×(n^2+1^2)^1/2
sin(ø+x)×(n^2+1^2)^1/2
As max value of sin(x+ø) is 1
So 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
Rajat
f=sinx+ncosx
f'= cosx -nsinx=0 (for maximum )
So tanx = 1/n
So, sinx= 1/√(1+n^2)
sinx+ncosx=1/√(1+n^2)+n^2/√(1+n^2)
= √(1+n^2)
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