Question

Prove that the max and min values of asinX + bcosX are respectively sqrt.(a^2+b^2) and -sqrt(a^2+b^2) .

Answer 1

The solution to the problem is as follows:
let y = asinx + bcosx 

dy/dx = acosx - bsinx 

= 0 for max/min 

bsinx = acosx 

sinx/cosx = a/b 

tanx = a/b 

then the hypotenuse of the corresponding right-angled triangle is √(a^2 + b^2) 

the max/min of y occurs when tanx = a/b 

then sinx = a/√(a^2 + b^2) and cosx = b/√(a^2 + b^2) 

y = a( a/√(a^2 + b^2)) + b( b/√(a^2 + b^2)) 

= (a^2 + b^2)/√(a^2 + b^2) 

= √(a^2 + b^2)

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