Question

Prove that max & min values of asinx+bcosx are =/- (a^2+b^2)^1/2

Answer 1

See:
http://www.jiskha.com/display.cgi?id=1373620166

Answer 2

or

let y = asinx + bcosx
dy/dx = acosx - bsinx
= 0 for max/min
bsinx = acosx
sinx/cosx = a/b
<b>tanx = a/b</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)

Answer 3

I want Maximum and minimum values of acosx+bsinx+c proof

Answer 4

Same is the proof...here derivative of c wrt x is zero...so max occurs at tanx= a/b...substitute , u will get answer

Answer 5

I was looking for this. Thank You

Answer 6

how can you find minimum value of it