Prove that max & min values of asinx+bcosx are =/- (a^2+b^2)^1/2
or
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)
I was looking for this. Thank You
how can you find minimum value of it
I want Maximum and minimum values of acosx+bsinx+c proof
To prove that the maximum and minimum values of the function f(x) = asinx + bcosx are equal to ±(a^2 + b^2)^1/2, we can use the concept of vector projection and the Cauchy-Schwarz inequality.
1. Let's start by applying the Cauchy-Schwarz inequality to the vectors u = (a, b) and v = (sinx, cosx):
|u·v| ≤ |u| * |v|
The dot product u·v is equal to au + bv, and the lengths |u| and |v| are given by the square root of the sums of the squares of their components:
|u·v| ≤ √(a^2 + b^2) * √(sin^2x + cos^2x)
2. We know that sin^2x + cos^2x = 1, so the inequality becomes:
|u·v| ≤ √(a^2 + b^2)
3. Rearranging the inequality, we have:
-√(a^2 + b^2) ≤ au + bv ≤ √(a^2 + b^2)
4. Now, let's rewrite the left-hand side of the inequality by factoring out the common magnitude of the vector u:
-√(a^2 + b^2) ≤ √(a^2 + b^2) * (A * sinx + B * cosx)
where A = a / √(a^2 + b^2) and B = b / √(a^2 + b^2)
5. Simplifying the inequality, we get:
-1 ≤ A * sinx + B * cosx
6. Using the fact that sin^2x + cos^2x = 1, we can rewrite the right-hand side of the inequality as follows:
-1 ≤ √(A^2 + B^2) * (sin(x + θ))
where θ is the phase shift and satisfies tan(θ) = B / A
7. Since -1 ≤ sin(x + θ) ≤ 1 for all values of x, we conclude that the maximum and minimum values of f(x) = asinx + bcosx are ±√(a^2 + b^2).
Therefore, we have proved that the maximum and minimum values of the function f(x) = asinx + bcosx are equal to ±(a^2 + b^2)^1/2.
See:
http://www.jiskha.com/display.cgi?id=1373620166