Prove that max & min values of asinx+bcosx are = +/- (a^2+b^2)^1/2
The idea is to convert
f(x)=asin(x)+bcos(x)
into the form
g(x)=k*cos(a+φ)
where k and φ are constants dependent on a and b.
This way, it is easy to show that the max/min values of the f(x) is ±k.
Start with expanding g(x) using compound angles.
g(x)
=k(cos(φ)sin(x)+sin(φ)cos(x))
=kcos(φ) sin(x) + ksin(φ) cos(x)
Now assume f(x)=g(x) and compare coefficients of g(x) and f(x) to conclude that:
a=kcos(φ) and
b=ksin(φ)
Using identity: sin²(u)+cos²(u)=1,
a²+b²
=k²(cos²(φ)+sin²(φ))
=k²
or
k=√(a²+b²)
Therefore:
f(x)=k*cos(x+φ)
which by trigonometric definition of cosine has a range of (-k,k), or
-√(a²+b²) ≤ f(x) ≤ √(a²+b²)
To find the maximum and minimum values of the function f(x) = a*sin(x) + b*cos(x), we can use the properties of sine and cosine functions alongside some algebraic manipulation.
Let's start by expressing the function f(x) in a different form that allows us to take advantage of trigonometric identities. We'll use the fact that sin(x) = cos(π/2 - x) and cos(x) = sin(π/2 - x) to rewrite f(x) as follows:
f(x) = a*sin(x) + b*cos(x)
= a*cos(π/2 - x) + b*sin(π/2 - x)
Now, let's introduce a new angle θ such that θ = π/2 - x. This allows us to rewrite f(x) entirely in terms of the sine function:
f(x) = a*cos(π/2 - x) + b*sin(π/2 - x)
= a*sin(θ) + b*cos(θ)
To find the extrema of this function, we can use the fact that the maximum value of sin(θ) + cos(θ) is √2 and the minimum value is -√2. Both of these values occur when sin(θ) = cos(θ) = 1/√2. Therefore, we can express the maximum and minimum values of f(x) in terms of a and b:
Maximum value of f(x) = a*sin(θ) + b*cos(θ) = √2 * (a/√2) + √2 * (b/√2) = √2 * (a + b) = √(a^2 + b^2)
Minimum value of f(x) = a*sin(θ) + b*cos(θ) = -√2 * (a/√2) - √2 * (b/√2) = -√2 * (a + b) = -√(a^2 + b^2)
Hence, we have proven that the maximum and minimum values of the function f(x) = a*sin(x) + b*cos(x) are ±√(a^2 + b^2), respectively.
To find the maximum and minimum values of the expression asin(x) + bcos(x), we can use the concept of the trigonometric identities and properties of the sine and cosine functions.
Step 1: Consider the expression asin(x) + bcos(x) and rewrite it using a single trigonometric function. We can use the identity sin(x + y) = sin(x)cos(y) + cos(x)sin(y) to rewrite the expression:
asin(x) + bcos(x) = (a * sin(x)) + (b * cos(x))
= sqrt(a^2 + b^2) * sin(x + phi)
where phi is the angle whose cosine is a/sqrt(a^2 + b^2), and sin(x + phi) is an equivalent expression for the original expression.
Step 2: We know that the maximum and minimum values of sin(x) are +1 and -1, respectively. Therefore, the maximum and minimum values of the expression sqrt(a^2 + b^2) * sin(x + phi) will also be sqrt(a^2 + b^2) and -sqrt(a^2 + b^2), respectively.
Therefore, the maximum value of the expression asin(x) + bcos(x) is sqrt(a^2 + b^2), and the minimum value is -sqrt(a^2 + b^2).
Hence, we have proved that the maximum and minimum values of asin(x) + bcos(x) are equal to +/- sqrt(a^2 + b^2).