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

Answer 1

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)