find the range of y=acosx + bsinx

Question

it is a trigonometrical function .plz provide me answer in detail.

Answer 1

y=acos(x)+bsin(x) can be expressed as:

y=rsin(θ)cos(x)+rcos(θ)sin(x)=rsin(θ+x) and since the range of sine is [-1,1] the range for y is [-r,r].

We need to relate r to a and b.

a=rsin(θ) and b=rcos(θ), so r²=a²+b², because sin²(θ)+cos²(θ)=1, r=√(a²+b²), making the range for y [-√(a²+b²),√(a²+b²)].

For example, if y=3cos(x)+4sin(x), the range for y is [-5,5].

[θ=tan^-1(a/b) and corresponds to a phase shift in x.]