Question

write 2sin(2x)+(rt12)cos(2x) in the form of y= Acos(2x-theta) by finding A>0 and theta is a real number between [0, 2pi]

I found that a=2 and theta =pi/2. But I don't think that's right

Answer 1

let 2x = k
then rewording the question:
write 2sin(k)+√12cos(k) in the form of y= Acos(k-theta)

Acos(k-Ø) = A(coskcosØ + sinksinØ)

2sin(k)+√12cos(k) = AcoskcosØ + AsinksinØ
matching up terms
2sink = AsinksinØ
2 = AsinØ
and
√12cosk = AcoskcosØ
√12 = AcosØ

AsinØ/(AcosØ) = 2/√12 = 2/2√3 = 1/√3
tanØ = 1/√3
Ø = 30° = π/6

and in 2 = AsinØ
2 = A(1/2)
A = 4

so 2sin(2x) + √12cos(2x) = 4cos(2x - π/6)

pick a strange value of x, say x = 2.47, and set your calculator to radians
LS = 2sin(4.94) + √12cos(4.94) = appr -1.166739...
RS = 4cos(4.94 + π/6) = appr -1.166739..

I am reasonable certain that my answer is correct