Inactive Tutor answered • 03/20/13
Let's use the method you listed above, but we need to correct the formula: (and note "T+a") is not the a that is in the first term, it is an angle, which I will represent with delta, δ):
a cos θ + b sin θ = c cos ( θ - δ )
The original equation, if we switch the order of the first two terms, will help us find a, b and c.
4 cos θ + 2 sin θ = sqrt(5)
From the above equation, a=4, b=2, and c=sqrt(a2+b2)=(sqrt(42+22))= sqrt(16+4)=sqrt(20).
Substituting these values into the method:
4 cos θ + 2 sin θ = sqrt(20) * cos ( θ - δ ) = sqrt(5)
The last equality, sqrt(5), comes from the original equation which looks just like the method on the left hand side of the equation.
Dividing both sides by sqrt(20):
cos ( θ - δ ) = sqrt(5) / sqrt(20) = sqrt(5/20) = sqrt(1/4) = 1/2.
So now we have cos ( θ - δ ) = 1/2. Taking the inverse cosine on both sides:
θ - δ = cos-1(0.5).
Adding delta to both sides:
θ = cos-1(0.5) + δ.
We need to determine δ to solve this equation.
A handy relationship is that tan δ = b/a, so δ = tan-1(b/a).
b=2 and a=4, so δ = tan-1(1/2) = tan-1(0.5).
The phase shift δ in the cosine curve cos ( θ - δ ) is tan-1(0.5), which is ~.464 radians or ~26.56 degrees.
So θ = cos-1(0.5) + 26.56. So for every angle where cos-1(angle)=0.5, that is our solution (provided it lies within the 0 to 360 degree range stipulated at the beginning of the problem.
We know that this happens when the angle is -60 and +60 degrees, and that pair repeats every 360 degrees. Adding -60 to 26.56 gives us an angle out of range. 60 works, as does -60+360=300. 60+360=420 is out of range too. So 60 and 300 are valid angles, which result in theta values of:
θ = 60 + 26.56 = 86.56 degrees and θ = 300 + 26.56 = 326.56 degrees.
The curve will be a standard cosine curve shifted to the right 26.56 degrees and multiplied by sqrt(20), such that its range will not be +1 to -1 but +(sqrt(20)) to -(sqrt(20))... a 'taller' curve if you will by almost 4.5 times.
