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# find the values of theta..... hardest trig Q..

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).

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.

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