Any chance of a hand?

brownacid

A friend asked me to solve an equation for him, I've worked it out to be A = 14.02. The equation is

-0.38 = sin(2A)-0.85cos(2A)

Solve for A.

By messing with the trig I can simplify it down to be
sin2A = 0.47

I then said x = 2A and said arcsin(0.47) = 28.04, therefore A = 14.02. I don't think I can just make that leap and solve x though. Also when I plug the value back in for A it doesn't give me the right answer. I put it into matlab and solved numerically and got A to be 11.75 so I reckon thats the right answer but could anyone give a hand and tell me how I can solve it analytically?

Cheers!

MathsManiac

Not sure how you did your initial manipulation. I would use one or other of the following methods:

Let x=sin2A, giving cos2A=sqrt(1 - x^2).
Substituting this in and doing a bit of algebraic manipulation gives the quadratic equation: 1.7225x^2 + 0.76x - 0.5781.

This gives x = 0.3993, or x = -0.8405.

The first of these gives a solution for 2A in the first quadrant; in degrees it's 2A = 23.534, with an arbitrary multiple of 360 added, giving A = 11.767 + 180n.

The second x value yields a solution in the third quadrant for 2A which leads to A = 118.596 + 180n.

Alternative method:
Try to write the right-hand side as a compound angle. To do this, scale the equation in such a way that the squares of the coefficients of sin2A and cos2A add up to 1. In this case, multiply across by 0.7619. This gives:
-0.2895 = 0.7619sin2A - 0.6476cos2A.

Now, if you let B be the acute angle whose sin is 0.7619, then you have:
-0.2895 = sinBsin2A - cosBcos2A
which can be written as:
cos(B+2A) = 0.2895.
This gives B+2A = 73.16976 or 286.83024 (each with +360n)
Subtracting off the known value of B and dividing by 2 yields the same answers as method 1:
A = 11.767 + 180n and A = 118.596 + 180n.

bnt

Solving numerically (HP 35S calculator) I also get A = 11.767 if I assume the angles are in Degrees, or A=0.2054 radians). (The Degrees / Radians thing can be a problem with Trig on calculators & computers.)