Isolation of variable through the use of inverse

Question

Here is my equation: cosA=b/c. Now, I want to isolate A on the left-hand side. It is my understanding that cosA is a calculated value. "A" has a degree value, but cos has no numerical value until we associate this function with an angle, in this case, A. Let us say that the value of cosA is 4/5. Does not the equation then become 4/5=b/c? And if I multiply both sides by the inverse of this cos value, does not the equation become 1=arccos(b/c)? From my original source, the recommended solution is to multiply both sides by the inverse, but I am not understanding how this isolates A. To me, cosA is fundamentally different from, say, xy, in which case both components have an independent existence.  Thanks. Cary

Answer 1

Hi Cary,

 

Cosine is an operation that is performed on an input, and so is inverse cosine. 

 

If if I had an equation such as:

 

    3*x = 15

 

I want to get x by itself on one side of the equation. It's not "alone" right now, because it's being multiplied by 3. In order to "undo" that multiplication, I would divide by 3. Anything that I do to either side of the equation I must do to both sides, so that the equation still holds true. So:

 

    3*x = 15

      3       3

 

    x = 5

 

In your example,

 

    cos(A) = 4/5

 

In this case, A is not "alone", so we want to undo the cos function that is being applied to it. In the same way that multiplication and division undo each other, trig functions and their inverses undo each other. We want to take the inverse cosine of the left hand side (which does NOT mean multiplying by inverse cosine), in order to cancel the cosine, leaving A by itself. But, again, we have to take the inverse cosine of BOTH sides of the equation. So:

 

    cos^-1(cos(A)) = cos^-1(4/5)

 

Taking the inverse cosine of a cosine cancels them both, leaving

 

    A = cos^-1(4/5)

 

Which you can plug into your calculator... I hope this helps!

Answer 2

Cary,

 

Cos(A) is just an example of a function f(A), where the angle A can be defined either in degrees or in radians (2Pi radians in the circumference of a circle).  

 

The ratio of b/c is just the value of Cos(A) for some angles. In a right triangle, it is the ratio of the adjacent side to the hypotenuse.  The solution for angle A is multi-valued.  For example. Cos (45^o)= (1/2)√2 = 0.707 and so is the Cos(315^o) or Cos(-45^o).

 

The solution can be represented as:    arc cosine(cos(A)) = A = arc cosine(b/c).  Then you use tables (and interpolate between given values) to match the angle (in radians probably) to the calculated arc cosine(b/c) value.  Or you can use the cos(A) tables and match A to the calculated b/c ratio.

 

Cosine(A) is different as you say. It is a transcendental function whose value is approximated by an infinite power series in A.  That's why you don't solve for A "algebraically" but must use tabular approximations and do the matching process.

 

That is, cos(A) = 1 - A²/2! + A⁴/4! - A⁶/6! + etc.  for all real A in radians and ! is the factorial function.

Tabulated data is calculated using a finite set of these alternating terms (more than I've shown here).