Trying to solve a Trig Equation

**levdev** · April 30th 2012, 09:12 AM

Hi There,

I've just joined the forum, and this is my first post. I'm an engineer working on problem with some angle offsets in a device, and would like to be able to solve the following equation. a, b and c are known variables, I am trying to solve for SF, is this possible?

c = Sin^-1(a/SF) - Sin^-1(b/SF)

**pickslides** · April 30th 2012, 02:18 PM

Look here

c = Sin^(-1)(a/S) - Sin^(-1)(b/S) solve for S - Wolfram|Alpha

**BobP** · May 1st 2012, 07:36 AM

The result from WA looks horrendous, maybe it simplifies ? However, try this instead.

First note that $\sin^{-1}(a/SF) = \cos^{-1}(\sqrt{SF^2 - a^2}/SF),$ and similarly for $\sin^{-1}(b/SF).$

Take the sine of the equation and making use of the expansion for $\sin(A - B),$ arrive at

$SF^{2}\sin(c) = a\sqrt{SF^{2}-b^{2}}-b\sqrt{SF^{2}-a^{2}}\qquad \dots$ (1).

Next, take the cosine of the original equation and having used the expansion for $\cos(A - B),$ arrive at

$SF^{2}\cos(c) = \sqrt{SF^{2}-a^{2}}.\sqrt{SF^{2}-b^{2}}+ab \qquad \dots$ (2).

Finally, square (1), make the obvious substitution for the product of the two square roots using (2) and then simplify to arrive at

$SF^2 = \frac{a^{2}+b^{2}-2ab\cos(c)}{\sin^{2}(c)}.$

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