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Context:

In a previous question , I've stated I'm making a program that will be used for calculating stuff with Statics of a particle.

I've come across another scenario in which there's three forces acting on the particle, but two of the forces are in terms of each other.

Here's a Quick diagram.

$Q$ and $F$ are in terms of each other, such that:

$$P\cos a + R(\lambda\cos b + \mu\cos c) = 0\tag1$$ $$P\sin a + R(\lambda\sin b + \mu\sin c) = W\tag2$$ and $$Q = R\lambda$$ $$F = R\mu$$

I've already formulated an equation for R, which is

$$R = \frac{W}{\lambda(\sin b - \cos b\tan a) + \mu(\sin c - \cos c\tan a)}\tag3$$

I'm in the midst of doing one for P, and what I've got so far is

$$P = \frac{W}{\sin a - \frac{\lambda\sin b\cos a + \mu\sin c\cos a}{\lambda\cos b + \mu\cos a}}\tag4$$

However, I'm just wondering if there is any way to simplify it further. I've tried what I know about trig identities but not a lot of them seem to work because of the $\lambda$ and $\mu$ coefficients.

Of course the $\cos a$ can be factored out to get $$P = \frac{W}{\sin a - \frac{\cos a(\lambda\sin b + \mu\sin c)}{\lambda\cos b + \mu\cos a}}$$

but I'm not really sure what I can do after that, as I'm left with $$\frac{\lambda\sin b + \mu\sin c}{\lambda\cos b + \mu\cos a}$$

I thought about using the identities $$\sin b + \sin c = 2\sin \left(\frac{b + c}{2}\right)\cos \left(\frac{b - c}{2}\right)$$ and $$\cos b + \cos c = -2\sin \left(\frac{b + c}{2}\right)\sin \left(\frac{b - c}{2}\right)$$

but as I previously stated, the coefficients made that difficult.

I have a feeling that perhaps this isn't actually possible to simplify further into one simple expression, but there's also a possibility I'm bad at maths so I'm asking here just to make sure.

Thanks for reading, and I appreciate any help.

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    $\begingroup$ Your previous question was quite unclear and SimpleArt had to ask, "In the end, are you trying to solve for b?" This new question is just as unclear. You used about $8$ variables. For the benefit of your readers, kindly state which variable you need to isolate or need to solve for. $\endgroup$ – Tito Piezas III Mar 26 '16 at 8:02
  • $\begingroup$ Sorry but I thought it would already be clear. My problem here is that I want to know how I can simplify the equation $$P = \frac{W}{\sin a - \frac{\cos a(\lambda\sin b + \mu\sin c)}{\lambda\cos b + \mu\cos a}}$$ Mainly just this fraction $$\frac{\lambda\sin b + \mu\sin c}{\lambda\cos b + \mu\cos a}$$ I'm not having any issues solving for any variables, it's just simple simplification as the equation has a nested fraction and looks very ugly. $\endgroup$ – NotAPro Mar 27 '16 at 0:21
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From $(3)$ and $(4)$, I am assuming you want to solve for $P$ and $R$. The trick is to remove details that may be distracting. Express your $(1)$ and $(2)$ simply as,

$$P\cos a + R\, u_1 = 0$$ $$P\sin a + R\, u_2 = W$$

Solve for $P,R$ and you have,

$$P = \frac{u_1 W}{u_1\sin a -u_2\cos a },\quad R = \frac{-\cos a \,W}{u_1\sin a -u_2\cos a }$$

From $(1),(2)$ in your post above we know $u_1,u_2$. Substitute and simplify using the rule,

$$\sin(x-y) = \sin x \cos y -\cos x \sin y$$

an insight courtesy of Mathematica, and we get the more aesthetic forms,

$$P = W\frac{\lambda\cos b + \mu\cos c}{\lambda\sin (a-b) + \mu\sin (a-c)}$$

$$R = -W\frac{\cos a }{\lambda\sin (a-b) + \mu\sin (a-c)}$$

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