Simplifying a complex trigonometric expression

Question

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.

Answer 1

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,

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

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