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\begin{cases} P\cos a + Q\cos b + F\cos c = 0 \\ P\sin a + Q\sin b + F\sin c = W \end{cases}

I am trying to find $a$ and $b$.

My initial attempt was using the identity $\sin a = \sqrt{1 - \cos^2 a}$ to try and sub the $\cos a$ in the first equation into the second equation, like so:

$$\cos a = \frac{-(Q\cos b + F\cos c)}{P}$$

$$\cos^2 a = \frac{Q^2\cos^2 b + 2QF\cos b\cos c + F^2\cos^2 c}{P^2}$$

$$\sin a = \sqrt{1 - \frac{Q^2\cos^2 b + 2QF\cos b\cos c + F^2\cos^2 c}{P^2}}$$

$$ \left(P\sqrt{1 - \frac{Q^2\cos^2 b + 2QF\cos b\cos c + F^2\cos^2 c}{P^2}} + Q\sin b\right)^2 = (W - F\sin c)^2$$

I was going to continue but I realised after I expanded the quadratic on the LHS that I'd end up with another square root term involving $b$, which I'd have to square again, and the cycle would continue.

So I'm not really sure what approach I should be making, or even if I was going in the right direction.

Would appreciate help with this, and thanks for reading.

EDIT:

Using the identity $\sin^2 a + \cos^2 a = 1$, I've formulated a very convoluted quadratic expression that manages to eliminate $a$.

$$cos^2 a = \frac{Q^2\cos^2 b + 2QF\cos b\cos c + F^2\cos^2 c}{P^2} \tag{1}$$ $$\sin^2 a = \frac{W^2 - 2WQ\sin b - 2WF\sin c + 2FQ\sin b\sin c + Q^2\sin^2 b + F^2\sin^2 c}{P^2} \tag{2}$$

Now the only problem is rearranging for $a$ or $c$.

NotAPro
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2 Answers2

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We have $$-F\cos c=P\cos a+Q\cos b\tag0$$ Squaring the both sides gives $$F^2\cos^2c=P^2\cos^2a+Q^2\cos^2b+2PQ\cos a\cos b\tag1$$ Similarly, $$W^2-2WF\sin c+F^2\sin^2c=P^2\sin^2a+Q^2\sin^2b+2PQ\sin a\sin b\tag2$$ Now $(1)+(2)$ with $\cos^2\theta+\sin^2\theta=1$, $$W^2-2WF\sin c+F^2=P^2+Q^2+2PQ\cos(a-b),$$ i.e. $$a-b=\arccos\left(\frac{W^2-2WF\sin c+F^2-P^2-Q^2}{2PQ}\right)\tag4$$

Now, let $d$ be the RHS of $(4)$.

Then, from $(0)$ we have $$P\cos a+Q\cos(a-d)+F\cos c=0,$$ i.e. $$(P+Q\cos d)\cos a+Q\sin d\sin a+F\cos c=0$$ which we can write as $$s\cos a+t\sin a+F\cos c=0$$ where $s=P+Q\cos d,t=Q\sin d$, and so $$\sqrt{s^2+t^2}\cos\left(a-\arctan\frac ts\right)+F\cos C=0$$ Thus, $a,b$ can be written as $$\color{red}{a=\arctan\frac ts+\arccos\left(\frac{-F\cos c}{\sqrt{s^2+t^2}}\right),\qquad b=\arctan\frac ts+\arccos\left(\frac{-F\cos c}{\sqrt{s^2+t^2}}\right)-d}$$ where $$d=\arccos\left(\frac{W^2-2WF\sin c+F^2-P^2-Q^2}{2PQ}\right),\qquad s=P+Q\cos d,\qquad t=Q\sin d.$$

mathlove
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For simplicity, we rewrite

$$P\cos(a)+Q\sin(b)=A,\\P\sin(a)+Q\cos(b)=B.$$

Then we can eliminate one of the unknowns with

$$Q^2=Q^2\cos^2(b)+Q^2\sin^2(b)=(A-P\cos(a))^2+(B-P\sin(a))^2\\ =A^2+B^2+P^2-2PA\cos(a)-2PB\sin(a)).$$

This can be put in the well-known form

$$\alpha\cos(a)+\beta\sin(a)=\gamma$$

which is addressed for instance here Solving trigonometric equations of the form $a\sin x + b\cos x = c$.

$$a=\pm\arccos\left(\frac\gamma{\sqrt{\alpha^2+\beta^2}}\right)+\arctan\left(\frac\beta\alpha\right).$$

Similary fo $b$.

Because of the squarings, we may have introduced alien solutions. You can detect them by plugging the values in the initial equations.

Community
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