Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have come up with the following system, I want to solve it for $a$ and $c$:

$ a \sin (x_0) - c \sin(x_0 - L) = 0\\ c \cos(x_0 - L) - a \cos(x_0) = 1 $

In this system $x_0$ and $L$ are arbitrary.

p.s. why can't $L$ be $n\pi$?

share|improve this question
    
so you want to solve it for a and c am I right? –  iostream007 May 29 '13 at 10:23
    
Yes, sorry forgot to mention it –  user54297 May 29 '13 at 10:25

3 Answers 3

I assume you are solving for $a$ and $c$. Then the solution may be written in matrix form as

$$\begin{align}\left ( \begin{array}\\a\\c\end{array}\right) &= \frac{1}{\sin{x_0} \cos{(x_0-L)} - \cos{x_0} \sin{(x_0-L)}} \left ( \begin{array}\\\cos{(x_0-L)} & \sin{(x_0-L)}\\\cos{x_0} & \sin{x_0}\end{array}\right)\left ( \begin{array}\\0\\1\end{array}\right) \\ &= \frac{1}{\sin{L}}\left ( \begin{array}\\\sin{(x_0-L)}\\\sin{x_0}\end{array}\right)\end{align}$$

Note that I used the sine addition formula to simplify the denominator term. This explains why $L \ne n \pi$: the denominator would be zero.

Then

$$a = \frac{\sin{(x_0-L)}}{\sin{L}}$$

$$c = \frac{\sin{x_0}}{\sin{L}}$$

share|improve this answer

$ a \sin (x_0) - c \sin(x_0 - L) = 0\\ c \cos(x_0 - L) - a \cos(x_0) = 1 $

from eqn (1)

$$a \sin (x_0) =c \sin(x_0 - L)\implies a=\dfrac {c \sin(x_0 - L)}{\sin (x_0)}$$

in eqn (2)

$$c \cos(x_0 - L) - \cos(x_0)\cdot\dfrac{c \sin(x_0 - L)}{\sin (x_0)} = 1$$ $$c[\cos(x_0 - L)\cdot\sin (x_0)-\sin(x_0 - L)\cdot\cos(x_0)]=\sin {(x_0)}$$ $$c[\sin (x_0-(x_0 - L))]=\sin {(x_0)}$$ $$c=\dfrac{\sin {(x_0)}}{\sin (L)}$$

Now $$a=\dfrac {c \sin(x_0 - L)}{\sin (x_0)}$$ $$a=\dfrac {\dfrac{\sin {(x_0)}}{\sin (L)}\cdot\sin(x_0 - L)}{\sin (x_0)}$$ $$a=\dfrac{\sin(x_0-L)}{\sin L}$$

share|improve this answer

Writing $P:= \sin x_0$, $Q := \sin(x_0-L)$, $R := \cos x_0$, $S := \cos(x_0-L)$, we solve this far-less-intimidating system ... $$\begin{align} \phantom{-}a P - c Q &= 0 \\ - a R + c S &= 1 \end{align}$$ ... to get ... $$a = \frac{Q}{PS-QR} \qquad c = \frac{P}{PS-QR}$$

Then, we invoke trig's angle-difference formula to simplify: $$\begin{align} PS-QR &= \sin x_0 \cos(x_0-L)-\sin(x_0-L)\cos x_0 \\[4pt] &= \sin\left(x_0-(x_0-L)\right) \\[4pt] &= \sin L \end{align}$$ so that $$a = \frac{\sin(x_0-L)}{\sin L} \qquad c = \frac{\sin x_0}{\sin L}$$ where $\sin L \neq 0$ (whence $L \neq n \pi$ for any $n$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.