# Question on trigonometric linear equations

Doing an exercise on complex analysis where it began by asking me to solve some equations for $u_x$ and $u_y$ I got stuck and looked up the answer.

$$u_x\cos\theta+u_y\sin\theta=u_r,$$ $$-u_xr\sin\theta+u_yr\cos\theta=u_\theta.$$ Solving these simultaneous linear equations for $u_x$ and $u_y$, we find that $$u_x=u_r\cos\theta-u_\theta\frac{\sin\theta}{r}\quad\text{ and }\quad u_y=u_r\sin\theta+u_\theta\frac{\cos\theta}{r}.$$

(Here's the original image of this quoted text.)

The answer says 'solving these simultaneous linear equations'...but surely these are not linear equations when they use cos and sin?

And how do you solve these equations anyway, i can see you can divide across the bottom equation but I can see what to do after that.

• $\theta$, $u_\theta$, $u_r$, and $r$ are treated as constants. Note, the solution is different for different $\theta$, $u_\theta$, and $u_r$, $r$. – David Mitra Feb 11 '12 at 18:43
• They're not "linear in $\theta$" since $\sin$ and $\cos$ are involved. They're linear in $u_x$ and $u_y$. – Michael Hardy Feb 11 '12 at 20:28

The equations are linear in $u_x$ and $u_y$—that is, if $\theta$ is a constant, $\sin\theta$ and $\cos\theta$ are constants, so they're just like having numbers there.
If you multiply the first equation by $\sin\theta$ and the second equation by $\frac{\cos\theta}{r}$, you'll have $$u_x\cos\theta\sin\theta+u_y\sin^2\theta=u_r\sin\theta$$ $$-u_x\cos\theta\sin\theta+u_y\cos^2\theta=u_\theta\frac{\cos\theta}{r}$$ Adding these two equations gives $$u_y(\sin^2\theta+\cos^2\theta)=u_r\sin\theta+u_\theta\frac{\cos\theta}{r}$$ and since $\sin^2\theta+\cos^2\theta=1$, $$u_y=u_r\sin\theta+u_\theta\frac{\cos\theta}{r}.$$
You can find $u_x$ from this $u_y$, or by applying a similar technique to the original equations.