# Single equation, single unknown. How to solve?

• Mar 13th 2010, 01:30 PM
s3n4te
Single equation, single unknown. How to solve?
I have this equation:

$\displaystyle a+b+c*cos(\theta_4)=cos(d-\theta_4)$

I was wondering how do I solve for $\displaystyle \theta_4$?

a,b,c,and d are known constants.
• Mar 13th 2010, 08:23 PM
Soroban
Hello, s3n4te!

It can be solved, but I'm sure you won't like it . . .

Quote:

Solve for $\displaystyle \theta\!:\;\;a+b+c\cos(\theta) \:=\:\cos(d-\theta)$

We have: .$\displaystyle \cos(d-\theta) - c\cos\theta \;=\;a+b$

. . $\displaystyle \cos d\cos\theta + \sin d\sin\theta - c\cos\theta \;=\;a+b$

. . $\displaystyle (\cos d - c)\cos\theta + \sin d\sin\theta \;=\;a+b$ .[1]

Consider: .$\displaystyle r^2 \:=\:(\cos d - c)^2 + (\sin d)^2 \;=\;\cos^2\!d - 2c\cos d + c^2 + \sin^2\!d \;=\;1 - 2c\cos d + c^2$

. . Then: .$\displaystyle r \;=\;\sqrt{1-2c\cos d + c^2}$

Divide equation [1] by $\displaystyle r\!:\;\;\frac{\cos d - c}{r}\cos\theta + \frac{\sin d}{r}\sin\theta \:=\:\frac{a+b}{r}$

Let $\displaystyle \alpha$ be an angle such that: .$\displaystyle \begin{Bmatrix}\sin\alpha &=& \dfrac{\cos d - c}{r} \\ \\[-3mm] \cos\alpha &=& \dfrac{\sin d}{r} \end{Bmatrix}$
. . Note that: .$\displaystyle \tan\alpha \:=\:\frac{\cos d - c}{\sin d}$

Our equation becomes: .$\displaystyle \sin\alpha\cos\theta + \cos\alpha\sin\theta \:=\:\frac{a+b}{r}$

Then: .$\displaystyle \sin(\theta + \alpha) \;=\;\frac{a+b}{r} \quad\Longrightarrow\quad \theta + \alpha \;=\;\sin^{-1}\!\left(\frac{a+b}{r}\right)$

Therefore: .$\displaystyle \theta \;=\;\sin^{-1}\!\left(\frac{a+b}{r}\right) - \alpha \quad\Longrightarrow\quad \boxed{\;\theta \;=\;\sin^{-1}\!\left(\frac{a+b}{r}\right) - \tan^{-1}\!\left(\frac{\cos d - c}{\sin d}\right)}$

• Mar 14th 2010, 10:44 AM
s3n4te
Thank you very much for the tip!
• Mar 14th 2010, 12:37 PM
s3n4te
I have a question though.

Can't you just make

$\displaystyle sin(\alpha) =cos(d)-c$ and $\displaystyle cos(\alpha)=sin(d)$

Which would make $\displaystyle \theta =sin^{-1}(a+b)-tan^{-1}(\frac{cos(d)-c}{sin(d)})$

What is the point of using r?