# Single equation, single unknown. How to solve?

• March 13th 2010, 02:30 PM
s3n4te
Single equation, single unknown. How to solve?
I have this equation:

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

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

a,b,c,and d are known constants.
• March 13th 2010, 09:23 PM
Soroban
Hello, s3n4te!

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

Quote:

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

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

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

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

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

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

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

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

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

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

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

• March 14th 2010, 11:44 AM
s3n4te
Thank you very much for the tip!
• March 14th 2010, 01:37 PM
s3n4te
I have a question though.

Can't you just make

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

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

What is the point of using r?