1. solving cosine ratio equations

I have an identily of the form:

cos(a1 – b)/cos(a2 – b) =(x1/x2)*cos(a1)/cos(a2)

Given a1,a2,x1 and x2, how can I solve it for b?

Thanks in advance for any help

2. Hello jrr

Welcome to Math Help Forum!
Originally Posted by jrr
I have an identily of the form:

cos(a1 – b)/cos(a2 – b) =(x1/x2)*cos(a1)/cos(a2)

Given a1,a2,x1 and x2, how can I solve it for b?

Thanks in advance for any help
It's not clear what the symbols are between the $\cos$ and the brackets, so I shall ignore them for now. (If it turns out to be a power, like $\cos^2(a_1-b)$ the same method applies.)

Expand each of the cosine expressions on the LHS using $\cos(A-B) = \cos A \cos B + \sin A\sin B$, and then divide top and bottom by $\cos b$. Then solve the resulting equation for $\tan b$.

Here's the start:
$\frac{\cos(a_1-b)}{\cos(a_2-b)}=\frac{x_1}{x_2}\cdot\frac{\cos a_1}{\cos a_2}$

$\Rightarrow\frac{\cos a_1\cos b+\sin a_1\sin b}{\cos a_2\cos b+\sin a_2\sin b}=\frac{x_1}{x_2}\cdot\frac{\cos a_1}{\cos a_2}$

$\Rightarrow\frac{\cos a_1+\sin a_1\tan b}{\cos a_2+\sin a_2\tan b}=\frac{x_1}{x_2}\cdot\frac{\cos a_1}{\cos a_2}$
Can you now make $\tan b$ the subject of this equation, and hence find $b$?

3. Thanks

That makes it starightforward now

jrr

4. I re-arranged to get

tan(b)= (x1-x2).cos(a1).cos(a2)/(x2.cos(a2).sin(a1)-x1.cos(a1).sin(a2))

but I was wondering if this could be simplified any further?

5. Hello jrr
Originally Posted by jrr
I re-arranged to get

tan(b)= (x1-x2).cos(a1).cos(a2)/(x2.cos(a2).sin(a1)-x1.cos(a1).sin(a2))

but I was wondering if this could be simplified any further?
Yes, you can divide top-and-bottom by $\cos a_1\cos a_2$ to get:
$\tan b = \frac{x_1-x_2}{x_2\tan a_1 - x_1\tan a_2}$