cos2B+cosB
sin2B-sinB
=
cosb+1
sinB
start on TOP
Hello,
Let's see the denominator :
$\displaystyle \sin(2b)=2\cos(b)\sin(b)$
Hence $\displaystyle \sin(2b)-\sin(b)=\sin(b)(2 \cos(b)-1)$
Do you see the thing ? Do you see which formula for cos(2b) you will have to use ?
$\displaystyle \cos(2b)=2 \cos^2(b)-1$
So $\displaystyle \cos(2b)+\cos(b)=2 \cos^2(b)-1+\cos(b)$
Now this is the tricky thing :
$\displaystyle =2 \cos^2(b)+2 \cos(b)-\cos(b)-1=2 \cos(b)(\cos(b)+1)-(\cos(b)+1)$$\displaystyle =(\cos(b)+1)(2 \cos(b)-1)$
How did I know it ? Because I know where I want to go
--------> $\displaystyle \frac{\cos(2b)+\cos(b)}{\sin(2b)-\sin(b)}=\frac{(\cos(b)+1){\color{red} (2 \cos(b)-1)}}{\sin(b) {\color{red} (2 \cos(b)-1)}}$
$\displaystyle =\frac{\cos(b)+1}{\sin(b)}$
Seeing that I thanked you for your hard work, please help with this one
http://www.mathhelpforum.com/math-he...h-problem.html