1. ## trigo proving 4

$\displaystyle \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}=-cot^2\frac{A}{2}$

I started with the left hand side :
$\displaystyle \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}$

$\displaystyle =\frac{2cos2AcosA+2cos2A}{2cos2AcosA+cos2A}$

Then i bring out the common factor .

$\displaystyle =\frac{2cosA+2}{2cosA+1}$

I tried the half angle formula ..

$\displaystyle \frac{2(2cos^2\frac{A}{2}-1+1)}{2(a-sin^2\frac{A}{2}+1)}$

but to no avail .

2. Hello thereddevils
Originally Posted by thereddevils
$\displaystyle \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}=-cot^2\frac{A}{2}$

I started with the left hand side :
$\displaystyle \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}$

$\displaystyle =\frac{2cos2AcosA+2cos2A}{2cos2AcosA+cos2A}$

Then i bring out the common factor .

$\displaystyle =\frac{2cosA+2}{2cosA+1}$

I tried the half angle formula ..

$\displaystyle \frac{2(2cos^2\frac{A}{2}-1+1)}{2(a-sin^2\frac{A}{2}+1)}$

but to no avail .
Thanks for showing us your working. But I think there must be something wrong with the question. For any angle $\displaystyle A < 30^o$, the cosines of $\displaystyle A, 2A$ and $\displaystyle 3A$ are all positive. So the result can't possibly equal $\displaystyle -\cot^2(\tfrac12A)$, can it?

When you've got as far as $\displaystyle \frac{2\cos A+2}{2\cos A+1}$, did you think about using the half-angle formula using $\displaystyle t = \tan(\tfrac12A)$? Then $\displaystyle \cos A = \frac{1-t^2}{1+t^2}$

So $\displaystyle \frac{2\cos A+2}{2\cos A+1}= \frac{\dfrac{2(1-t^2)}{1+t^2}+2}{\dfrac{2(1-t^2)}{1+t^2}+1} = ...$ ?

Hello thereddevilsThanks for showing us your working. But I think there must be something wrong with the question. For any angle $\displaystyle A < 30^o$, the cosines of $\displaystyle A, 2A$ and $\displaystyle 3A$ are all positive. So the result can't possibly equal $\displaystyle -\cot^2(\tfrac12A)$, can it?

When you've got as far as $\displaystyle \frac{2\cos A+2}{2\cos A+1}$, did you think about using the half-angle formula using $\displaystyle t = \tan(\tfrac12A)$? Then $\displaystyle \cos A = \frac{1-t^2}{1+t^2}$

So $\displaystyle \frac{2\cos A+2}{2\cos A+1}= \frac{\dfrac{2(1-t^2)}{1+t^2}+2}{\dfrac{2(1-t^2)}{1+t^2}+1} = ...$ ?

$\displaystyle \frac{cosA+cos3A+2cos2A}{cosA+cos3A+cos2A}=-cot^2\frac{A}{2}$
Isn't it $\displaystyle \frac{cosA+cos3A+2cos2A}{cosA+cos3A-2cos2A}=-cot^2\frac{A}{2}$ ?
$\displaystyle \frac{\cos A+1}{\cos A-1}=\frac{2\cos^2\frac{A}{2}}{-2\sin^2\frac{A}{2}}=-\cot^2\frac{A}{2}$