$\cot(A) =\sin(2A)/(1-\cos(2A))$
I've tried converting the top and bottom by
every single one of their equivalent formulas but didn't have any luck.
Thank you
Asked
Active
Viewed 82 times
-2
-
Use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html – lab bhattacharjee Nov 27 '17 at 00:46
-
Possible duplicate of How to get $ \cot(\theta/2)$ from $ \frac {\sin \theta} {1 - \cos \theta} $?. – robjohn Nov 27 '17 at 00:54
2 Answers
2
$$\begin {align}\frac {\sin(2A)}{1-\cos(2A)}&=\frac{2\sin A \cos A}{1-\cos^2A+\sin^2A}\\ &=\frac {2 \sin A \cos A}{2\sin^2A}\\ &=\cot A \end {align}$$
Ross Millikan
- 374,822
-
On your second step I thought that Cos(2A) was equivalent to cos^2A-sin^2A – Jeremy Hernandez Nov 27 '17 at 00:41
-
@JeremyHernandez: yes, that is what I used. The $\cos(2A)$ has a minus sign in front. – Ross Millikan Nov 27 '17 at 00:42
-
Oh ok, so you moved the minus sign to the front of the identity. Thank you very much. – Jeremy Hernandez Nov 27 '17 at 00:44
-
This answer leaves me wondering what "every single one of their equivalent formulas" meant in the original question. These are the first formulas one would try. The formulas used here seem to be the obvious ones to try first. Perhaps there was an error in dealing with the negative signs, but since the efforts weren't shown, we don't know what went wrong. – David K Nov 27 '17 at 00:50
-
@DavidK I am sorry i left this out. Where I got stuck : 2sinAcosA/1-cos2A where cos2A is equal to cos^2-sin^2. So in the denominator i had 1 - cos^2-sin^2 and i wasnt sure how to convert that into an identity. – Jeremy Hernandez Nov 27 '17 at 00:57
-
So it was a sign error. I see now that the previous comments gave a clue about this. Now that we have seen that it should have been $1-\cos^2A+\sin^2A$ (because that's equal to $1 - (\cos^2A-\sin^2A)$), perhaps this will help you in future exercises. – David K Nov 27 '17 at 01:03