R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Aug 23, 2011 #1 I have listed my problem below and i am having trouble with exactly where cot^(-1)(0) exists on the unit circle. I know that cot = 1/(tan t). Would cot^(-1) = tan t = (sin t)/(cos t)? Anyway here is the problem. STUCK RIGHT HERE cot^(-1)(0). Attachments HW 1 11.jpg 86.9 KB · Views: 1
I have listed my problem below and i am having trouble with exactly where cot^(-1)(0) exists on the unit circle. I know that cot = 1/(tan t). Would cot^(-1) = tan t = (sin t)/(cos t)? Anyway here is the problem. STUCK RIGHT HERE cot^(-1)(0).
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Aug 23, 2011 #2 Ryan Rigdon said: I have listed my problem below and i am having trouble with exactly where cot^(-1)(0) exists on the unit circle. Click to expand... I am not sure what the question really is. But \(\displaystyle \text{arccot}(0)=\frac{\pi}{2}.\) The range of \(\displaystyle y=\text{arccot}(x)\) is \(\displaystyle (0,\pi)\).
Ryan Rigdon said: I have listed my problem below and i am having trouble with exactly where cot^(-1)(0) exists on the unit circle. Click to expand... I am not sure what the question really is. But \(\displaystyle \text{arccot}(0)=\frac{\pi}{2}.\) The range of \(\displaystyle y=\text{arccot}(x)\) is \(\displaystyle (0,\pi)\).
R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Aug 25, 2011 #3 Thanx pka i was able to finish my problem. Even though i got timed out. I had to do a new one here it is. Thought you might like to see it. Attachments hw 1 11.jpg 97.3 KB · Views: 3
Thanx pka i was able to finish my problem. Even though i got timed out. I had to do a new one here it is. Thought you might like to see it.