T Tinkermom New member Joined Sep 17, 2010 Messages 17 Nov 21, 2010 #1 Evaluate tan(2tan^-1 2) I missed a day in school, and am lost on inverse functions. Is tan^-1 of 2 still 2? If so, then would it be 2 x 2? Then would that be tan of 4?
Evaluate tan(2tan^-1 2) I missed a day in school, and am lost on inverse functions. Is tan^-1 of 2 still 2? If so, then would it be 2 x 2? Then would that be tan of 4?
D DrSteve Banned Joined Nov 14, 2010 Messages 87 Nov 21, 2010 #2 You need to use the double angle formula \(\displaystyle \tan 2A=\frac{2\tan A}{1-\tan^2 A}\) to get \(\displaystyle \frac{2tan(tan^{-1}2)}{1-tan^2(tan^{-1}2)} = \frac{2*2}{1-2^2} = -\frac{4}{3}\).
You need to use the double angle formula \(\displaystyle \tan 2A=\frac{2\tan A}{1-\tan^2 A}\) to get \(\displaystyle \frac{2tan(tan^{-1}2)}{1-tan^2(tan^{-1}2)} = \frac{2*2}{1-2^2} = -\frac{4}{3}\).
D Deleted member 4993 Guest Nov 21, 2010 #3 Tinkermom said: Evaluate tan(2tan^-1 2) I missed a day in school, and am lost on inverse functions. Is tan^-1 of 2 still 2? If so, then would it be 2 x 2? Then would that be tan of 4? Click to expand... Let: \(\displaystyle tan^{-1}(2) \ = \ \theta\) then \(\displaystyle tan(\theta) \ = \ 2\) ...........................(1) Then your problem becomes: \(\displaystyle tan(2\theta)\) Now express tan(2?) in terms of tan(?) - then use (1) to evaluate tan(2?)
Tinkermom said: Evaluate tan(2tan^-1 2) I missed a day in school, and am lost on inverse functions. Is tan^-1 of 2 still 2? If so, then would it be 2 x 2? Then would that be tan of 4? Click to expand... Let: \(\displaystyle tan^{-1}(2) \ = \ \theta\) then \(\displaystyle tan(\theta) \ = \ 2\) ...........................(1) Then your problem becomes: \(\displaystyle tan(2\theta)\) Now express tan(2?) in terms of tan(?) - then use (1) to evaluate tan(2?)
T Tinkermom New member Joined Sep 17, 2010 Messages 17 Nov 21, 2010 #4 Oh...ok, I see it now. Thank you very much.