X xc630 Junior Member Joined Sep 1, 2005 Messages 164 Feb 1, 2006 #1 Hi I need some help proving this identity (1- tan^2x) / (1+tan^2x) = cos 2x I tried replacing (1+tan^2x) with sec^2x and cross multiplied but I didn't get very far. I would appreciate any help. Thanks.
Hi I need some help proving this identity (1- tan^2x) / (1+tan^2x) = cos 2x I tried replacing (1+tan^2x) with sec^2x and cross multiplied but I didn't get very far. I would appreciate any help. Thanks.
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Feb 1, 2006 #2 xc630 said: I tried replacing (1+tan^2x) with sec^2x Click to expand... Good. Finish it: tan^2(x) = ?
xc630 said: I tried replacing (1+tan^2x) with sec^2x Click to expand... Good. Finish it: tan^2(x) = ?
X xc630 Junior Member Joined Sep 1, 2005 Messages 164 Feb 1, 2006 #3 Tan^2(x) = sec^2(x) -1 but I don't undersatand where you are going with it
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Feb 1, 2006 #4 I was thinking of two possible answers. You have chosen one of them. Go with it. 1 - tan^2(x) = 1 - ( . . .
I was thinking of two possible answers. You have chosen one of them. Go with it. 1 - tan^2(x) = 1 - ( . . .
X xc630 Junior Member Joined Sep 1, 2005 Messages 164 Feb 1, 2006 #5 Ok so I have: 1-tan^2(x)= cos 2x (sec^2x) Then 1-tan^2(x) = 2cos^2(x)-1 (sec^2x) Where should I go from here?
Ok so I have: 1-tan^2(x)= cos 2x (sec^2x) Then 1-tan^2(x) = 2cos^2(x)-1 (sec^2x) Where should I go from here?
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Feb 1, 2006 #6 We do not know that equality holds, so deal with one side at a time. LHS = [1 - (sec^2(x) - 1)]/(sec^2(x)) Show this equals the RHS.
We do not know that equality holds, so deal with one side at a time. LHS = [1 - (sec^2(x) - 1)]/(sec^2(x)) Show this equals the RHS.
T tsh44 Junior Member Joined Sep 4, 2005 Messages 67 Feb 1, 2006 #7 O thanks I got 2cos^2(x)-1 which equals 2cosx