H Hilda New member Joined Mar 23, 2016 Messages 1 Mar 23, 2016 #1 hey, I'm having a little bit of trouble doing identities can anyone help me please? sin2x/1+cos2x=tanx
hey, I'm having a little bit of trouble doing identities can anyone help me please? sin2x/1+cos2x=tanx
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,550 Mar 24, 2016 #2 Hilda said: hey, I'm having a little bit of trouble doing identities can anyone help me please? sin2x/1+cos2x=tanx Click to expand... As posted, the equation is as follows: . . . . .\(\displaystyle \dfrac{\sin(2x)}{1}\, +\, \cos(2x)\, =\, \tan(x)\) Was this what you meant? When you reply, please include a clear listing of your efforts so far. Thank you!
Hilda said: hey, I'm having a little bit of trouble doing identities can anyone help me please? sin2x/1+cos2x=tanx Click to expand... As posted, the equation is as follows: . . . . .\(\displaystyle \dfrac{\sin(2x)}{1}\, +\, \cos(2x)\, =\, \tan(x)\) Was this what you meant? When you reply, please include a clear listing of your efforts so far. Thank you!
D Deleted member 4993 Guest Mar 25, 2016 #3 Hilda said: hey, I'm having a little bit of trouble doing identities can anyone help me please? sin2x/1+cos2x=tanx Click to expand... If you meant: \(\displaystyle \dfrac{sin(2x)}{1+cos(2x)} \ = \ tan(x)\) You should have written your problem as: sin(2x)/[1+cos(2x)] = tan(x) If it is then: Can you expand sin(2x) = ?
Hilda said: hey, I'm having a little bit of trouble doing identities can anyone help me please? sin2x/1+cos2x=tanx Click to expand... If you meant: \(\displaystyle \dfrac{sin(2x)}{1+cos(2x)} \ = \ tan(x)\) You should have written your problem as: sin(2x)/[1+cos(2x)] = tan(x) If it is then: Can you expand sin(2x) = ?