December 3rd, 2016, 08:18 AM  #1 
Newbie Joined: Dec 2016 From: Ireland Posts: 4 Thanks: 0  Trig identities
I'm a little confused as to how sin2A/cos2A is equal to (1+ 1/cos2A) Can anyone explain? 
December 3rd, 2016, 08:53 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,542 Thanks: 2146 Math Focus: Mainly analysis and algebra 
I think you have made an error in your terms because they are not equal for all A, but only when $\sin 2A = 1+\cos 2A$

December 3rd, 2016, 10:47 AM  #3 
Newbie Joined: Dec 2016 From: Ireland Posts: 4 Thanks: 0  Hm I may have worded it wrong I have attached the solution that is in my textbook and im just a bit confused as to how they made the jump from sinA over cosA to become (1 + 1 over cosA)

December 3rd, 2016, 12:06 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 16,775 Thanks: 1236 
The sin2A/cos2A is not simply replaced with (1 + 1/cos2A). Instead, sin2A/cos2A as an expression added to sin2A is replaced with (1 + 1/con2A) as an expression that sin2A is multiplied by. It's like doing P + P/Q = P(1 + 1/Q), but with sin2A and cos2A instead of P and Q respectively. 
December 3rd, 2016, 12:33 PM  #5  
Newbie Joined: Dec 2016 From: Ireland Posts: 4 Thanks: 0  Quote:
Quote:
Last edited by skipjack; December 3rd, 2016 at 01:38 PM.  

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