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 December 3rd, 2016, 07:18 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, 07:53 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,232 Thanks: 2411 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, 09:47 AM   #3
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 Originally Posted by v8archie I think you have made an error in your terms because they are not equal for all A, but only when $\sin 2A = 1+\cos 2A$
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)
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 December 3rd, 2016, 11:06 AM #4 Global Moderator   Joined: Dec 2006 Posts: 18,704 Thanks: 1529 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. Thanks from Aoife73
December 3rd, 2016, 11:33 AM   #5
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 Originally Posted by v8archie I think you have made an error in your terms because they are not equal for all A, but only when $\sin 2A = 1+\cos 2A$
Quote:
 Originally Posted by skipjack 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.
Ohh, that makes sense; I hadn't copped that; thank you!

Last edited by skipjack; December 3rd, 2016 at 12:38 PM.

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