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December 3rd, 2016, 08:18 AM   #1
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Trig identities

I'm a little confused as to how sin2A/cos2A is equal to (1+ 1/cos2A)
Can anyone explain?
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December 3rd, 2016, 08:53 AM   #2
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I think you have made an error in your terms because they are not equal for all A, but only when $\sin 2A = 1+\cos 2A$
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December 3rd, 2016, 10:47 AM   #3
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Quote:
Originally Posted by v8archie View Post
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, 12:06 PM   #4
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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.
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December 3rd, 2016, 12:33 PM   #5
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Quote:
Originally Posted by v8archie View Post
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 View Post
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 01:38 PM.
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