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•  » Square roots with Trig identities.. Can anyone help me? :)

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MathsIsFun
2005-11-15 16:17:50

Gee, good work there Flowers.

Flowers4Carlos
2005-11-15 15:59:24

ahhh... a heart!  yes i recognize those!!  they are the ones that are always fluttering about chris evans.

angelpumpingas
2005-11-15 15:41:16

Omg... Thank you so much for your help. ^_^ It is greatly appreciated....

And <3 is just a symbol for a heart... I use it more than I should, really... But oh well.

Thanks again!!

Flowers4Carlos
2005-11-15 15:07:45

hi yaz angel!!

what does "<3" in your signature mean???  i see that expression being used a lot lately.

(1+sinθ)^(1/2)      (1+sinθ)
------------------ =  -----------          the left side is easier to work with
(1-sinθ)^(1/2)         |cosθ|

(1+sinθ)^(1/2)
------------------       multiply everything by the "conjugate" of the denominator
(1-sinθ)^(1/2)

(1+sinθ)^(1/2)*(1+sinθ)^(1/2)            (1+sinθ)^(2/2)                 1+sinθ
------------------------------------ = ------------------------------ = ------------------
(1-sinθ)^(1/2)*(1+sinθ)^(1/2)       [(1-sinθ)(1+sinθ)]^(1/2)     (1-sin²θ)^(1/2)

1+sinθ
------------------     remember the identity   cos²θ = 1-sin²θ
(1-sin²θ)^(1/2)

1+sinθ                 1+sinθ
------------------ = ----------------   since √a ≥ 0, then √cos²θ > 0 ⇒ |cosθ|
(cos²θ)^(1/2)      (cosθ)^(2/2)

1+sinθ      1+sinθ
--------- = ---------
√cos²θ      |cosθ|

angelpumpingas
2005-11-15 05:36:15

I'm really confused on problems with square roots... Can someone please explain to me how to do this problem? It would be greatly appreciated. ^_^ I'm sure it's something simple, but I'm so confused.

√1+sinθ/1-sinθ = 1+sinθ/|cosθ|