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Mathematics 14 Online
OpenStudy (anonymous):

Help proving this identity step-by-step? And give reasons please? Will give medal & fan. (sin(2x)/sin(x)) - (cos(2x)/cos(x)) = sec(x)

OpenStudy (anonymous):

Does sin(2x)/sin(x) = sin(x)?

OpenStudy (anonymous):

Oh wait no that would be sin^2 x...

OpenStudy (loser66):

sin (2x) = 2sinx cos x--> the first term becomes?

OpenStudy (anonymous):

(2sin(x)cos(x)/sin(x)) - (cos(2x)/cos(x)

OpenStudy (loser66):

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OpenStudy (anonymous):

2cos(x) - (cos(2x)/cos(x)) ?

OpenStudy (loser66):

now, the second term : cos (2x) = 2cos^2 x -1, so that it is \(\dfrac{2cos^2x-1}{cosx}=\dfrac{2cos^2x}{cosx}-\dfrac{1}{cosx}\) Simplify to get \(2cosx -\dfrac{1}{cosx}\) Now, combine with the first term : \[2cosx - 2cos x +\dfrac{1}{cosx}= \dfrac{1}{cos x}= secx\]

OpenStudy (loser66):

The key is memorize the identities. :)

OpenStudy (anonymous):

Ohhh okay (: I see, thank you very much!

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