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Please help! Trig Identities?

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Simplify: (1+cos2x)/sin2x

asked Jun 13, 2013 in TRIGONOMETRY by homeworkhelp Mentor

1 Answer

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(1+cos2x)/sin2x

Apply double angle formulas: cos(2x) = 1 - 2sin^2 (x) and sin(2x) = 2sin(x)cos(x)

= (1+ 1 - 2sin^2 (x) )/2sin(x)cos(x)

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

= 2(1 - sin^2 (x) )/2sin(x)cos(x)

Cancel common terms.

= (1 - sin^2 (x) )/sin(x)cos(x)

Apply pythagorean identity:cos^2 (x) + sin^2 (x) =1 ⇒cos^2 (x) = 1 - sin^2 (x).

= cos^2 (x)/sin(x)cos(x)

= (cosx)(cosx)/sin(x)cos(x)

Cancel common terms.

= cosx/sinx

= secx

answered Jun 13, 2013 by anonymous
cosx/sinx=cotx  not secx
Typo error... Thanks for correcting me

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