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July 26, 2014

Posted by **Josh G.** on Sunday, October 21, 2012 at 1:30am.

6cos^2-3

and

(cosx +sinx)(cosx-sinx)

- Trigonometry -
**Bosnian**, Sunday, October 21, 2012 at 2:00amcos ( 2 x ) = 2 cos ^ 2 ( x ) - 1

6 cos ^ 2 ( x ) - 3 =

3 [ 2 cos ^ 2 ( x ) - 1 ] =

3 sin ( 2 x )

- Trigonometry -
**Reiny**, Sunday, October 21, 2012 at 10:27amsecond one:

(cosx +sinx)(cosx-sinx)

=(cosx +sinx)(cosx-sinx) * (cosx + sinx)/(cosx + sinx)

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

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

or

1/cos 2x + sin 2x/cos 2x

= sec 2x + tan 2x

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