Asked by Josh G.
use the double angle formula to rewrite.
6cos^2-3
and
(cosx +sinx)(cosx-sinx)
6cos^2-3
and
(cosx +sinx)(cosx-sinx)
Answers
Answered by
Bosnian
cos ( 2 x ) = 2 cos ^ 2 ( x ) - 1
6 cos ^ 2 ( x ) - 3 =
3 [ 2 cos ^ 2 ( x ) - 1 ] =
3 sin ( 2 x )
6 cos ^ 2 ( x ) - 3 =
3 [ 2 cos ^ 2 ( x ) - 1 ] =
3 sin ( 2 x )
Answered by
Reiny
second 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
(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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