Help Plz

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Xeris

New member
Joined
Jul 19, 2005
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Can anyone help me with this problem I just cant seem to figure it out. Its been to long since Ive done this stuff. I posted this in the algebra part but no one seemed to be able to anser it so I thouht that I would try here.

Find f(n^r/4) if f( X) = (-2 sin (2X))/3 + (5 cos (4X))/6

Thanks.
 
f[x]= [-2Sin(2x)] / 3 +[5Cos(4x) /6
f[x]= [-4 sin (2x) +5 cos (4x)] / 6 not neccessary but convenient

let x = n^r/4
f[n^r/4]= [ -4 sin( n^r/2) + 5 cos ( n^r) ] /6 answer

you could could convert cos(n^r] = cos^2[n^r/2) - sin^2[n^r/2)
and have only one angle if you want to
Arthur
 
the answer is suppose to be -3/2 see I already got that far I just didn't know how to simplify it down. from where you left me how would I get the squares out of the sin and cos.
 
Double posting is VERY STRONGLY discouraged. If you already posted the question, allow the appropriate 24 hours for a reply. Resist the temptation to post in a different forum. It just wastes time and energy.
 
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