cos(5pie/3)= A. root 3/2 B. root 2/2 C. -root 2/2 D. 1/2

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cos(5pie/3)= A. root 3/2 B. root 2/2 C. -root 2/2 D. 1/2

Mathematics
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cos(5π/3) = cos(6π/3 - π/3) = cos(6π - π/3) = cos(2π)cos(π/3) + sin(2π)sin(π/3) = 1 • cos(π/3) + 0 • sin(2π/3) = cos(π/3)
my cos says root 1/2
cos(5π/3) = cos(6π/3 - π/3) = cos(6π - π/3) = cos(2π)cos(π/3) + sin(2π)sin(π/3) = 1 • cos(π/3) + 0 • sin(π/3) = cos(π/3)

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Other answers:

so D?
hint: \[\Large \begin{gathered} \cos \left( {\frac{{5\pi }}{3}} \right) = \cos \left( {\pi + \frac{{2\pi }}{3}} \right) = \hfill \\ \hfill \\ = \cos \pi \cos \left( {\frac{{2\pi }}{3}} \right) - \sin \pi \sin \left( {\frac{{2\pi }}{3}} \right) = ... \hfill \\ \end{gathered} \]
|dw:1437321949506:dw|
is it D?
yes, D is right
hint: \[\begin{gathered} \cos \left( {\frac{{5\pi }}{3}} \right) = \cos \left( {\pi + \frac{{2\pi }}{3}} \right) = \hfill \\ \hfill \\ = \cos \pi \cos \left( {\frac{{2\pi }}{3}} \right) - \sin \pi \sin \left( {\frac{{2\pi }}{3}} \right) = ... \hfill \\ \end{gathered} \]
you can do it as (2π-π/3), it seems easier this way a bit.... (to me at least)
Thank you
yw
and I still had a typo, once agian
cos(5π/3) = cos(6π/3 - π/3) = cos(2π - π/3) ^ I said that was 6π (when it is 2π) = cos(2π)cos(π/3) + sin(2π)sin(π/3) = 1 • cos(π/3) + 0 • sin(2π/3) = cos(π/3)
in any case, if you have questions ask, if not then good luck

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